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Infinite series for $1/\pi$. Is it known? Indirect method (associated with a certain problem of electrostatics) indicates that $$\sum\limits_{j=1}^\infty \frac{(2j-3)!!\,(2j-1)!!}{(2j-2)!!\,(2j+2)!!}=\frac{2}{3\pi}.$$ Is this result known?

Using the standard power series for the complete elliptic integral of the second kind $$E(k) = \frac{\pi}{2} \sum_{j=0}^\infty \left(\frac{(2j)!}{2^{2j}(j!)^2}\right)^2 \frac{k^{2j}}{1-2j},$$ we find \begin{align*} \sum\limits_{j=1}^\infty \frac{(2j-3)!!\,(2j-1)!!}{(2j-2)!!\,(2j+2)!!} k^{2j}&=\sum_{j=1}^\infty\frac{-j}{j+1} \left(\frac{(2j)!}{2^{2j}(j!)^2}\right)^2 \frac{k^{2j}}{1-2j} \\ &= -\frac{1}{k^2} \int_0^k\mathrm{d}k\,k^2 \frac{\mathrm{d}}{\mathrm{d}k}\left(\frac{2}{\pi}E(k)\right)\\ &= \frac{2}{3}\frac{k^2-1}{k^2}\frac{2}{\pi}K(k) - \frac{k^2-2}{3k^2}\frac{2}{\pi}E(k). \end{align*} In the limit $k\to 1$ only the second term survives with $E(1)=1$ and therefore \begin{align*} \sum\limits_{j=1}^\infty \frac{(2j-3)!!\,(2j-1)!!}{(2j-2)!!\,(2j+2)!!} &=\frac{2}{3\pi}. \end{align*}

Some Log integrals related to Gamma value Two years ago I evaluated some integrals related to $\Gamma(1/4)$. First example: $$(1)\hspace{.2cm}\int_{0}^{1}\frac{\sqrt{x}\log{(1+\sqrt{1+x})}}{\sqrt{1-x^2}} dx=\pi-\frac{\sqrt {2}\pi^{5/2}+4\sqrt{2}\pi^{3/2}}{2\Gamma{(1/4)^{2}}}.$$ The proof I have is based on the following formula concerning the elliptic integral of first kind (integrating both sides with carefully). $$i \cdot K(\sqrt{\frac{2k}{1+k}})=K(\sqrt{\frac{1-k}{1+k}})-K(\sqrt{\frac{1+k}{1-k}})\cdot\sqrt{\frac{1+k}{1-k}}$$ for $0<k<1$. \begin{align} (2)\hspace{.2cm}\int_{0}^{1}\frac{\sqrt{2x-1}-2x \arctan{(\sqrt{2x-1})}}{\sqrt{x(1-x)}(2x-1)^{3/2}}dx=\frac{\sqrt{2}\pi^{5/2}}{\Gamma{(1/4)}^2}-\frac{\sqrt{2\pi}\Gamma{(1/4)}^2}{8}. \end{align} \begin{align} (3)\hspace{.2cm}\int_{0}^{\pi/2}\frac{\sin{x}\log{(\tan{(x/2))}+x}}{\sqrt{\sin{x}}(\sin{x}+1)}dx=\pi-\frac{\sqrt{2\pi}\Gamma{(1/4)}^{2}}{16}-\frac{\sqrt{2}\pi^{5/2}}{2\Gamma{(1/4)}^{2}}. \end{align} Could you find a solution to (2) and (3) employing only Beta function or other method? I've tried with Mathematica, Mapple, etc and seems that this evaluations are not so well known. Question is an improvement of (1) that has been proved in an elementary approach.

I also played around with this integral. My solution is a bit shorter than the OPs: First use the trick by @Claude and define $$\tag{1} I(a)=\int_0^1 \mathrm dx \sqrt\frac{x}{1-x^2}\log(a+\sqrt{1+x}), $$ such that $$\tag{2} I(1) = I(0) + \int_0^1 \mathrm da \, I'(a). $$ Partial fraction decomposition of $I'(a)$ gives \begin{align} I'(a) &= \int_0^1 \mathrm dx \frac{\sqrt\frac{x}{1-x^2}}{a+\sqrt{1+x}} \,\frac{a-\sqrt{1+x}}{a-\sqrt{1+x}}\tag{3a}\\ &=\int_0^1 \mathrm dx \frac{\sqrt\frac{x}{1-x}}{1+x-a^2} + \int_0^1 \mathrm dx \frac{a\sqrt\frac{x}{1-x^2}}{a^2-x-1}\tag{3b}. \end{align} Now we exchange the integration order in the second term only and also move $I(0)$ into the second term, to get the result \begin{align} I(1) &= \int_0^1 \mathrm da \int_0^1 \mathrm dx \frac{\sqrt\frac{x}{1-x}}{1+x-a^2} &+& \int_0^1 \mathrm dx \int_0^1 \mathrm da \frac{a\sqrt\frac{x}{1-x^2}}{a^2-x-1}+I(0) \tag{4a}\\ &= \int_0^1 \mathrm da \, \pi\left(1-\sqrt{\tfrac{1-a^2}{2-a^2}}\right) &+& \int_0^1 \mathrm dx \sqrt{\tfrac{x}{1-x^2}} \log\sqrt x\tag{4b}\\ &= \pi-\pi^{3/2}\frac{\Gamma\left(\tfrac{3}{4}\right)}{\Gamma\left(\tfrac{1}{4}\right)} &+&\,\, \frac{(\pi-4)\sqrt\pi \,\Gamma\left(\tfrac{3}{4}\right) }{2\Gamma\left(\tfrac{1}{4}\right)}\tag{4c}\\ &=\pi-\frac{(\pi+4)\sqrt\pi \,\Gamma\left(\tfrac{3}{4}\right) }{2\Gamma\left(\tfrac{1}{4}\right)}.\tag{4d} \end{align} Note that in (4c) the Beta function integral as motivated by @Carlo was used.

Euclidean volume of the unit ball of matrices under the matrix norm The matrix norm for an $n$-by-$n$ matrix $A$ is defined as $$|A| := \max_{|x|=1} |Ax|$$ where the vector norm is the usual Euclidean one. This is also called the induced (matrix) norm, the operator norm, or the spectral norm. The unit ball of matrices under this norm can be considered as a subset of $\Bbb R^{n^2}$. What is the Euclidean volume of this set? I'd be interested in the answer even in just the $2$-by-$2$ case.

I worked out the answer for the 2 by 2 case as well. First, when dealing with 2 by 2 matrices in general, a convenient variable change is: $$a\rightarrow\frac{w+x}{\sqrt{2}},d\rightarrow\frac{w-x}{\sqrt{2}},c\rightarrow\frac{y-z}{\sqrt{2}},b\rightarrow\frac{y+z}{\sqrt{2}}.$$ Then $a^2+b^2+c^2+d^2 = w^2+x^2+y^2+z^2$. And the determinant $(ad-bc) = \frac{1}{2}(x^2+y^2-w^2-z^2)$. (Aside: this set of coordinates lets you see for instance that the set of rank 1 matrices in the space of 2D matrices realized as $\mathbb{R}^4$ is a cone over the Clifford torus, since $x^2+y^2 = w^2+z^2$ on a sphere $x^2+y^2+w^2+z^2=r^2$ implies $x^2+y^2 = r^2/2$ and $w^2+z^2 = r^2/2$, which are scaled equations for a flat torus) Let $r_1^2 = x^2+y^2, r_2^2 = w^2+z^2$. (These are radial coordinates of a coordinate system consisting of two orthogonal 2D cylindrical coordinate systems). Then the norm squared is: $$\frac{1}{2}\left(r_1^2+r_2^2 + \sqrt{ (r_1^2+r_2^2)^2 - (r_1^2-r_2^2)^2 }\right)$$ When this is less than one, this corresponds to the region plotted below: Note that each point in the $r_1,r_2$ picture corresponds to a different "torus", $x^2+y^2=r_1^2, w^2+z^2=r_2^2$. We can now integrate over the shaded in region, $\int_{region} dw dx dy dz$. This 4-D integral can be reduced to 2D using $r_1$ and $r_2$, since $dx dy = 2\pi r_1 dr_1, dw dz = 2\pi r_2 dr_2$: $$(4\pi^2) \int_{region} dr_1 dr_2 r_1 r_2. $$ Now, note that we can rewrite $r_2$ in terms of $r_1$. In particular, after some manipulation of our norm, the shaded-in region is defined by $r_2^2 \leq 2-2\sqrt{2}r_1+r_1^2=(\sqrt{2}-r_1)^2$. Hence $r_2\leq \sqrt{2}-r_1$, and we can evaluate the $r_2$ integral: $$4\pi^2 \int_{r_1=0}^\sqrt{2} dr_1 r_1 \int_{r_2=0}^{\sqrt{2}-r_1} r_2 dr_2 \\ = 4\pi^2 \int_{r_1=0}^\sqrt{2} dr_1 r_1 (\sqrt{2}-r_1)^2/2\\ = (4\pi^2) (1/6).$$ This yields $2\pi^2/3$, as Armin found.

A lower bound of a particular convex function Hello, I suspect this reduces to a homework problem, but I've been a bit hung up on it for the last few hours. I'm trying to minimize the (convex) function $f(x) = 1/x + ax + bx^2$ , where $x,a,b>0$. Specifically, I'm interested in the minimal objective function value as a function of $a$ and $b$. Since finding the minimizer $x^*$ is tricky (requires solving a cubic), I figured I'd try and find a lower bound using the following argument: if $b=0$, the minimizer is $x=1/\sqrt{a}$ and the minimal value is $2\sqrt{a}$. If $a=0$, the minimizer is $x=(2b)^{-1/3}$ and the minimal value is $\frac{3\cdot2^{1/3}}{2}b^{1/3}$. Therefore, one possible approximate solution is the convex combination $(\frac{a}{a+b})\cdot2\sqrt{a} + (\frac{b}{a+b})\cdot\frac{3\cdot2^{1/3}}{2}b^{1/3}$. Numerical simulations suggest that the above expression is a lower bound for the minimal value. Does this follow from some nice result about parameterized convex functions? It seems like it shouldn't be hard to prove. I guess in a nutshell I just want to prove that for all $x,a,b>0$ we have $(\frac{a}{a+b})\cdot2\sqrt{a} + (\frac{b}{a+b})\cdot\frac{3\cdot2^{1/3}}{2}b^{1/3} \leq 1/x + ax + bx^2$. Thanks! EDIT: It also appears that if I take the convex combination $(\frac{a^{3/5}}{a^{3/5}+b^{2/5}})\cdot2\sqrt{a} + (\frac{b^{2/5}}{a^{3/5}+b^{2/5}})\cdot\frac{3\cdot2^{1/3}}{2}b^{1/3}$ then I get a tighter lower bound, and in fact the lower bound is within a factor of something like $3/2$ of the true minimal solution.

As Nishant Chandgotia sugessted: simply write $f(x) = \left(p\cdot \frac{1}{x} + ax\right) + \left((1-p)\frac{1}{x}+bx^2 \right)$ for some parametr $p\in[0,1]$. For the first term, minimizer is equal to $ p^{\frac{1}{2}}a^{-\frac{1}{2}}$ and the minimal value is $p^{\frac{1}{2}} 2a^{\frac{1}{2}}$. For the second therm, minimizer is equal to $(1-p)^{\frac{1}{3}}(2b)^{-\frac{1}{3}} $ and the minimal value is $(1-p)^{\frac{2}{3}} \cdot\frac{3\cdot 2^{\frac{1}{3}} }{2} b^{\frac{1}{3}}$ The best estimate is achieved when both minimizers are equal, which means, in therms of $p$, that \begin{equation} \left(\frac{p}{a}\right)^{3} = \left( \frac{1-p}{2b}\right)^{2} \end{equation} Note, that this equation has a solution in interval $0 < p < 1$ by Mean-value theorem, unfortunately not expressible in nice way. Any way, we get for any $p\in[0,1]$ the following estimate: \begin{equation} f(x) \geqslant p^{\frac{1}{2}} \cdot 2a^{\frac{1}{2}} + (1-p)^{\frac{2}{3}} \cdot\frac{3\cdot 2^{\frac{1}{3}} }{2} b^{\frac{1}{3}} \end{equation} Finally, we check that $p^{\frac{1}{2}} + (1-p)^{\frac{2}{3}} \geqslant 1$. This inequality implies, that estimate reamins valid after using any convex combination instead of weights $p^{\frac{1}{2}}, (1-p)^{\frac{2}{3}}$, i.e. \begin{equation} f(x) \geqslant \alpha \cdot 2a^{\frac{1}{2}} + (1-\alpha) \cdot\frac{3\cdot 2^{\frac{1}{3}} }{2} b^{\frac{1}{3}} \end{equation} This can be seen immediately, however, by the inequality $f \geqslant \alpha f_0 + (1-\alpha) f_1$. My goal was to complete presented ideas and to show when exact optimum is attained.

General integer solution for $x^2+y^2-z^2=\pm 1$ How to find general solution (in terms of parameters) for diophantine equations $x^2+y^2-z^2=1$ and $x^2+y^2-z^2=-1$? It's easy to find such solutions for $x^2+y^2-z^2=0$ or $x^2+y^2-z^2-w^2=0$ or $x^2+y^2+z^2-w^2=0$, but for these ones I cannot find anything relevant.

Let´s take any $x>3$ and choose a,b such that $a<b$,$a-b$ even and $ab=x^2-1$. Then with $y=(b-a)/2$, $z=(b+a)/2$ we have $x^2 + y^2 = z^2 + 1$. Particular cases∶ if $x$ even then $x^2 + ((x^2-2)/2)^2 = (x^2/2)^2 + 1$; if $x$ odd then $x^2 + ((x^2-5)/4)^2 = ((x^2+3)/4)^2 + 1$.

Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$? Possibly the most striking proof of Archimedes's inequality $\pi < 22/7$ is an integral formula for the difference: $$ \frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 \frac{dx}{1+x^2}, $$ where the integrand is manifestly positive. This formula is "well-known" but its origin remains somewhat mysterious. I ask: Who discovered this integral, and in what context? The earliest reference I know of is Problem A-1 on the 29th Putnam Exam (1968). According to J.H.McKay's report in the American Math. Monthly (Vol.76 (1969) #8, 909-915), the Questions Committee consisted of N.D.Kazarinoff, Leo Moser, and Albert Wilansky. Is one of them the discoverer, and if so which one? The printed solution, both in the Monthly article and in the book by Klosinski, Alexanderson, and Larson, says only "The standard approach, from elementary calculus, applies. By division, rewrite the integrand as a polynomial plus a rational function with numerator of degree less than 2. The solution follows easily." But surely there's more to be said, because this integral is a minor miracle of mathematics: $\bullet$ Not only is the integrand manifestly positive, but it is always small: $x-x^2 \in [0,1/4]$ for $x \in [0,1]$, and the denominator $1+x^2$ is at least 1, so $(x-x^2)^4/(x^2+1) < 1/4^4 = 1/256$. A better upper bound on the integral is $\int_0^1 (x-x^2)^4 dx$, which comes to $1/630$ either by direct expansion or by recognizing the Beta integral $B(5,5)=4!^2/9!$. Hence $\frac{22}{7} - \pi < 1/630$, which also yields Archimedes's lower bound $\pi > 3\frac{10}{71}$. $\bullet$ The "standard approach" explains how to evaluate the integral, but not why the answer is so simple. When we expand $$ \frac{(x-x^2)^4}{1+x^2} = x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac4{x^2+1}, $$ the coefficient of $x/(x^2+1)$ vanishes, so there's no $\log 2$ term in the integral. [This much I can understand: the numerator $(x-x^2)^4$ takes the same value $(1\pm i)^4 = -4$ at both roots of the denominator $x^2+1$.] When we integrate the polynomial part, we might expect to combine fractions with denominators of 2, 3, 4, 5, 6, and 7, obtaining a complicated rational number. But only 7 appears: there's no $x$ or $x^3$ term; the $x^4$ coefficient 5 kills the denominator of 5; and the terms $-4x^5-4x^2$ might have contributed denominators of 6 and 3 combine to yield the integer $-2$. Compare this with the next such integrals $$ \int_0^1 (x-x^2)^6 \frac{dx}{1+x^2} = \frac{38429}{13860} - 4 \log 2 $$ and $$ \int_0^1 (x-x^2)^8 \frac{dx}{1+x^2} = 4\pi - \frac{188684}{15015}, $$ which yield better but much more complicated approximations to $\log 2$ and $\pi$... This suggests a refinement of the "in what context" part of the question: Does that integral for $(22/7)-\pi$ generalize to give further approximations to $\pi$ (or $\log 2$ or similar constants) that are useful for the study of Diophantine properties of $\pi$ (or $\log 2$ etc.)?

This integral has a series counterpart $$\sum_{k=0}^\infty \frac{240}{(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)}=\frac{22}{7}-\pi$$ https://math.stackexchange.com/a/1657416/134791 (UPDATE Peter Bala New series for old functions https://oeis.org/A002117/a002117.pdf, 2009, formula 5.1) Equivalently, $$\sum_{k=1}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{22}{7}-\pi$$ which may be seen as the first truncation of $$\sum_{k=0}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{10}{3}-\pi$$ Therefore, this series shows the following path to $\frac{22}{7}$ $$\frac{10}{3}-\frac{240}{1·2·3·5·6·7}=\frac{10}{3}-\frac{2·5!}{\frac{7!}{4}}=\frac{10}{3}-\frac{8·5!}{5!6·7}=\frac{1}{3}\left(10-\frac{4}{7}\right)=\frac{1}{3}·\frac{66}{7}=\frac{22}{7}$$ also illustrating how only $7$ remains.

Number of Permutations? Edit: This is a modest rephrasing of the question as originally stated below the fold: for $n \geq 3$, let $\sigma \in S_n$ be a fixed-point-free permutation. How many fixed-point-free permutations $\tau$ are there such that $\sigma \tau^{-1}$ is also fixed-point free? As the original post shows, this number is a function of $\sigma$; can one give a formula based on the character table of $S_n$? Given two permutation of $1, \ldots, N$. Where $3\le N\le 1000$ Example For $N=4$ First is $\begin{pmatrix}3& 1& 2& 4\end{pmatrix}$. Second is $\begin{pmatrix}2& 4& 1& 3\end{pmatrix}$. Find the number of possible permutations $X_1, \ldots, X_N$ of $1, \ldots, N$ such that if we write all three in $3\times N$ matrix, each column must have unique elements. $\begin{pmatrix}3 & 1 & 2 & 4\\ 2 & 4 & 1 & 3\\ X_1 & X_2 & X_3 & X_4\end{pmatrix},$ here $X_1$ can't be 3 or 2, $X_2$ can't be 1 or 4, $X_3$ can't be 2 or 1, $X_4$ cant be 4 or 3, Answer to above sample is 2 and possible permutation for third row is $\begin{pmatrix}1 & 3 & 4 & 2\end{pmatrix}$ and $\begin{pmatrix}4 & 2 & 3& 1\end{pmatrix}$. Example 2 First is $\begin{pmatrix}2 & 4 & 1 & 3\end{pmatrix}$. Second is $\begin{pmatrix}1 & 3 & 2 & 4\end{pmatrix}$. Anwser is 4. Possible permutations for third row are $\begin{pmatrix}3&1&4&2\end{pmatrix}$, $\begin{pmatrix}3&2&4&1\end{pmatrix}$, $\begin{pmatrix}4&1&3&2\end{pmatrix}$ and $\begin{pmatrix}4&2&3&1\end{pmatrix}$.

Let me copy here an answer from Russian forum dxdy.ru that I obtained using the approach outlined in my paper. Two given rows of a $3\times N$ matrix define a permutation of order $N$. Let $c_i$ ($i=1,2,\dots,N$) be the number of cycles of length $i$ in this permutation (in particular, $c_1$ is the number of fixed points, which is 0 iff given permutations form a derangement). Then the number of different third rows that form derangements with respect to each of the first two rows equals $$\sum_{j=0}^n (-1)^j\cdot (n-j)!\cdot [z^j]\ F(z),$$ where $[z^j]$ is the operator of taking the coefficient of $z^j$ and $$F(z) = (1+z)^{c_1}\cdot \prod_{i=2}^n \left( \left(\frac{1+\sqrt{1+4z}}2\right)^{2i} + \left(\frac{1-\sqrt{1+4z}}2\right)^{2i} \right)^{c_i}.$$ Particular cases: * *For $c_1=n$ (i.e., two given rows are equal), we get just the number derangements. *For $c_n=1$, we get menage numbers A000179(n). *For $n=2m$ and $c_2=m$, we get A000316(m) = A000459(m)$\cdot 2^m$. This question inspired me to add the following new sequences to the OEIS: A277256, A277257, and A277265.

Convexity of a certain sublevel set Consider the polynomial of degree $4$ in variable $r$ $$ r^4 + (x^2 + y^2)\ r^2 - 2 x y\ r + x^2 y^2 $$ The discriminant of this polynomial in $r$ is the following expression (obtained using Mathematica) $$ \Delta := 16 x^2 y^2 \left( -x^6 + x^8 - 27 x^2 y^2 + 33 x^4 y^2 - 4 x^6 y^2 + 33 x^2 y^4 + 6 x^4 y^4 - y^6 - 4 x^2 y^6 + y^8 \right) $$ Consider the subset $D$ of $\mathbb{R}^2$ defined by the polynomial inequality $\frac{\Delta}{16 x^2 y^2} \le 0$ $$ D : -x^6 + x^8 - 27 x^2 y^2 - 4 x^6 y^2 - 4 x^2 y^6 + 33 x^4 y^2 + 33 x^2 y^4 + 6 x^4 y^4 - y^6 + y^8 \le 0 $$ I would like to prove that $D$ is convex. Any help would be appreciated. Thanks!

You can parametrize the zero locus of $\Delta$ (other than the origin) in the first quadrant by $$ (x,y) = \left(\frac{(3{-}t)\sqrt{(3{+}t)(1{-}t)}}{8}, \frac{(3{+}t)\sqrt{(3{-}t)(1{+}t)}}{8}\right) \qquad\qquad -1\le t\le 1 $$ and then compute that the curvature is $$ \frac{(x'y''-y'x'')}{((x')^2+(y')^2)^{3/2}} = \frac{2^{3/2}(3{+}t^2)}{\bigl(3{-}t^2\bigr)^{5/2}}. $$ Since this is always positive, it follows that the curve is strictly convex (since it is symmetric under $x$ and $y$ reflections.

Prime divisors of $p^n+1$ Let $p$ be a rational prime and $n$ be a positive integer. It can be easily deduced from Zsigmondy's theorem that $p^n+1$ has a prime divisor greater than $2n$ except when $(p,n)=(2,3)$ or $(2^k-1,1)$ for some positive integer $k$. Hence we know that there exists an odd prime divisor of $p^n+1$ greater than $n$ if and only if $(p,n)\neq(2,3)$ or $(2^k-1,1)$ for any positive integer $k$. Question: (1). For which $(p,n)$ does there exist at least two odd prime divisors of $p^n+1$ coprime to $n$? (2). For which $(p,n)$ does there exist at least two odd prime divisors of $p^n+1$ greater than $n$?

$x^n+1$ factors over $\mathbb{Z}[x]$ unless $n$ is a power of two. For $n=15$ the factorization is $ (x + 1) \cdot (x^{2} - x + 1) \cdot (x^{4} - x^{3} + x^{2} - x + 1) \cdot (x^{8} + x^{7} - x^{5} - x^{4} - x^{3} + x + 1)$. The factors $(p^{2} - p + 1)$ and $(p^{4} - p^{3} + p^{2} - p + 1)$ are odd and with congruence reasons you can get at least two prime factors larger than $15$.

The elliptic curve for $x_1^9+x_2^9+\dots+x_6^9 = y_1^9+y_2^9+\dots+y_6^9$ I. Theorem: "If there are $a,b,c,d,e,f$ such that, $$a+b+c = d+e+f\tag1$$ $$a^2+b^2+c^2 = d^2+e^2+f^2\tag2$$ $$3u^3-3uv+w=-def\tag3$$ where $u=a+b+c,\; v = ab+ac+bc,\;w = abc$, then, $$(a + u)^k + (b + u)^k + (c + u)^k + (d - u)^k + (e - u)^k + (f - u)^k = \\ (a - u)^k + (b - u)^k + (c - u)^k + (d + u)^k + (e + u)^k + (f + u)^k\tag4$$ for $k=1,2,3,9$." A rational point implies that, $$\small14(a^6+b^6+c^6-d^6-e^6-f^6)^2-9(a^4+b^4+c^4-d^4-e^4-f^4)(a^8+b^8+c^8-d^8-e^8-f^8) = t^2$$ The first solution to $\sum\limits^6 x_i^9 = \sum\limits^6 y_i^9$ was found by Lander in 1967. In a 2010 paper, Bremner and Delorme realized that it had the form of $(4)$, was good for $k = 1,2,3,9$, a rational point on a homogeneous cubic, and thus one can find an infinite more. (The first soln had $a,b,c,d,e,f = 9, 14, -19, 17, -18, 5$.) II. An alternative method is to directly solve $(1),(2)$ with simple identities such as, $$a,b,c = 3 + 3 m + n - 3 m n + x,\; -6 m - 2 n + x,\; -3 + 3 m + n + 3 m n + x$$ $$d,e,f = 3 - 3 m + n + 3 m n + x,\quad 6 m - 2 n + x,\,\; -3 - 3 m + n - 3 m n + x$$ which incidentally also obeys, $$-a+nb+c = -d+ne+f$$ Then substitute it into $(3)$, and end up only with a quadratic in $m$ whose discriminant $D$ must be made a square. After some algebra, let $c_1 = \tfrac{1}{8}(n^2+3),\; c_2 = \tfrac{1}{2}(n^3-9n)$, then one is to find rational $x$ such that, $$Poly_1:= 7c_1x+c_2$$ $$Poly_2:= -7x^3-21c_1x+c_2$$ and, $$D:=Poly_1 Poly_2 = \text{square}\tag5$$ a situation similar to the linked post for sixth powers. If a rational point $x$ can be found for some constant $n$, it is a simple matter to transform it into an elliptic curve. (The first soln used $n = 1/2$.) Question: Is it possible to find a non-trivial polynomial solution to $(5)$ as a non-zero square $y$? P.S. To clarify re comments, a polynomial solution would be $x,y$ as rational functions of $n$. Or $x,y,n$ as rational functions of a variable $v$.

For the equation, \begin{equation*} y^2=(-7x^3-21c_1x+c_2)(7c_1x+c_2) \end{equation*} where $c_1=(n^2+3)/8$ and $c_2=(n^3-9n)/2$, and assuming $n$ is rational, we have a rational point $(0,c_2)$, so the quartic is birationally equivalent to an elliptic curve. Using an ancient Ms-Dos version of Derive, it is easy to use Mordell's method to find the curve \begin{equation*} v^2=u^3+49((n^2+3)u+4(3n^6-5n^4+145n^2+49))^2 \end{equation*} which has $2$ points of order $3$ at \begin{equation*} ( \, \, 0 \, , \, \pm \, 28(3n^6-5n^4+145n^2+49)\, \, ) \end{equation*} and numerical experimentation suggests that $\mathbb{Z}/3\mathbb{Z}$ is the torsion subgroup. The reverse transformation is \begin{equation*} x=\frac{8n(n^2-9)u}{(252n^6-420n^4+7n^2(u+1740)+3(7u+v+1372))} \end{equation*} Further numerical experimentation suggests that, for $1 \le n \le 19$, the curve has rank one for $n=6,7,8,9,12,15,16,17$, rank two for $n=18$ and rank zero otherwise - the curve is singular for $n=3$. The presence of rank $0$ curves means (I think!!) that there cannot be a polynomial solution in $n$, valid for all $n$. It might be possible to find a parametric subset of n-values by investigating the curves with strictly positive rank. The problem is that the curves do not have a point of order $2$, so finding generators can take time. Allan MacLeod

Common roots of polynomial and its derivative Suppose $f$ is a uni-variate polynomial of degree at most $2k-1$ for some integer $k\geq1$. Let $f^{(m)}$ denote the $m$-th derivative of $f$. If $f$ and $f^{(m)}$ have $k$ distinct common roots then, Is it true that $f$ has to be a zero polynomial? Here $m<k$ is a positive integer. This statement is true for $m=1$ but is it true for larger $m$ also?

Assume $a,b,c \in \mathbb{R}$ solve $$2(a^3+b^3+c^3)-3(a^2b+ab^2+b^2c+bc^2+a^2c+ac^2)+12abc=0,$$ e.g. $(a,b,c)=(-1,1,3)$. Then $$ \begin{eqnarray} f(x)&:=&(x-a)(x-b)(x-c)(3x^2-2(a+b+c)x+3(ab+bc+ca)-2(a^2+b^2+c^2))\\ &=&3x^5-5(a+b+c)x^4+10(ab+bc+ca)x^3\\ &&+(2(a^3+b^3+c^3)-3(a^2b+ab^2+\dots)-18abc)x^2+\dots \end{eqnarray} $$ satisfies $$f^{(2)}(x)=60(x-a)(x-b)(x-c)$$ and is therefore a counterexample for $m=2$ and $k=3$.

Is this a rational function? Is $$\sum_{n=1}^{\infty} \frac{z^n}{2^n-1} \in \mathbb{C}(z)\ ?$$ In a slightly different vein, given a sequence of real numbers $\{a_n\}_{n=0}^\infty$, what are some necessary and sufficient conditions for $\sum a_nz^n$ to be in $\mathbb{C}(z)$ with all poles simple?

$\sum a_n z^n$ is a rational function iff $a_n$ is a sum of polynomials times exponentials. This is a straightforward corollary of partial fraction decomposition. So, suppose $\frac{1}{2^n - 1}$ can be expressed as such a sum. Taking $n \to \infty$ shows that the largest $r$, in absolute value, such that $r^n$ appears in this sum is $r = \frac{1}{2}$, and moreover (after multiplying both sides by $2^n$) that its polynomial coefficient must be the constant polynomial $1$. That is, the sum must begin $$\frac{1}{2^n - 1} = \frac{1}{2^n} + \text{smaller terms}.$$ The next largest $r$ such that $r^n$ can apppear in this sum is determined by the asymptotic behavior of $\frac{1}{2^n - 1} - \frac{1}{2^n} = \frac{1}{2^n(2^n - 1)} \approx \frac{1}{4^n}$, and the same $n \to \infty$ and multiplying by $4^n$ argument as above shows that it must be $r = \frac{1}{4}$ with polynomial coefficient $1$. So the sum must continue $$\frac{1}{2^n - 1} = \frac{1}{2^n} + \frac{1}{4^n} + \text{smaller terms}.$$ But it's clear that in fact we have $$\frac{1}{2^n - 1} = \sum_{k \ge 1} \frac{1}{2^{kn}}$$ so this argument never terminates, and it follows that $\frac{1}{2^n - 1}$ cannot be expressed as a finite sum of polynomials times exponentials. This is a less complex-analytic version of the argument that proceeds by showing that the generating function has infinitely many poles.

rational numbers and triangular numbers This question is an offshoot of Ratio of triangular numbers. Suppose $ka(a+1)=nb(b+1)$, where $k,n >1$ are relative prime integers, and $a,b \geq 0$ are integers. Which $k,n$ pairs have no solution other than the trivial one $a = 0, b = 0$? (Checked for $k,n<21$. Found no solutions for $4,9$; $9,16$; and - surprise - $16,19$. The rest have solutions in well-defined families. E.g. for $k+1=n$: $a(0)=0; a(1)=2k+1; a(n)=(4k+2)a(n-1)-a(n-2)+2k;$ $b(0)=0; b(1)=2k; b(n)=(4k+2)b(n-1)-b(n-2)+2k$.)

There should always be solutions unless $kn$ is a square. The equation is equivalent to $$k (2a+1)^2 - n (2b+1)^2 = k - n.$$ Let $(x_0, y_0)$ be the fundamental solution of the Pell equation $x^2 - 4 k n y^2 = 1$. Then $$ a = \frac{x_0-1}{2} + n y_0, \quad b = \frac{x_0-1}{2} + k y_0 $$ give you a solution. For $k = 16$, $n = 19$, a solution is $$a = 6981194415, \qquad b = 6406383360.$$ (See also Aeryk's comment to the question.) Now assume that $kn$ is a square. We can obviously reduce to the case that $k$ and $n$ are coprime. Then $k = k_1^2$ and $n = n_1^2$, and we have the equation $$ (k_1 (2a + 1))^2 - (n_1 (2b+1))^2 = k_1^2 - n_1^2. $$ The left hand side factors as $(k_1(2a+1) + n_1(2b+1))(k_1(2a+1) - n_1(2b+1))$. If $k_1^2 - n_1^2 = rs$ is the corresponding factorization of the right hand side, then we get $$2a+1 = \frac{r+s}{2k_1} \qquad\text{and}\qquad 2b+1 = \frac{r-s}{2n_1}$$ (assuming $r \ge s$). $r = k_1 + n_1$, $s = k_1 - n_1$ gives the trivial solution (when $k_1 > n_1$). There can be other factorizations that work, but there may be none. For example, with $k = 6^2$, $n = 1$, we have $6^2 - 1 = 35 \cdot 1$, which gives $a = 1$, $b = 8$. In any case, there are only finitely many solutions, whereas in the nonsquare case, there will always be infinitely many.

How many integer solutions of $a^2+b^2=c^2+d^2+n$ are there? Are there any references to study the integer solutions (existence and how many) of Diophantine equations like $a^2+b^2=c^2+d^2+2$, $a^2+b^2=c^2+d^2+3$, $a^2+b^2=c^2+d^2+5$...? Actually, I can prove that there are integer solutions (a,b,c,d) for any integers n. But I don't know how to count them. Thanks a lot.

Here's an elementary way to see that there are always plenty of solutions. Find $r$ and $s$ such that $r+s=n$ and neither $r$ nor $s$ is twice an odd number --- for large $n$, there will be many ways to do this. Then, there always exist $a,b,c,d$ such that $a^2-c^2=r$ and $b^2-d^2=s$, whence $a^2+b^2=c^2+d^2+n$.

A conjecture harmonic numbers I will outlay a few observations applying to the harmonic numbers that may be interesting to prove (if it hasn't already been proven). From the Online Encyclopedia of Positive Integers we have: $a(n)$ is the number of permutations $p$ of $\{1,\ldots,n\}$ such that the minimum number of block interchanges required to sort the permutation $p$ to the identity permutation is maximized. $1, 1, 5, 8, 84, 180, 3044, 8064, 193248, 604800,\ldots$ (https://oeis.org/A260695) Consider the following harmonic numbers: $$1 + 1/2 = (1 + 2)\cdot 1/2!$$ $$1 + 1/2 + 1/3 + 1/4 = (1 + 2 + 3 + 4)\cdot 5/4!$$ $$1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 = (1 + 2 + 3 + 4 + 5 + 6)\cdot 84/6!$$ $$1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8)\cdot 3044/8!$$ and so on. Therefore, the following generalization suggests itself: $$1 + 1/2 + 1/3 + 1/4 + \cdots + 1/2n = (2n^2 + n)\cdot a(2n - 1)/(2n)!$$ If this hasn't been proven, then I will leave it as a conjecture. Cheers, Robert

We can use the characterization by Christie. Let $\pi \in S_n$. Add a fixed point $0$ to $\pi$, and let $c$ be the cycle $(0, 1, \ldots, n)$. Then the smallest number of block interchanges to sort $\pi$ is equal to $\frac{n + 1 - t}{2}$, where $t$ is the number of cycles in decomposition of $c \pi^{-1} c^{-1} \pi$. When $n$ is odd, the maximum value is obtained at $t = 2$, and we are counting $\pi$ such that $c \pi^{-1} c^{-1} \pi$ decomposes into two cycles. Note that $\pi^{-1} c^{-1} \pi$ is itself a cycle, and for any cycle $d$ the equation $d = \pi^{-1} c^{-1} \pi$ has a single solution for $\pi$ (under $\pi(0) = 0$). According to a result of Zagier (a different presentation here), the product $cd$ of two random $2n$-cycles $c, d$ decomposes into exactly two cycles with probability $2s_{2n + 1, 2} / (2n + 1)! = 2H_{2n} / (2n + 1)$, where $s_{2n + 1, 2}$ is the Stirling number of the first kind. Since $c$ is fixed, we immediately have $a(2n - 1) = (2n - 1)! \frac{2H_{2n}}{2n + 1} = (2n)! \frac{H_{2n}}{n(2n + 1)}$. For $a(2n)$ we want to count the number of $(2n + 1)$-cycles $d$ such that $cd$ is a $(2n + 1)$-cycle. Using the same formula, we have an even simpler relation $a(2n) = \frac{(2n)!}{n + 1}$.

The Diophantine equation $ (xz+1)(yz+1)=az^{3}+1$ has no solutions in positive integers with $z > a^2+2a$ Problem. Let $a$ be a positive integer that is not a perfect cube. Show that the Diophantine equation $(xz+1)(yz+1)=az^{3}+1$ has no solutions in positive integers $x, y, z$ with $z > a^{2}+2a$. If this result is true, we can prove that for a given integer $a$, $a \not= m^3$, there are finitely many Fermat pseudoprimes of the form $ap^{3}+1$ to any base $b>1$ where $p$ runs through the primes. Theorem 1. Let $n = ap^{3}+1$, $a \not = m^3$, $p > a^2+2a$ is prime. If there exists an integer $b$ such that $b^{n-1} \equiv 1 \ ($mod $\ n)$ and $b^{a} \not\equiv 1 \ ($mod $\ n)$ then $n$ is prime. Proof. It can be shown that if $p \ | \ \phi(n)$ then $n=(sp+1)(tp+1)$ for some integer $sp+1$ and prime $tp+1$. Because the Diophantine equation $(sp+1)(tp+1)=ap^3+1$ has no solutions in positive integers $s, t$ when $p > a^2 +2a$, we must have $s = 0$. Therefore if $p \ | \ \phi(n)$, $n = tp+1$ is prime. Assume $n$ is composite, $b^{n-1} \equiv 1 \ ($mod $\ n)$ and $b^{a} \not\equiv 1 \ ($mod $\ n)$, we have $ord_n b \ | \ \phi(n)$. If $p \ | \ ord_n b$, then $p \ | \ \phi(n)$, a contradiction because $n$ is assumed composite therefore $ p \ \not | \ ord_n b$. We also have $ord_n b \ | \ n-1 = ap^3$. Because $ p \ \not | \ ord_n b$, we must have $ ord_n b \ | \ a$. Hence $b^a \ \equiv 1 \ (mod \ n)$ which contradicts the hypothesis therefore $n$ must be prime. Remark. For all $ n > b^a$, we have $b^a \ \not\equiv 1 \ (mod \ n)$ therefore there are finitely many Fermat pseudoprimes of the form $ap^3+1$ for a given positive integer $a$ that is not a perfect cube. A lengthy and incomplete solution to this problem can be found here https://math.stackexchange.com/questions/3842292/prove-that-the-diophantine-equation-xz1yz1-az3-1-has-no-solutions-in

The given equation $(xz+1)(yz+1)=az^3+1$ can be rewritten as $az^2-xyz-(x+y)=0$. We shall show that for any solution $(x,y,z)$, we have $z \le a^2+2a. \ $ Note that $z \ | \ x+y$, therefore $z \le x+y. \ $ Treating $x, y$ as constants, the only positive solution for $z \ $ is \begin{equation} z = \frac{xy+\sqrt{x^2y^2+4a(x+y)}} {2a} \end{equation} In order for $z$ to be rational, the discriminant must be a perfect square. Therefore $w^2 = x^2y^2+4a(x+y)$. We see that $w > xy$ and $w \equiv xy \ ( $mod$ \ 2)$. We can write $w = xy + 2t$, $t > 0$. Substituting $w$ above, $(xy+2t)^2 = x^2y^2+4a(x+y)$. Expanding and simplifying, $txy - ax - ay +t^2 = 0$. Multiplying through by $t$ and factoring, $(tx-a)(ty-a)=a^2 - t^3$. We must have $t \le a-1$ otherwise $RHS<0$ and $LHS \ge 0$. Because $a$ is not a perfect cube, $a^2 - t^3 \not = 0$. The remainder of the proof utilizes the result: If $ab = c \ $ where $a,b, c \not = 0$ are integers then $a+b \le c+1$ if $c>0$ and $a+b \le -(c+1)$ if $c < 0$. We now consider two cases: Case $1: \ $ $ a^2 - t^3 >0 \ ;$ Using the result above on the factored equation, we have $(tx-a)+(ty-a) \le a^2 - t^3+1 \le a^2$. Hence, $z \le x+y \le (a^2+2a)/t \le a^2+2a \\\\$. Case $2: \ $ $ a^2 - t^3 < 0 \ ;$ As in case $1$, we have $(tx-a)+(ty-a) \le t^3 - a^2-1 \ $, $x+y \le t^2 - (a^2-2a+1)/t < t^2$. Hence , $z \le x+y < t^2 \le (a-1)^2 < a^2 +2a$

Show that these matrices are invertible for all $p>3$ I am working on a paper which will extend a result in my thesis and have boiled one problem down to the following: show that the symmetric matrix $M_p$, whose definition follows, is invertible for all odd primes $p$. Letting $p>3$ be prime and $\ell = \frac{p-1}{2}$, we define $$M_p = \begin{pmatrix} 2ij - p - 2p\left\lfloor\frac{ij}{p}\right\rfloor\end{pmatrix}_{1\leq i,j\leq \ell}$$ Examples: * *For $p=5$ we have $M_5 = \begin{pmatrix} -3 & -1 \\ -1 & 3 \end{pmatrix}$ and $\det(M_5) = -1\cdot 2\cdot 5$. *For $p=7$ we have $M_7 = \begin{pmatrix} -5 & -3 & -1 \\ -3 & 1 & 5 \\ -1 & 5 & -3 \end{pmatrix}$ and $\det(M_7) = 2^2 \cdot 7^2$. *For $p=11$ we have $M_{11} = \begin{pmatrix} -9 & - 7 & -5 & -3 & -1 \\ -7 & -3 & 1 & 5 & 9 \\ -5 & 1 & 7 & -9 & -3 \\ -3 & 5 & -9 & -1 & 7 \\ -1 & 9 & -3 & 7 & -5 \end{pmatrix}$ and $\det(M_{11}) = -1\cdot 2^4\cdot 11^4$. Though this (seemingly) nice formula that we see above fails for primes greater than 19, though the determinant has been checked to be non-zero for primes less than 1100. (My apologies if this question is not as motivated or as well discussed as is desired. If there are any questions or if further clarification is needed just let me know!)

Experimentally, we have the following formula for $p$ prime: $$\det(M_p)=(-1)^{(p^2-1)/8}(2p)^{(p-3)/2}h_p^-\;,$$ where $h_p^-$ is the minus part of the class number of the $p$-th cyclotomic field, itself essentially equal to a product of $\chi$-Bernoulli numbers. I have not tried to prove this, but since there are many determinant formulas for $h_p^-$ in the literature, it should be possible.

Prove positivity of rational functions We say a rational function $F(z)$ is positive if the coefficients of its Maclaurin expansion, in the variable $z$, are non-negative. In this context, let $$F_r(z):=\frac{1 - 2z + z^r - (1 - z)^r}{(1 - z)^{r - 1}(1 - 2z)}.$$ Is the following true? Note: $F_2(z)=0$ and $F_3(z)$ is easier to manage. QUESTION. For $r\geq4$, each of the rational functions $F_r(z)$ is positive. Example. After simplifications, $F_4(z)=\frac{2z}{(1-z)^2}$.

Notice that $$F_r(z) = \frac{1}{(1-z)^{r-1}} - \sum_{k=0}^{r-1} \left(\frac{z}{1-z}\right)^k$$ and therefore for $r\geq 4$ and $n\geq 1$, we have \begin{split} [x^n]\ F_r(z) &= \binom{n+r-2}{r-2} - \sum_{k=1}^{r-1} \binom{n-1}{k-1} \\ & = \binom{n+r-2}{r-2} - \binom{n-1}{r-2} - \binom{n-1}{r-3} - \sum_{k=1}^{r-3} \binom{n-1}{n-k} \\ &\geq \binom{n+r-2}{r-2} - \binom{n}{r-2} - \sum_{k=1}^{r-3} \binom{n-1+r-3-k}{n-k} \\ &= \binom{n+r-2}{r-2} - \binom{n}{r-2} - \binom{n+r-4}{r-3} \\ &\geq \binom{n+r-2}{r-2} - \binom{n+r-4}{r-2} - \binom{n+r-4}{r-3} \\ &= \binom{n+r-3}{r-3}\\ &> 0. \end{split} ADDED. The above bound implies a stronger statement: for $r\geq 2$ the function $$F_r(z) + 1 - \frac{1}{(1-z)^{r-2}}$$ is non-negative.

Solve this sextic I'm working with the expression $m = 32an^6+96an^5+120an^4+80an^3+28an^2+4bn^2+4an+4bn+2a+2c $. What exactly is the closed radical form of this, if one were to write $n$ in terms of $m$? There's no general formula for sixth powers but I want to know if this particular class can be given in radicals for all $a, b \in \mathbb{R}$.

Mathematica finds a closed-form expression for the solutions $n$ as a function of $m$ of the equation $$m = 32an^6+96an^5+120an^4+80an^3+28an^2+4bn^2+4an+4bn+2a+2c.$$ The expressions for general $a,b,c$ are lengthy. By way of example, for $a=1$, $b=2$, $c=-3$ the two real solutions are $$n=-\tfrac{1}{2}\pm\tfrac{1}{2}\sqrt{f_m^{1/3}-f_m^{-1/3}},\;\;f_m=\sqrt{m^2+12 m+37}+m+6.$$ More generally, two of the solutions are $$n=-\frac{1}{2}\pm\frac{1}{8}\sqrt{\frac{2^{2/3} \left(\sqrt{4 s^3+z^2}+z\right)^{2/3}- 2^{4/3} s}{3a \left(\sqrt{4 s^3+z^2}+z\right)^{1/3}}},$$ $$z=27648 a^2 (b-2a)+27648 a^2 (m-2c),\;\;s=192 a(2b-a).$$ I have not checked the parameter range where these two solutions are real.

Some Log integrals related to Gamma value Two years ago I evaluated some integrals related to $\Gamma(1/4)$. First example: $$(1)\hspace{.2cm}\int_{0}^{1}\frac{\sqrt{x}\log{(1+\sqrt{1+x})}}{\sqrt{1-x^2}} dx=\pi-\frac{\sqrt {2}\pi^{5/2}+4\sqrt{2}\pi^{3/2}}{2\Gamma{(1/4)^{2}}}.$$ The proof I have is based on the following formula concerning the elliptic integral of first kind (integrating both sides with carefully). $$i \cdot K(\sqrt{\frac{2k}{1+k}})=K(\sqrt{\frac{1-k}{1+k}})-K(\sqrt{\frac{1+k}{1-k}})\cdot\sqrt{\frac{1+k}{1-k}}$$ for $0<k<1$. \begin{align} (2)\hspace{.2cm}\int_{0}^{1}\frac{\sqrt{2x-1}-2x \arctan{(\sqrt{2x-1})}}{\sqrt{x(1-x)}(2x-1)^{3/2}}dx=\frac{\sqrt{2}\pi^{5/2}}{\Gamma{(1/4)}^2}-\frac{\sqrt{2\pi}\Gamma{(1/4)}^2}{8}. \end{align} \begin{align} (3)\hspace{.2cm}\int_{0}^{\pi/2}\frac{\sin{x}\log{(\tan{(x/2))}+x}}{\sqrt{\sin{x}}(\sin{x}+1)}dx=\pi-\frac{\sqrt{2\pi}\Gamma{(1/4)}^{2}}{16}-\frac{\sqrt{2}\pi^{5/2}}{2\Gamma{(1/4)}^{2}}. \end{align} Could you find a solution to (2) and (3) employing only Beta function or other method? I've tried with Mathematica, Mapple, etc and seems that this evaluations are not so well known. Question is an improvement of (1) that has been proved in an elementary approach.

I was hoping to finish it but I am stuuck with a last integral. Then, this is just a comment. Consider $$I(a)=\int_{0}^{1}\frac{\sqrt{x}\log{(a+\sqrt{1+x})}}{\sqrt{1-x^2}}\, dx$$ $$I'(a)=\int_{0}^{1} \frac{\sqrt{x}}{\sqrt{1-x^2} \left(a+\sqrt{x+1}\right)}\,dx$$ $$I'(a)=\pi-\pi \sqrt{1+\frac{1}{a^2-2}}-2 a K(-1)+2 a\, \Pi \left(\left.\frac{1}{a^2-1}\right|-1\right)$$ $$I(1)=\pi-K(-1) -\sqrt{\frac{\pi }{2}} \Gamma \left(\frac{3}{4}\right)^2+2\color{red}{\int_0^1 a\, \Pi \left(\left.\frac{1}{a^2-1}\right|-1\right)\,da}$$ $$I(0)=\frac 12\int_0^1 \frac{\sqrt{x} \log (x+1)}{\sqrt{1-x^2}}\,dx$$ $$I(0)=-\sqrt{\pi } (\gamma +\log (2))\frac{ \Gamma \left(\frac{3}{4}\right)}{\Gamma \left(\frac{1}{4}\right)}+$$ $$\frac{\pi ^{3/2} }{4 \sqrt{2}} \text{HypergeometricPFQRegularized}^{(\{0,0,0\},\{0,1\},0)}\left( \left\{\frac{1}{2},\frac{1}{2},\frac{1}{2}\right\},\left\{2,\frac {1}{2}\right\},\frac{1}{2}\right)$$

Equality of the sum of powers Hi everyone, I got a problem when proving lemmas for some combinatorial problems, and it is a question about integers. Let $\sum_{k=1}^m a_k^t = \sum_{k=1}^n b_k^t$ be an equation, where $m, n, t, a_i, b_i$ are positive integers, and $a_i \neq a_j$ for all $i, j$, $b_i \neq b_j$ for all $i, j$, $a_i \neq b_j$ for all $i, j$. Does the equality have no solutions? For $n \neq m$, it is easy to find solutions for $t=2$ by Pythagorean theorem, and even for $n = m$, we have solutions like $1^2 + 4^2 + 6^2 + 7^2 = 2^2 + 3^2 + 5^2 + 8^2$. For $t > 2$, similar equalities hold: $1^2 + 4^2 + 6^2 + 7^2 + 10^2 + 11^2 + 13^2 + 16^2 = 2^2 + 3^2 + 5^2 + 8^2 + 9^2 + 12^2 + 14^2 + 15^2$ and $1^3 + 4^3 + 6^3 + 7^3 + 10^3 + 11^3 + 13^3 + 16^3 = 2^3 + 3^3 + 5^3 + 8^3 + 9^3 + 12^3 + 14^3 + 15^3$, and we can extend this trick to all $t > 2$. The question is, if we introduce one more restriction, that is, $|a_i - a_j| \geq 2$ and $|b_i - b_j| \geq 2$ for all $i, j$, is it still possible to find solutions for the equation? For $t = 2$ we can combine two Pythagorean triples, say, $5^2 + 12^2 + 25^2 = 7^2 + 13^2 + 24^2$, but how about the cases for $t > 2$ and $n = m$?

An even harder problem than $t>2$ and $n=m$ is the Prouhet–Tarry–Escott problem. Now I leave it to you and google to find lots of examples ;-) http://en.wikipedia.org/wiki/Prouhet-Tarry-Escott_problem

Evaluation of the following series... $S = 1/(2\times3) + 1/(5\times6) + 1/(7\times8) + 1/(10\times11) + ... $ EDIT, Will Jagy, December 8, 2010: to anyone considering working on this, please first see http://mathoverflow.tqft.net/discussion/817/could-a-few-moderators-please-remove-one-of-my-questions/#Item_9 which gives the story behind this peculiar sum. Note that the OP is no longer interested in the results, as they arose from one kind of error and cannot be applied because of a different sort of misunderstanding. The double sum version below was provided recently by Harald Hanche-Olsen. ORIGINAL. I'm curious one of you is able to find the exact evaluation of the following series: $$\begin{aligned} S &= 1/(2\times3) +1/(5\times6) + 1/(7\times8) + 1/(10\times11) + \cdots \\\\&= \sum_{n=1}^\infty\sum_{k=1}^{n}\frac1{(n^2+2k-1)(n^2+2k)} \end{aligned}$$ I'm not exactly sure on how to state the 'general term' of the series. Perhaps I can illustrate it with an example: $ 1/(1\times2) + 1/(3\times4) + 1/(5\times6) + 1/(7\times8) + \ldots + 1/((2n - 1) \times 2n) + \ldots = \log(2)$. Now, to answer Nate Eldredge: let $a_0=2$ and $a_{k+1}=a_{k} + 1 $ unless $ a_{k} + 1$ is a square, in which case let $a_{k + 1} = a_{k} + 2$. Now, multiply $a_{k}$ with $a_{k+1}$. That's a term. Let me show the first few terms: $ S = 1/(2\times3)$ [now skip 4] $ + 1/(5\times6) + 1/(7\times8)$ [now skip 9] $ + 1/(10\times11) + 1/(12\times13) + 1/(14\times15)$ [now skip 16] $ + 1/(17\times18) + \ldots $ So all the squares (1,4,9,16,25, etc) are 'skipped' in the terms. I hope this clarifies it a bit... Thanks a lot in advance, Max Muller PS: If someone has any ideas as to how the general term of this series can be written in a more concise manner, please let me know! For the Meta-users, see also the relevant discussion on this question.

As you are interested in $ \zeta(3) $ you might prefer this variant of your construction. This is also a small part of what Pietro Majer would have done in the direction indicated by Charles Matthews. Define $ f(x)$ for $ | x | \leq 1 $ by $$ f(x) = \frac{x^4}{2 \cdot 3 \cdot 4} + \frac{x^7}{5 \cdot 6 \cdot 7} + \frac{x^{11}}{9 \cdot 10 \cdot 11} + \frac{x^{14}}{12 \cdot 13 \cdot 14} +\cdots + \frac{x^{30}}{28 \cdot 29 \cdot 30} + \frac{x^{33}}{31 \cdot 32 \cdot 33} + \cdots $$ Then I took the third derivative, power series has all coefficients 1 and simplifies as $$ f'''(x) = (x + x^4 ) + ( x^8 + x^{11} + \cdots + x^{23}) + ( x^{27} + x^{30} + \cdots) + \cdots $$ $$ f'''(x) = \sum_{n=1}^\infty \; \; x^{n^3} \left( \frac{1 - x^{3 n^2 + 3 n}}{1 - x^3} \right) $$ or $$ \left( \frac{1 }{1 - x^3} \right) \cdot \left( \sum_{n=1}^\infty \; \; x^{n^3} - x^{n^3 + 3 n^2 + 3 n} \right) $$ $$ \left( \frac{1 }{1 - x^3} \right) \cdot \left( \sum_{n=1}^\infty \; \; x^{n^3} - \left( \frac{1}{x} \right) \sum_{n=1}^\infty \; \; x^{(n + 1)^3} \right) $$ $$ \left( \frac{1 }{1 - x^3} \right) \cdot \left( \sum_{n=1}^\infty \; \; x^{n^3} - \left( \frac{1}{x} \right) \sum_{m=2}^\infty \; \; x^{m^3} \right) $$ $$ \left( \frac{1 }{1 - x^3} \right) \cdot \left( 1 + \sum_{n=1}^\infty \; \; x^{n^3} - \left( \frac{1}{x} \right) \sum_{m=1}^\infty \; \; x^{m^3} \right) $$ $$ \left( \frac{1 }{1 - x^3} \right) \cdot \left( 1 - \left( \frac{1-x}{x} \right) \sum_{m=1}^\infty \; \; x^{m^3} \right) $$ $$ f'''(x) = \left( \frac{1 }{1 - x^3} \right) - \left( \frac{1}{x ( 1 + x + x^2)} \right) \sum_{m=1}^\infty \; \; x^{m^3} $$ And so on more or less forever, with no explicit value for $f(1)$ likely.

An optimization problem The following problem optimization problem arose in a project I am working on with a student. I would like to minimize the quantity: $$ M=\frac{1}{12} + \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right)^2 Q(x)^2 \ dx - 4 \left[ \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) Q(x) \ dx\right]^2 $$ over all continuously differentiable functions $Q$ subject to the conditions that $Q(\tfrac{1}{2})=0$ and $$ \int_0^\frac{1}{2} Q(x)^2 \ dx = 1.$$ Is it possible to explicitly solve such a problem? (The presence of the constant $1/12$ should ensure that $M$ is positive for all $Q$.)

I tried calculus of variations. It's been a while since I've done calculus of variations, so I could be messing it up completely. Also, this isn't completely rigorous. There are ways of making calculus of variations rigorous, but I don't know them, and they're a lot harder than just doing calculations. By my comment above, the $Q(\frac{1}{2})=0$ constraint is worthless, so we will ignore it. Assume $Q$ is the minimum function (this is the first, and most important, non-rigorous step, since we are assuming a minimum exists). Now, let's plug in $Q+\epsilon$, where $Q$ and $\epsilon$ are both functions of $x$. We have $$M=\frac{1}{12} + \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right)^2 (Q+\epsilon)^2 \ dx - 4 \left[ \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) (Q+\epsilon) \ dx\right]^2 .$$ Extracting the first-order terms in $\epsilon(x)$, we get $$\Delta M = 2\int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right)^2 Q \epsilon \ dx -8 \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) Q \ dx \ \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) \epsilon \ dx .$$ If we change $Q$ by adding an infinitessimal $\epsilon$, and keeping the normalization condition on $Q$, then $\Delta M$ has to be 0, or otherwise $Q$ wouldn't be a minimum. We still have to take care of the normalization condition $\int_0^\frac{1}{2} Q(x)^2 \ dx = 1$. We do this using Lagrange multipliers, and so we get the expression $$2\int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right)^2 Q \epsilon \ dx -8 \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) Q \ dx \ \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) \epsilon \ dx + \lambda \int_0^\frac{1}{2} Q \epsilon\ dx .$$ This has to be zero for all functions $\epsilon(x)$. To simplify things further, let $$C = \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) Q \ dx,$$ since it's a constant independent of $\epsilon$. Now, we have $$2 \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right)^2 Q \epsilon \ dx -8 C \ \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) \epsilon \ dx + \lambda \int_0^\frac{1}{2} Q \epsilon\ dx $$ or, putting everything in the same integral sign, $$ \int_0^\frac{1}{2} 2 \left( \tfrac{1}{2}-x \right)^2 Q \epsilon -8 C \left( \tfrac{1}{2}-x \right) \epsilon + \lambda Q \epsilon\ dx $$ and for this to be $0$ for all functions $\epsilon$, we must have $$Q = \frac{ (\frac{1}{2}-x)}{\alpha(\frac{1}{2}-x)^2+\beta}$$ for some $\alpha$, $\beta$. Note that, if $\beta \neq 0$, we get $Q(\frac{1}{2}) = 0$, as desired. A Maple or Mathematica program should at least let you calculate $\alpha$ and $\beta$ numerically.

Counting card distributions when cards are duplicated If we have a deck of 48 different cards and 4 players each get 12 cards, it is well known how to calculate the number of possible distributions: take fac(48) and divide 4 times by fac(12). In a german card came (Doppelkopf) there are 24 different card types, but 2 copies of each type, that is 48 cards at all. How many distributions when there are 4 players? My first approach was to assume there are 48 different cards, and then to divide by 2 and by 2 (... 24 times). But this underestimates the real number of distributions. Because: Say player 1 has both spade kings. This has been ruled out already by dividing by fak(12). But if player 1 and player 2 both have a spade king, then we must divide by 2. How to calculate the number of different distibutions? Is there a closed formula as in the single card deck case?

Label the players as 1,2,3, and 4. We first count the total possible number of hands that 1 can be dealt. Let $a$ be the number of singletons 1 has and let $b$ be the number of duplicates 1 has. This yields a total number of $$\sum \binom{24}{a} \binom{24-a}{b}$$ possibilities, where the sum ranges over all $a$ and $b$ such that $a+2b=12$. We now count the total number of possibilities for 1 and 2 (by conditioning on what 1 has first). We let $c$ denote the number of singletons that both 1 and 2 have. We let $d$ denote the number of singletons that 2 has but 1 does not have. Finally we let $e$ denote the number of duplicates that 2 has. Then the total number of possibilites for $1$ and $2$ is $$\sum \binom{24}{a} \binom{24-a}{b} \binom{a}{c} \binom{24-a-b}{d} \binom{24-a-b-d}{e}$$ possibilities, where the sum ranges over all $a, b,c,d, e$ such that $a+2b=12$, $c \leq a$, and $c+d+2e=12$. We end by computing the total number of possibilities for 1,2, and 3. This actually gives all possibilities since then the hand for player 4 is determined. Let $f$ and $g$ denote the number of singletons and duplicates for 3 respectively. The total answer is then $$\sum \binom{24}{a} \binom{24-a}{b} \binom{a}{c} \binom{24-a-b}{d} \binom{24-a-b-d}{e} \binom{24-a-b-d-e}{g}\binom{24-b-c-e-g}{f}$$ where the sum ranges over all $a, b,c,d, e, f$ and $g$ such that $a+2b=12$, $c \leq a$, $c+d+2e=12$, and $f+2g=12$.

General integer solution for $x^2+y^2-z^2=\pm 1$ How to find general solution (in terms of parameters) for diophantine equations $x^2+y^2-z^2=1$ and $x^2+y^2-z^2=-1$? It's easy to find such solutions for $x^2+y^2-z^2=0$ or $x^2+y^2-z^2-w^2=0$ or $x^2+y^2+z^2-w^2=0$, but for these ones I cannot find anything relevant.

Of course for the equation $X^2+Y^2=Z^2+t$ There is a particular solution: $X=1\pm{b}$ $Y=\frac{(b^2-t\pm{2b})}{2}$ $Z=\frac{(b^2+2-t\pm{2b})}{2}$ But interessuet is another solution: $X^2+Y^2=Z^2+1$ If you use the solution of Pell's equation: $p^2-2s^2=\pm1$ Making formula has the form: $X=2s(p+s)L+p^2+2ps+2s^2=aL+c$ $Y=(p^2+2ps)L+p^2+2ps+2s^2=bL+c$ $Z=(p^2+2ps+2s^2)L+p^2+4ps+2s^2=cL+q$ number $L$ and any given us. The most interesting thing here is that the numbers $a,b,c$ it Pythagorean triple. $a^2+b^2=c^2$ This formula is remarkable in that it allows using the equation $p^2-2s^2=\pm{k}$ Allows you to find Pythagorean triples with a given difference. $a=2s(p+s)$ $b=p(p+2s)$ $c=p^2+2ps+2s^2$ $b-a=\pm{k}$ Pretty is not expected relationship between the solutions of Pell's equation and Pythagorean triples.

Generalized tic-tac-toe We begin with $2n+1$ cards, each with a distinct number from $-n$ to $+n$ on it, face up in between the two players of the game. The players take turns selecting a card and keeping it. The first player to collect three cards that sum to zero wins the game. If the cards are exhausted and neither player has won, a draw is declared. Tic-tac-toe, or noughts and crosses, is of course the special case $n=4$, by using the essentially unique $3\times3$ magic square: $$\begin{matrix} 3 & -4 & 1 \\\ -2 & 0 & 2 \\\ -1 & 4& -3\end{matrix}$$ Has the case of general $n$ been studied?

First player wins for $n$ at least five. First turn, name $0$. They name a number, say $-a$. Choose two numbers $b$ and $c$ such that neither $b$, $c$, nor $b+c=a$. Then name $b$, forcing them to name $-b$, then $c$, forcing them to name $-c$, then $-b-c$, winning. You can always choose two such numbers, since each positive number is missed by one of the following triples: $1+2=3, 1+3=4, 1+4=5, 2+3=5$. As quid points out, this is more complicated than I originally made it seem. If $c\neq a+b$ but $a+b$ is in the interval, then the second player can name $a+b$ in response to $c$ and win. To avoid this, if $1 <a\leq n-2$, choose $b=1$ and $c=a+1$. Neither $1$, $a+1$, nor $a+2=a$ so this works. If $a\geq n-1$, choose $b=2$ and $c=1$. Since $n\geq 5$, neither $1$, $2$, nor $3=a$ so this works, and $a+b=a+2>n$. If $a=1$, choose $b=2$ and $c=3$, so $c=a+b$ and neither $2$, $3$, nor $5=a$.

Determinant of non-symmetric sum of matrices Given three real, symmetric matrices $A\succ0$ and $B$, $C⪰ 0$. How can it be shown that: $$\det(A^2+AB+AC) \leq \det(A^2 +BA +AC+BC) ? \qquad (\star)$$ Where $A^2$ is symmetric and positive definite. Eigenvalues of $BA$, $AC$, and $BC$ are all $> 0$, but symmetry is lost. Thank you!

None of the conjectured inequalities hold. This answer contains three counterexamples. The first one is to $(\star)$, while the second and third ones (below the line) refer to previous inequalities conjectured by the OP. \begin{equation*} A=\begin{bmatrix} 5 & 5\\\\ 5 & 5\end{bmatrix},\quad B=\begin{bmatrix} 8 &4 \\\\ 4 & 2\end{bmatrix},\quad C=\begin{bmatrix} 8 &6 \\\\ 6 & 5\end{bmatrix}. \end{equation*} Then $A^2+AB+AC$ is a rank-1 matrix, so its determinant is zero, while $A^2+AB+AC+BC=\begin{bmatrix}268 & 203\\\\ 224 & 169\end{bmatrix}$, so its determinant is $-180$. The structure of this counterexample and of the other ones below is to setup matrices so that the left hand side becomes a rank-1 matrix, which will have determinant zero. Then, one can adjust the other terms to violate inequalities in any direction. EDIT I'm adding explicit matrices where $A,B,C \succ 0$, but still we have a counterexample to quell the OP's insistence ;-) \begin{equation*} A=\begin{bmatrix}11&12&7\\\\ 12 & 14 & 8\\\\ 7 & 8 & 11\end{bmatrix},\ B=\begin{bmatrix}19 &14&7\\\\ 14& 14&6\\\\ 7&6&3\end{bmatrix},\ C=\begin{bmatrix}17&17&16\\\\ 17&19&17\\\\ 16&17&17\end{bmatrix} \end{equation*} For these matrices, $\det(A^2+AB+AC) \approx 2.35\times 10^5$, while $\det(A^2+AB+AC+BC) \approx 4.67 \times 10^4$. A particularly cute counterexample for your last question (edit: where $\det(A+B) \ge \det(A)+\det(B)$ holds for non symmetric matrices with positive eigenvalues) is the following: \begin{equation*} A = \begin{bmatrix} 0.5 & 0 & 0& 0\\\\ 1 & 0.5 & 0 & 0\\\\ 1 & 1 & 0.5 & 0\\\\ 1 & 1 & 1 & 0.5 \end{bmatrix},\quad B = A^T. \end{equation*} Then, $\det(A+B)=0$, but $\det(A)+\det(B) = 1/8$. The updated question, whether $\det(XY+YZ) \ge \det(XY)+\det(YZ)$ holds is also false. Here is a nice counterexample. \begin{eqnarray*} X = \begin{bmatrix} 2 & 2\\\\ 2 & 2 \end{bmatrix},\quad Y = \begin{bmatrix} 5 & 5\\\\ 5 & 10 \end{bmatrix},\quad Z = \begin{bmatrix} 10 & 4\\\\ 4 & 2 \end{bmatrix}. \end{eqnarray*} With this choice, $\det(XY+YZ) = -300$, while $\det(XY)+\det(YZ)=100$.

An entropy inequality Let $X,Y$ be probability measures on $\{1,2,\dots,n\}$, and set $K=\sum_i\sqrt{X(i)Y(i)}$ so that $Z:=\frac{1}{K}\sqrt{XY}$ is also a probability measure on $\{1,2,\dots,n\}$. How can we prove the inequality $$H(X)+H(Y)\geq 2K^2 H(Z),$$ where $H(X)=-\sum_{i=1}^n X(i)\log X(i)$ is the entropy function. The problem originates from this math stack exchange post, and cardinal's rewording of it in the comments. Despite having being asked over two years ago, with numerous bounties posted, the problem was never solved, and for that reason I am posting it here. I checked the inequality numerically on matlab for millions of choices of $X$ and $Y$, with $n$ up to size $100$, and it always held, which suggests that finding a counter example is unlikely. Remark: By Cauchy Schwarz, $1\geq K^2,$ so the above inequality would be implied by $H(X)+H(Y)\geq 2H(Z).$ However it is worth noting that this inequality does not hold, so the factor of $K^2$ is important.

Remark: This is to give an alternative proof of the inequality in fedja's nice answer: $$ (1+a)\log(1+b)+(1+b)\log(1+a)\ge 2(1+c)\log(1+c) $$ where $a, b, c>0$ with $ab=c^2$. Proof. WLOG, assume that $a \le b$. Let $x = \frac{a}{c} \in (0, 1]$. It suffices to prove that $$f(x) := (1 + cx)\ln(1 + c/x) + (1 + c/x)\ln(1 + cx) - 2(1 + c)\ln(1 + c) \ge 0.$$ We have $$f'(x) = c\ln(1 + c/x) - \frac{c(1+cx)}{x(x + c)} - \frac{c\ln(1 + cx)}{x^2} + \frac{c(x+c)}{x(1+cx)}$$ and $$f''(x) = \frac{2c}{x^2(x+c)} - \frac{c^2(1+cx)}{x^2(x+c)^2} + \frac{2c\ln(1+cx)}{x^3} - \frac{2c^2}{x^2(1+cx)} - \frac{c^2(x+c)}{x(1+cx)^2}.$$ Let $g(x) := x^3 f''(x)$. We have \begin{align*} g'(x) &= \frac{c^2(1-x^2)}{(x+c)^3(1+cx)^3}\\ &\qquad \times [cx^4 + (c^2 + 3)x^3 + (2c^5 - 6c^3 + 10c)x^2 + (c^2 + 3)x + c]\\ & \ge 0. \end{align*} Also, $g(0) = 0$. Thus, $g(x) \ge 0$ on $(0, 1]$. Thus, $f''(x) \ge 0$ on $(0, 1]$. Also, $f'(1) = 0$. Thus, $f'(x) \le 0$ on $(0, 1]$. Also, $f(1) = 0$. Thus, $f(x) \ge 0$ on $(0, 1]$. We are done.

Decomposition of $SU(3)$ representation $6\times 15$ into irreducibles? The 6 and 15 dimensional representations of $SU(3)$ are irreducible. The 90 dimensional tensor product representation $6\times 15$ decomposes into a sum of irreducible representations. What factors occur and with what multiplicity? Note: by 6 I mean the 2 index symmetric representation and not its complex conjugate (which is also 6 dimensional). Similarly for 15.

This question is borderline between what is on topic and what isn't; if you want to do a number of computations like this you should pick up a book on representation theory. My standard recommendations for $SL_n$ rep theory are Chapter 8 of Fulton's Young Tableaux or Appendix II (by Fomin) in Stanley's Enumerative Combinatorics Volume 2. You then just need someone to tell you that the finite dimensional representation theories of $SU$ and $SL$ are the same. Perhaps someone will recommned a book that works directly in $SU$. But I can easily imagine someone working in some other area just needing one answer and not wanting to open the book, so here is the computation. Irreducible $SU$ representations are indexed by partitions. The $6$ and $15$ dimensional representations you speak of are $\mathrm{Sym}^2$ and $\mathrm{Sym}^4$ of the standard representation, so they are represented by the partitions $(2)$ and $(4)$. Tensor products where one factor is a single horizontal row are computed by the Pieri rule. There are three components, each of multiplicity $1$, corresponding to the partitions $(6)$, $(5,1)$ and $(4,2)$. The first one is $\mathrm{Sym}^6$, with dimension $28$. The other two don't have simple descriptions, but their characters are the Schur functions $$s_{51}(x,y,z) = x^5 y + x^4 y^2 + x^3 y^3 + x^2 y^4 + x y^5 + x^5 z + 2 x^4 y z + 2 x^3 y^2 z + 2 x^2 y^3 z + 2 x y^4 z + y^5 z + x^4 z^2 + 2 x^3 y z^2 + 2 x^2 y^2 z^2 + 2 x y^3 z^2 + y^4 z^2 + x^3 z^3 + 2 x^2 y z^3 + 2 x y^2 z^3 + y^3 z^3 + x^2 z^4 + 2 x y z^4 + y^2 z^4 + x z^5 + y z^5$$ and $$s_{42}(x,y,z) = (x^2 + x y + y^2) (x^2 + x z + z^2) (y^2 + y z + z^2).$$ Their dimensions are $35$ and $27$.

Binomial coefficient identity It seems to be nontrivial (to me) to show that the following identity holds: $$ \binom {m+n}{n} \sum_{k=0}^m \binom {m}{k} \frac {n(-1)^k}{n+k} = 1. $$ This quantity is related to the volume of the certain polytope.

This can be proved by induction on $m$ (for all $n$). That is, we want to prove by induction on $m \geq 0$ that $$ \binom{m+n}{n} \sum_{k=0}^m \binom{m}{k} \frac{n(-1)^k}{n+k} = 1 $$ for all $n \geq 1$. When $m = 0$ the left side is 1 for all $n$. If the above equation holds for $m$, then we want to show $$ \binom{m+1+n}{n} \sum_{k=0}^{m+1} \binom{m+1}{k} \frac{n(-1)^k}{n+k} \stackrel{?}{=} 1 $$ for all $n$. In the sum, split off $k = 0$ from the rest and for $k \geq 1$ rewrite $\binom{m+1}{k}$ as $\binom{m}{k-1} + \binom{m}{k}$: \begin{eqnarray*} \sum_{k=0}^{m+1} \binom{m+1}{k} \frac{n(-1)^k}{n+k} &=& 1 + \sum_{k=1}^{m+1} \binom{m}{k-1}\frac{n(-1)^k}{n+k} + \sum_{k=1}^{m+1}\binom{m}{k}\frac{n(-1)^k}{n+k} \\ & = & 1+\sum_{k=0}^{m} \binom{m}{k}\frac{n(-1)^{k+1}}{n+k+1} + \sum_{k=1}^{m}\binom{m}{k}\frac{n(-1)^k}{n+k}. \end{eqnarray*} Absorb the 1 into the second sum as a term at $k=0$ and massage the first sum to make it look like a sum of the type we care about with $n+1$ in place of $n$: \begin{eqnarray*} \sum_{k=0}^{m+1} \binom{m+1}{k} \frac{n(-1)^k}{n+k} & = & \sum_{k=0}^{m} \binom{m}{k}\frac{n(-1)^{k+1}}{n+k+1} + \sum_{k=0}^{m}\binom{m}{k}\frac{n(-1)^k}{n+k} \\ & = & -n\sum_{k=0}^{m} \binom{m}{k}\frac{(-1)^{k}}{(n+1)+k} + \sum_{k=0}^{m}\binom{m}{k}\frac{n(-1)^k}{n+k} \\ & = & \frac{-n}{n+1}\sum_{k=0}^{m} \binom{m}{k}\frac{(n+1)(-1)^{k}}{(n+1)+k} + \sum_{k=0}^{m}\binom{m}{k}\frac{n(-1)^k}{n+k}. \end{eqnarray*} By induction (on $m$) the first sum is $1/\binom{m+n+1}{n+1}$ and the second sum is $1/\binom{m+n}{n}$. Therefore $$ \binom{m+1+n}{n}\sum_{k=0}^{m+1} \binom{m+1}{k} \frac{n(-1)^k}{n+k} = \frac{-n}{n+1}\frac{\binom{m+1+n}{n}}{\binom{m+n+1}{n+1}} + \frac{\binom{m+1+n}{n}}{\binom{m+n}{n}}. $$ Writing the binomial coefficients as ratios of factorials and simplifying, the first term is $-n/(m+1)$ and the second term is $(m+n+1)/(m+1)$. Add them together and you get $1$.

What is known about primes of the form $x^2-2y^2$? David Cox's book Primes of The Form: $x^2+ny^2$ does a great job proving and motivating a lot of results for $n>0$. I was unable to find anything for negative numbers, let alone the case I am interested in, $n=-2$. What is the reason for this? Maybe I am missing something. Any references to results involving primes of this form will be greatly appreciated.

As $2 = x^2 -2y^2$ for $x = 2$ and $y=1$ we fix an odd prime number $p$, and Claim: There exist integers $x,y$ such that $p = x^2 -2y^2$ if and only if $p \equiv \pm1 \mod 8$. First if $p = x^2 -2y^2$ for some $x,y \in \mathbb{Z}$ then observing that the squares mod $8$ are $0,1,4$, we conclude that $p$ can only be $0,1,2,4,6,7$ mod $8$ but it is odd so $p \equiv \pm1 \mod 8$. For the other direction, suppose that $p \equiv \pm1 \mod 8$. By a lemma of Gauss, there exists some $t \in (\mathbb{Z}/p\mathbb{Z})^*$ such that $t^2 \equiv 2 \mod p$. Define: $$S = \{a \in \mathbb{Z} \mid 0 \leq a \leq \sqrt{p}\}$$ and observe that $|S| > \sqrt{p}$. Consequently, $|S^2| > p$ so by the pigeonhole principle, there are different $(x_1,y_1),(x_2,y_2) \in S^2$ for which $x_1 - ty_1 \equiv x_2-ty_2 \mod p$, or equivalently, $x_1 - x_2 \equiv t(y_1 - y_2) \mod p$. Set $x = |x_1 - x_2|, y = |y_1 - y_2|$ and note that $(x,y) \in S^2$ and that $x \equiv \pm t y \mod p$, so after squaring we get that $x^2 \equiv 2y^2 \mod p$, that is $p | x^2 - 2y^2$. On the other hand, $$|x^2 -2y^2| \leq |x|^2 +2|y|^2 < p + 2p = 3p.$$ Therefore, $x^2 -2y^2 \in \{-2p,-p,0,p,2p\}$. Case 1: $x^2-2y^2 = 0$ which means that $x^2 = 2y^2$. Taking into account the number of times $2$ divides both sides (alternatively, set $x = 2x_0$ and apply infinite descent), we see that equality is possible if and only if $x = y = 0$, but this contradicts our assumption that $(x_1,y_1) \neq (x_2,y_2)$. Case 2: $x^2 -2y^2 = 2p$. Clearly, $x$ is even. Set $a = x + y, b = x/2+y$ and observe that $a^2 -2b^2 = p$. Case 3: $x^2 -2y^2 = -p$. Set $a = x + 2y, b = x + y$ and observe that $a^2 -2b^2 = p$. Case 4: $x^2 -2y^2 = -2p$. Clearly, $x$ is even. Set $a = 2x + 3y, b =3x/2 + 2y$ and observe that $a^2 -2b^2 = p$. Case 5: $x^2 -2y^2 = p$, which is what we want.

Action of upper triangular matrices Let $M,N$ be two $n\times m$ matrices with $n\leq m$ and coefficients in an algebraically closed field of characteristic zero $K$, both of full rank $n$. Do there exist two upper triangular matrices $A\in SL(n)$ and $B\in SL(m)$ such that $A\cdot M \cdot B^{T} = \lambda N$ for $\lambda\in K\setminus\{0\}$ ?

I think the answer to your question is negative. Consider for instance $$M = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad N = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$ Assume that there exist $$A = \begin{pmatrix} a_{11} & a_{12} \\ 0 & a_{22} \end{pmatrix}, \quad B = \begin{pmatrix} b_{11} & b_{12} \\ 0 & b_{22} \end{pmatrix}$$ with $\det(A) = \det(B) = 1$ and such that $$A\cdot M\cdot B^{T} = N$$ Then $$A\cdot B^{T} = \begin{pmatrix} a_{11}b_{11}+a_{12}b_{12} & a_{12}b_{22}\\ a_{22}b_{12} & a_{22}b_{22} \end{pmatrix} = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$$ and hence either $a_{22} = 0$ or $b_{22}=0$ which contradict $\det(A) = \det(B) = 1$. More generally, your action stabilizes the locus of matrices of the form $\begin{pmatrix} m_{11} & m_{12}\\ m_{21} & 0 \end{pmatrix}$.

On the polynomial $\sum_{k=0}^n\binom{n}{k}(-1)^kX^{k(n-k)}$ Let $n = 2m$ be an even integer and let $F_n(X)$ be the polynomial $$F_n(X):=\sum_{k=0}^n\binom{n}{k}(-1)^kX^{k(n-k)}.$$ I observed (but cannot prove) that the polynomial $F_n$ is always divisible by $(1-X)^m$, and the quotient $F_n(X)/(1-X)^m$ always has positive coefficients. Similarly, with $m, n$ as above, let $G_n(X)$ be the polynomial $$G_n(X):=\sum_{k=0}^n\binom{n}{k}(-1)^k X^{(k+1)(n-k)}.$$ Since all the exponents of $X$ are even, the polynomial $G_n$ is essentially a polynomial of $X^2$. Now it turns out (still without proof) that $G_n$ is always divisible by $(1-X^2)^m$, and the quotient $G(X)/(1-X^2)^m$ always have positive coefficients. Are these polynomials well known? Could someone give a proof of any of the assertions? My ideas I spent some time on this, trying to attack by establishing recurrence relations among them, as well as looking at the generating functions (i.e. $\sum F_nT^n$). I also noted that they are both special values of the polynomial $$H_n(X,Y):=\sum_{k=0}^n\binom{n}{k}(-1)^kX^{k(n-k)}Y^k.$$ However, all my usual tools eventually failed, because of the strange exponent $k(n-k)$ (which is morally $k^2$). I only have seen this kind of exponent in Jacobi triple product before, but that is also not quite relevent... To give some examples: \begin{eqnarray*} F_2(X) &=& 2(1-X)\\ F_4(X) &=& 2(1-X)^2(1 + 2X + 3X^2)\\ F_6(X) &=& 2(1-X)^3(1 + 3X + 6X^2 + 10X^3 + 15X^4 + 15X^5 + 10x^6)\\ F_8(X) &=& 2(1-X)^4(1 + 4X + 10X^2 + 20X^3 + 35X^4 + 56X^5 + 84X^6 + 112X^7 + 133X^8 + 140X^9+126X^{10}+84X^{11}+35X^{12})\\ \\ G_2(X) &=& (1-X^2)\\ G_4(X) &=& (1-X^2)^2(1+2X^2)\\ G_6(X) &=& (1-X^2)^3(1+3X^2+6X^4+5X^6)\\ G_8(X) &=& (1-X^2)^4(1+4X^2+10X^4+20X^6+28X^8+28X^{10}+14X^{12}) \end{eqnarray*} The examples are calculated using sagecell. UPDATE Using the method of Joe Silverman, I am able to prove the following: $$ F_{2m}(X)=2(1-X)^m(a_{m,0} + a_{m,1}X + a_{m,2}X^2 + \cdots),$$ where the coefficients $a_{m,k}$ are given by: $$a_{m,k} = \frac{1}{2}\sum_{j = 0}^{\infty}(-1)^j\binom{2m}{j}\binom{m-1+k-2mj+j^2}{m-1}.$$ Here the convention is that $\binom{z}{y} = 0$ for all integers $z < y$. A remark: since the degree of $F_{2m}$ is equal to $m^2$, we must have $a_{m,k} = 0$ for all $k > m^2-m$. This, however, is not obvious from the above explicit formula... For $k \leq m^2-m$, the above formula can also be written as $$a_{m,k} = \sum_{j = 0}^{m}(-1)^j\binom{2m}{j}\binom{m-1+k-2mj+j^2}{m-1}.$$ Now the problem is to show that all the $a_{m,k}$ are positive. According to the comment of Zach Teitler, we have the following guess: For fixed $m$, the function $k\mapsto a_{m,k}$ is the Hilbert function of an ideal of $k[\mathbb{P}^{m-1}]$ generated by $2m$ general degree $2m-1$ forms. The positivity of all $a_{m,k}$ would follow from this, since the Hilbert functions, being dimensions, are positive.

Here is a proof for the positivity of coefficients. Suppose that $$F_{2m}\,(x)=2(1-x)^m\sum_ka_{m,k}\,x^k,$$ where $a_{m,k}=0$ if $k<0$ or $k>m^2-m$. We shall show that $a_{m,k}>0$ for all other $k$. First of all, a negative sign was missed in Joe Silverman's deduction from the third last step. The correct formula is $$F_n'(x)=-n(n-1)x^{n-2}F_{n-2}\,(x).$$ Using this, the derivative of $F_{2m}(x)$ is \begin{align} F_{2m}'\,(x) &=-2m(2m-1)x^{2m-2}F_{2m-2}\,(x)\\ &=-2(1-x)^{m-1}\sum_{k}2m(2m-1)a_{m-1,\,k-2m+2}\ \ x^{k}. \end{align} The derivative can be computed alternatively as \begin{align} F_{2m}'(x) &=-2m(1-x)^{m-1}\sum_{k}a_{m,k}x^k+2(1-x)^m\sum_{k}a_{m,k}\ k\,x^{k-1}\\ &=2(1-x)^{m-1}\sum_{k}(\,(k+1)a_{m,\,k+1}-(m+k)a_{m,k}\,)\ x^k. \end{align} Comparing the coefficient of $x^k$ in the sums of the above two formulas we find $$a_{m,k}=\frac{2m(2m-1)a_{m-1,\,k-2m+2}\,+(k+1)a_{m,\,k+1}}{m+k},\quad \text{for $m\ge1$}.$$ The initial value to use this recurrence for each $m$ is the leading coefficient of the sum part in $F_{2m}(x)$, that is, $$a_{m,\,m^2-m}=\frac{1}{2}\binom{2m}{m}=\binom{2m-1}{m},\quad\text{for $m\ge1$}.$$ which is a positive integer. The desired positivity follows from the recurrence immediately by induction on $m$. For instance, by the initial value formula, we have $a_{1,0}=1>0$. That is all we need to show for $m=1$. For $m=2$, we have $a_{2,2}=3$. Using the recurrence formula for $m=2$, we obtain $$ a_{2,1}=\frac{12a_{1,-1}+2a_{2,2}}{3}=2\quad\text{and}\quad a_{2,0}=\frac{12a_{1,-2}+a_{2,1}}{2}=1.$$ For $m=3$, the initial value formula gives $a_{3,6}=10$; the recurrence formula allows us to compute $a_{3,k}$ for $k=5,4,3,2,1,0$ one by one: $$a_{3,5}=\frac{30a_{2,1}+6a_{3,6}}{8}=15,\quad a_{3,4}=\dots.$$

Dividing a cake between $n-1$, $n$, or $n+1$ guests A housewife is waiting for guests and has prepared a cake. She doesn't know how many guests will come, but it will be $n-1$, $n$, or $n+1$. What is the minimal number $f(n)$ of pieces the cake should be cut to make it possible to divide between guests equally? For $n=2$, $f(n)=f(2)=4$: The problem was posed 16.10.2018 by Oleksandr Maksymets on page 76 of Volume 2 of the Lviv Scottish Book. The prize: Cooked duck or lunch + beer!

$f(7)=15$. $f(7)\ge15$ follows from a comment of Fedor Petrov on the original question, so it suffices to find a way to cut the cake into $15$ pieces so as to serve $6$, $7$, or $8$ guests. Let the size of the cake be $168$ (so that all the following computations involve only whole numbers). Let the $15$ pieces be of sizes $1,2,4,5,7,8,10,11,13,14,16,17,19,20,21$ (that is, every size not a multiple of $3$ up to $20$, and $21$). Then $$1+20=2+19=4+17=5+16=7+14=8+13=10+11=21,$$ $$4+20=5+19=7+17=8+16=10+14=11+13=1+2+21(=24),$$ $$7+21=8+20=4+5+19=11+17=2+10+16=1+13+14(=28).$$ Note that this disproves my conjecture $f(n)=[5n/2]-1$ which evaluates to $16$ when $n=7$.

Factorizing a bivariate polynomial I have a bivariate polynomial for each $n=0,1,2...$ $$ f_n(x,y)=\sum _{k=0}^n \frac{(-1)^k}{2 k+1} \binom{n}{k} \left(x ^2-y ^2\right)^{2 n-2 k}\left([y ( x^2 -1) +x(1 -y^2 )]^{2 k+1}\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad-[y ( x^2 -1) -x(1 -y^2 )]^{2 k+1}\right) $$ looking at low orders explicitly in Mathematica reveals that this polynomial contains $n+1$ factors of $(1-y^2)$. I know that one factor comes from the difference of two odd powers, but cannot see where the other $n$ factors come from. The factorized expressions appear quite complex, for example: $$ f_3=\frac{2}{35} \left(y ^2-1\right)^4 x \left(-35 y^6+35 \left(5 y ^2-1\right) x ^{10}+7 \left(15 y ^4-73 y ^2+3\right) x ^8+(y -1) (y +1) \left(5 y^4-388 y^2+5\right) x ^6-7 y ^2 \left(3 y ^4-73 y ^2+15\right) x ^4+35 y ^4 \left(y^2-5\right) x ^2+35 x ^{12}\right) $$ Assuming $f_n$ contains $(1-y^2)^{n+1}$, How can I find the formula for $f_n$ in its factorized form, for general $n$? If it is any use, the series coefficients are related to the following integral: $$ \sum _{k=0}^n \frac{(-1)^k}{2 k+1}\binom{n}{k} z^{2k+1}=\int_0^z(1-u^2)^n\mathrm{d}u $$

So, \begin{split} f_n(\xi,\eta) &= (\xi^2-\eta^2)^{2n+1} \left( F\big(\frac{\eta ( \xi^2 -1) +\xi(1 -\eta^2)}{\xi^2-\eta^2}\big) - F\big(\frac{\eta ( \xi^2 -1) -\xi(1 -\eta^2)}{\xi^2-\eta^2}\big)\right) \\ &= (\xi^2-\eta^2)^{2n+1} \left( F\big(\frac{\eta\xi + 1}{\xi+\eta}\big) - F\big(\frac{\eta\xi - 1}{\xi-\eta}\big)\right), \end{split} where $$F(x) := \int_0^x (1-u^2)^n\,du.$$ To show that $(1-\eta^2)^{n+1}$ is a factor $f_n(\xi,\eta)$, it's enough to show that $(1-\eta^2)^n$ divides \begin{split} \frac{\partial}{\partial \eta} \left( F\big(\frac{\eta\xi + 1}{\xi+\eta}\big) - F\big(\frac{\eta\xi - 1}{\xi-\eta}\big) \right) &= \bigg(1-\big(\frac{\eta\xi + 1}{\xi+\eta}\big)^2\bigg)^n \frac{\xi^2-1}{(\xi+\eta)^2} - \bigg(1-\big(\frac{\eta\xi - 1}{\xi-\eta}\big)^2\bigg)^n\frac{\xi^2-1}{(\xi-\eta)^2} \\ &= (1-\eta^2)^n (\xi^2-1)^{n+1} \left( \frac{1}{(\xi+\eta)^{2n+2}} - \frac{1}{(\xi-\eta)^{2n+2}}\right), \end{split} and we see that it does.

Computing the integral $\int_{-1}^1 dx \, |x| J_0(\alpha \sqrt{1 - x^2}) P_\ell(x)$ I would like to compute the following integral: $$ I_\ell(\alpha) := \int_{-1}^1 dx \, |x| J_0(\alpha \sqrt{1 - x^2}) P_\ell(x) \tag{1} \label{1} $$ where $\alpha \geq 0$, $J_0$ is the zeroth-order Bessel function of the first kind, $P_\ell(x)$ is the Legendre polynomial of order $\ell$, and $\ell$ is an arbitrary positive integer or zero. Since the integrand is odd if $\ell$ is odd, we have that $I_\ell(\alpha) = 0, \ell \text{ odd}$, so we just need to care about even $\ell$s. Mathematica reports remarkably simple results for some actual values of $\ell$: $$ I_0(\alpha) = \frac{2 J_1(\alpha )}{\alpha }\\ I_2(\alpha) = \frac{6 J_2(\alpha )-\alpha J_1(\alpha )}{\alpha ^2}\\ I_4(\alpha) = \frac{3 \alpha ^2 J_1(\alpha )-60 \alpha J_2(\alpha )+280 J_3(\alpha )}{4 \alpha ^3} $$ This seems to suggest that we have something along the lines of (purely heuristically, not necessarily true): $$ I_\ell(\alpha) = \sum\limits_k a_k \alpha^{b_k} J_k(\alpha) $$ where $b_k$ seem to be integers. Now, one idea I had in mind was to use the expansion (DLMF 10.60, written out in a more suitable form): $$ J_0\left(\alpha\sqrt{1 - x^2}\right)=\sum_{n=0}^\infty (4n+1) \frac{(2n)!}{2^{2n}(n!)^2} j_{2n}(\alpha) P_{2n} (x) $$ along with the following identities (see here and here): $$ P_k P_\ell = \sum\limits_{m=|k - \ell|}^{k + \ell} \begin{pmatrix}k & \ell & m\\ 0 & 0 & 0\end{pmatrix}^2 (2m + 1) P_m \\ |x| = \begin{cases} -P_1(x),\quad x \leq 0\\ P_1(x),\quad x > 0 \end{cases} \\ \int_0^1 dx\; P_m P_n = \begin{cases} \frac{1}{2n + 1}, & m=n\\ 0, & m \neq n,m,n \text{ both even or odd}\\ f_{m,n}, & m \text{ even},n\text{ odd}\\ f_{n,m} ,& m \text{ odd},n\text{ even} \end{cases} $$ where I'll call $g(m,n) \equiv \int_0^1 dx\; P_m P_n$ for brevity, and: $$ f_{m,n} \equiv \frac{(-1)^{(m+n+1)/2}m!n!}{2^{m+n-1} (m - n) (m + n + 1) \big[\big(\frac{1}{2}m\big)!\big]^2 \big\{\big[\frac{1}{2}(n - 1)\big]!\big\}^2 } $$ We can rewrite: $$ \int_{-1}^1 d\mu\; |\mu| P_{2n} P_\ell = [(-1)^\ell + 1] \int_0^1 d\mu\; P_1 P_{2n} P_\ell $$ and likewise: \begin{align*} \int_0^1 d\mu\; P_1 P_{2n} P_\ell &= \sum\limits_{m=|2n - \ell|}^{2n + \ell} \begin{pmatrix} 2n & \ell & m\\ 0 & 0 & 0 \end{pmatrix}^2 (2m + 1) \int_0^1 d\mu\; P_1 P_m\\ &= \sum\limits_{m=|2n - \ell|}^{2n + \ell} \begin{pmatrix} 2n & \ell & m\\ 0 & 0 & 0 \end{pmatrix}^2 (2m + 1) g(1, m) \end{align*} so that we have: \begin{align} I_\ell(\alpha) &= \sum_{n=0}^\infty \sum\limits_{m=|2n - \ell|}^{2n + \ell} (4n+1) \frac{(2n)!}{2^{2n}(n!)^2} j_{2n}\left(\alpha\right) [(-1)^\ell + 1] \begin{pmatrix} 2n & \ell & m\\ 0 & 0 & 0 \end{pmatrix}^2 (2m + 1) g(1, m) \tag{2} \label{2} \end{align} This is where I kind of got stuck, as I have no idea how to evaluate the double sum. An alternate method would be to use the expansion from here: $$ J_0(\alpha \sqrt{1 - x^2}) = e^{-\alpha x} \sum\limits_{n=0}^\infty \frac{P_n(x)}{n!}\alpha^n $$ but then I end up with integrals of the form: $$ \int_{-1}^1 dx\; |x| P_\ell (x) P_n(x) e^{-\alpha x} $$ which seems even more challenging to evaluate. One idea for this one would be to expand $e^{-\alpha x} = \sum_k \frac{1}{k!} (-1)^k \alpha^k x^k$, and then rewrite $x^k$ as a linear combination of Legendre polynomials, but this again yields an integral over three Legendre polynomials, so I'd probably just obtain eq. \ref{2} in a more roundabout way. Any hints would be appreciated!

Thanks to the comment by Johannes, the solution can indeed be obtained by using the following identities: \begin{equation} P_\ell(z) = \frac{1}{2^\ell} \sum\limits_{k=0}^{\left\lfloor \frac{\ell}{2}\right\rfloor} (-1)^k \begin{pmatrix} \ell \\ k \end{pmatrix} \begin{pmatrix} 2\ell - 2k \\ \ell \end{pmatrix} z^{\ell - 2k} \tag{3} \label{3} \end{equation} and: \begin{equation} \int_{0}^{\frac{1}{2}\pi}J_{\mu}\left(z\sin\theta\right)(\sin\theta)^{\mu+1}(% \cos\theta)^{2\nu+1}\mathrm{d}\theta=2^{\nu}\Gamma\left(\nu+1\right)z^{-\nu-1}% J_{\mu+\nu+1}\left(z\right) \tag{4} \label{4} \end{equation} Transforming the integral yields: \begin{align} I_\ell(\alpha) &= \int_{-1}^1 dx\, |x|\, J_0(\alpha\sqrt{1 - x^2}) P_\ell(x)\\ &= [(-1)^\ell + 1] \int_0^1 dx\, x\, J_0(\alpha \sqrt{1 - x^2}) P_\ell(x)\\ &= |\mathrm{substitution}\;x = \cos \phi|\\ &= [(-1)^\ell + 1] \int_0^\frac{\pi}{2} d\phi\, \sin \phi\, \cos \phi\, P_\ell(\cos \phi)\, J_0 (\alpha \sin \phi)\\ &= |\mathrm{expansion\;of}\;P_\ell|\\ &= [(-1)^\ell + 1] \sum\limits_{k=0}^{\left\lfloor \frac{\ell}{2}\right\rfloor} (-1)^k \begin{pmatrix} \ell \\ k \end{pmatrix} \begin{pmatrix} 2\ell - 2k \\ \ell \end{pmatrix} \int_0^\frac{\pi}{2} d\phi\, \sin \phi\, \cos \phi\, (\cos \phi)^{\ell - 2k} J_0 (\alpha \sin \phi)\\ &= [(-1)^\ell + 1] \sum\limits_{k=0}^{\left\lfloor \frac{\ell}{2}\right\rfloor} (-1)^k \begin{pmatrix} \ell \\ k \end{pmatrix} \begin{pmatrix} 2\ell - 2k \\ \ell \end{pmatrix} \int_0^\frac{\pi}{2} d\phi\, \sin \phi\, (\cos \phi)^{\ell - 2k + 1} J_0 (\alpha \sin \phi) \end{align} The integral in the above sum has the same form as the Bessel identity \ref{4}, with $\mu = 0$ and $\nu = \ell / 2 - k$, so that the final result is: \begin{equation} \boxed{ I_\ell(\alpha) = \frac{ [(-1)^\ell + 1] } { 2^\frac{\ell}{2} } \sum\limits_{k = 0}^{\left \lfloor \frac{\ell}{2} \right \rfloor} \frac{(-1)^k}{2^k} \begin{pmatrix} \ell \\ k \end{pmatrix} \begin{pmatrix} 2\ell - 2k \\ \ell \end{pmatrix} \Gamma\left[\frac{\ell}{2} - k + 1\right] \frac{ J_{\frac{\ell}{2} - k + 1} (\alpha) } { \alpha^{\frac{\ell}{2} - k + 1} } } \tag{5} \label{5} \end{equation}

Generating functions of Collatz iterates? Let $C(n) = n/2$ if $n$ is even and $3n+1$ otherwise be the Collatz function. We look at the generating function $f_n(x) = \sum_{k=0}^\infty C^{(k)}(n)x^k$ of the iterates of the Collatz function. The Collatz conjecture is then equivalent to: For all $n$: $$f_n(x) = p_n(x) + x^{d+1} \frac{1+4x+2x^2}{1-x^3}$$ where $d$ is the degree of the polynomial $p_n(x)$ with natural numbers as coefficients. I have computed some of these generating functions. Let $$F_n(x) = (f_n(x), f_{C^{(1)}(n)}(x),\cdots,f_{C^{(l)}(n)}(x))$$ where $l$ is the length of the Collatz sequence of $n$ ending at $1$. The vector $F_n(x)$ when plugging in for $x$ a rational number, seems to parametrize an algebraic variety. Assuming that the Collatz conjecture is true. Can it be explained if or why this vector parametrizes an algebraic variety? Here is an example for $n=3$: The variety is given by the equations: G^3 - H^3 - 4*G^2 + 4*G*H + H^2 + 4*G - 8*H = 0 A*C - B^2 + 10*B - 3*C = 0 F^2 - G*H - 4*F + G = 0 F*G - H^2 - 2*F + H = 0 F*H - G^2 + 2*G - 4*H = 0 E - H - 7 = 0 and it is parametrized by: A = (7*x^7 + 14*x^6 + x^5 + 2*x^4 - 13*x^3 - 5*x^2 - 10*x - 3)/(x^3 - 1) B = (7*x^6 + 14*x^5 + x^4 + 2*x^3 - 16*x^2 - 5*x - 10)/(x^3 - 1) C = (7*x^5 + 14*x^4 + x^3 - 8*x^2 - 16*x - 5)/(x^3 - 1) D = (7*x^4 + 14*x^3 - 4*x^2 - 8*x - 16)/(x^3 - 1) E = (7*x^3 - 2*x^2 - 4*x - 8)/(x^3 - 1) F = (-x^2 - 2*x - 4)/(x^3 - 1) G = (-4*x^2 - x - 2)/(x^3 - 1) H = (-2*x^2 - 4*x - 1)/(x^3 - 1) where $F_3(x) = (A,B,C,D,E,F,G,H)$ Here is some Sagemath script which does the computations. You can change the number $N=3$ in the script, but for $N=7$ it already takes a long time to compute the Groebner basis. Edit: Furthermore, the point $(n,C^{(1)}(n),\cdots,C^{(l)}(n))$ seem to always be a rational point of this variety. Example: G^3 - H^3 - 4*G^2 + 4*G*H + H^2 + 4*G - 8*H = 0 A*C - B^2 + 10*B - 3*C = 0 F^2 - G*H - 4*F + G = 0 F*G - H^2 - 2*F + H = 0 F*H - G^2 + 2*G - 4*H = 0 E - H - 7 = 0 ..... A = 3 B = 10 C = 5 D = 16 E = 8 F = 4 G = 2 H = 1 This last observation can be explained if the previous is true, because we can substitute $x=0$: $$\forall m \text{ we have : } f_m(0) = m$$ and hence: $$F_n(0) = (n,C^{(1)}(n),\cdots,C^{(l)}(n))$$ is a rational point on the variety.

Without assuming the Collatz conjecture, it can be shown that the generating functions satsify certain polynomial equations: Observe that for all $n$: $$f_{C(n)}(x) = \frac{f_n(x)-n}{x}$$ hence: $$f_{C^{(2)}(n)}(x) = \frac{f_{C(n)}(x)-C(n)}{x}$$ Solving for $x$ and equating the two identities and letting $x_k:=f_{C^{(k)}(n)}(x)$, we find the polynomial equation: $$\forall k=0,1,2,\cdots \text{ we have }: x_k x_{k+2}-C^{(k)}(n) x_{k+2}-x_{k+1}^2+C^{(k+1)}(n)x_{k+1} = 0$$ which according to Wolfram Alpha each of these equations represent a "one-sheeted hyperboloid": Wikipedia Mathworld

Minimal polynomial in $\mathbb Z[x]$ of seventh degree with given roots I am looking for a seventh degree polynomial with integer coefficients, which has the following roots. $$x_1=2\left(\cos\frac{2\pi}{43}+\cos\frac{12\pi}{43}+\cos\frac{14\pi}{43}\right),$$ $$x_2=2\left(\cos\frac{6\pi}{43}+\cos\frac{36\pi}{43}+\cos\frac{42\pi}{43}\right),$$ $$x_3=2\left(\cos\frac{18\pi}{43}+\cos\frac{22\pi}{43}+\cos\frac{40\pi}{43}\right)$$ $$x_4=2\left(\cos\frac{20\pi}{43}+\cos\frac{32\pi}{43}+\cos\frac{34\pi}{43}\right),$$ $$x_5=2\left(\cos\frac{10\pi}{43}+\cos\frac{16\pi}{43}+\cos\frac{26\pi}{43}\right),$$ $$x_6=2\left(\cos\frac{8\pi}{43}+\cos\frac{30\pi}{43}+\cos\frac{38\pi}{43}\right)$$ and $$x_7=2\left(\cos\frac{4\pi}{43}+\cos\frac{24\pi}{43}+\cos\frac{28\pi}{43}\right).$$ I see only that $\sum\limits_{k=1}^7x_k=-1$, but the computations for $\sum\limits_{1\leq i<j\leq7}x_ix_j$ and the similar are very complicated by hand and I have no any software besides WA, which does not help. Thank you for your help! Update. I got $$\sum\limits_{1\leq i<j\leq7}x_ix_j=-18.$$

In SageMath, you can enter the following: U.<zeta> = CyclotomicField(43) P.<x> = PolynomialRing(U) def c(j): # cos(j * pi / 43) return (zeta ** j + zeta ** (-j))/2 x1 = 2*(c(2) + c(12) + c(14)) x2 = 2*(c(6) + c(36) + c(42)) x3 = 2*(c(18) + c(22) + c(40)) x4 = 2*(c(20) + c(32) + c(34)) x5 = 2*(c(10) + c(16) + c(26)) x6 = 2*(c(8) + c(30) + c(38)) x7 = 2*(c(4) + c(24) + c(28)) (x-x1)*(x-x2)*(x-x3)*(x-x4)*(x-x5)*(x-x6)*(x-x7) And you get: x^7 + x^6 - 18*x^5 - 35*x^4 + 38*x^3 + 104*x^2 + 7*x - 49 that is: $x^{7} + x^{6} - 18 x^{5} - 35 x^{4} + 38 x^{3} + 104 x^{2} + 7 x - 49$.

Characteristic polynomial of a matrix related to pairs of elements generating $\mathbb{Z}/n\mathbb{Z}$ Fix $n\geq 2$. Let $A$ be the matrix whose rows and columns are indexed by pairs $(a,b)\in \mathbb{Z}/n\mathbb{Z}$ such that $a,b$ generate $\mathbb{Z}/n\mathbb{Z}$ (the number of such pairs is $\phi(n)\psi(n)$, where $\phi(n)$ is the Euler phi-function and $\psi(n)$ is the Dedekind $\psi$-function), with the $((a,b),(c,d))$ entry defined by $$ A_{(a,b),(c,d)}=\left\{ \begin{array}{rl} 1, & \mathrm{if}\ c=a\ \mathrm{and}\ d=b-a\\ 1, & \mathrm{if}\ c=a-b\ \mathrm{and}\ d=b\\ 0, & \mathrm{otherwise}. \end{array} \right. $$ The characteristic polynomial of $A$ factors quite a bit. What is the explanation for this behavior? Is $A$ diagonalizable (over $\mathbb{C}$)? What is the rank of $A$? This matrix arises in the congruence properties mod $n$ of the entries of Stern's triangle. Here are these polynomials for $n\leq 11$. I write $[k]$ for an irreducible polynomial (over $\mathbb{Q}$) of degree $k$. $$ n=2:\ \ (x+1)(x-1)(x-2) $$ $$ n=3:\ \ x^2(x-1)^3(x-2)(x^2+x+2) $$ $$ n=4:\ \ x^2(x+1)^3(x-1)^4(x-2)(x^2-x+2) $$ $$ n=5:\ \ x^4(x+1)^2(x-1)^7(x-2)(x^2+1)^2(x^2-x+2)(x^4-x^2+4) $$ $$ n=6:\ \ x^6(x+1)^4(x-1)^7(x-2)(x^2+2x+2)(x^2-2x+2)(x^2+x+2) $$ $$ n=7:\ \ x^{10}(x+1)^4(x-1)^{11}(x-2)(x^2+2)(x^4+3)^2[4]^2[4] $$ $$ n=8:\ \ x^8(x+1)^8(x-1)^{13}(x-2)(x^2+1)^2(x^2-x+2)(x^4-x^2+4)[4][4] $$ $$ n=9:\ \ x^{14}(x+1)^6(x-1)^{17}(x-2)(x^2-2x+2)^2(x^2+x+2)^3[6]^2[6]^2 $$ $$ n=10:\ \ x^{12}(x+1)^{10}(x-1)^{15}(x-2)(x^2-x+2)(x^2+1)^2 [4]^2[4][4][4][4] $$ $$ n=11:\ \ x^{20}(x+1)^{10}(x-1)^{21}(x-2)(x^2+2x+2)(x^2-2x+2) (x^2+x+2)[3]^2[4]^2[4]^2[8]^2[12]^2. $$ The factor $x-2$ is clear since the all 1's vector is an eigenvector with eigenvalue 2.

The corank is $\phi(n)\psi(n)/6$ for all $n>3$. In particular, your computations imply that the matrix is not diagonalizable for $n=6,7,9$. This follows from the fact that all nonzero entries are accumulated in the $6\times 6$ blocks with rows indexed by $$ (a,b),\; (a-b,a),\; (-b,a-b),\; (-a,-b),\; (b-a,a), \; (b,b-a) $$ and columns indexed by $$ (a-b,b),\; (-b,a),\; (-a,a-b),\; (b-a,-b),\; (b,-a),\; (-a,b-a) $$ (and all indices are paritioned into such 6-tuples!). Each such block has the form $$ \begin{pmatrix} 1& 1&\\ & 1& 1\\ & & 1& 1\\ & & & 1& 1\\ & & & & 1& 1\\ 1& & & & & 1\\ \end{pmatrix}, $$ hence it is of corank 1. This can be seen from the following observation. Move along your matrix, passing alternately from a one to another one in the same row, and to another one in the same column. A pair of moves performs the row change $(a,b)\mapsto (a-b,a)$, and a column change $(c,d)\mapsto (-d,c+d)$. Both these maps have order $6$, and (for $n>3$) they have no orbit of smaller length. For $n=2$ and $n=3$, there appear smaller cycles.

An infinite series that converges to $\frac{\sqrt{3}\pi}{24}$ Can you prove or disprove the following claim: Claim: $$\frac{\sqrt{3} \pi}{24}=\displaystyle\sum_{n=0}^{\infty}\frac{1}{(6n+1)(6n+5)}$$ The SageMath cell that demonstrates this claim can be found here.

Write the sum as $S_{o}=\sum_{n=1}^{\infty} \frac{1}{(3(2n+1)-2)(3(2n+1)+2)}=\sum_{n=\text{odd}} \frac{1}{(3n)^2-(2)^2}$. Hence, $S_o=S-S_e$. Where, for $S$ , $n$ are all natural numbers in the above expression and for $S_e$, $n$ takes only positive even numbers. Hence, from the identity $\frac{\pi\text{cot}(x)}{x}=\frac{1}{x^2}+\sum_{n=1}^{\infty} \frac{2}{x^2-n^2}$, we get $S=\frac{9-6\pi\text{cot}(\frac{2\pi}{3})}{72}$ And, $S_e=\frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{(3n)^2-1}$. This similarly gives $S_e=\frac{9-3\pi\text{cot}(\frac{\pi}{3})}{72}$. Hence, $S_o=\frac{3\pi\text{cot}(\frac{\pi}{3})-6\pi\text{cot}(\frac{2\pi}{3})}{72}=\frac{\sqrt{3}\pi}{24}$

Positive integer solutions to the diophantine equation $(xz+1)(yz+1)=z^4+z^3 +z^2 +z+1$ Let $P(z) = z^4 +z^3 +z^2 +z+1$ where $z$ is a positive integer. While working with the diophantine equation $(xz+1)(yz+1)=P(z)$, I was able to construct a seemingly infinite and complete solution set to the diophantine equation in positive integers $(x, y, z)$. First, define functions $r$ and $q$ of a positive integer $m$ as follows: \begin{align} r(m) &= m^2+m-1 \\ q(m) &= (r(m) +2)^2-2 \\ \end{align} And for a particular positive integer $m$, define sequences $A, B, C, D$ as follows; \begin{align} A_1 &= m+1\\ A_2 &= m^5 + 3m^4 + 5m^3 + 4m^2 +m\\ A_n &= q(m)A_{n-1} - A_{n-2} - r(m)\\ \end{align} \begin{align} B_1 &= m^3 + 2m^2 +2m\\ B_2 &= q(m)B_1+r(m)\\ B_n &= q(m)B_{n-1} - B_{n-2} + r(m) \\ \end{align} \begin{align} C_1 &= m^3 + m^2 +m+1 \\ C_2 &= q(m)C_1-r(m)\\ C_n &= q(m)C_{n-1} - C_{n-2} - r(m)\\ \end{align} \begin{align} D_1 &= m^5 + 2m^4 + 3m^3 + 3m^2 +m\\ D_2 &= q(m)D_1+ r(m)- m\\ D_n &= q(m)D_{n-1} - D_{n-2} + r(m) \\ \end{align} It appears all positive integer solutions $(x,y,z)$ are given by $(A_n, A_{n+1} , B_n)$, $(C_n, C_{n+1}, D_n), n = 1, 2, \dots $. How does one go about proving that $(A_n, A_{n+1} , B_n)$, $(C_n, C_{n+1}, D_n)$ are indeed solutions for all $n$ and showing that these are the only solutions? My approach would be to obtain closed formulae for $A_n, B_n, C_n$ and $ D_n$ then substitute into $(xz+1)(yz+1)=P(z)$ and check if the $LHS=RHS$. However, the closed forms of these sequences are too bulky. And also this method does not prove whether these are the only solutions. If sequences $A, B, C, D $ cover all solutions in positive integers $x, y, z$ then it becomes much more efficient to determine if $P(z) $ possesses any proper factor of the form $xz+1$ for a particular $z$ using these sequences as compared to trial divisions on $P(z) $ or prime factorizing $P(z) $. Also, is it possible to construct all solutions for irreducible polynomials $P(z) = z^n+z^{n-1}+ \cdots + z^2+z+1$ of degree $n>4?$

"OP" wants to know of a numerical solution which does not satisfy his parametric solution. It is given below. $(x,y,z)=(138,49331482518,2609165)$

Can $y^2-4$ be a divisor of $x^3-x^2-2 x+1$? Do there exist integers $x$ and $y$ such that $\frac{x^3-x^2-2 x+1}{y^2-4}$ is an integer? In other words, can any integer representable as $x^3-x^2-2 x+1$ have any divisor representable as $y^2-4$? This is the simplest non-trivial example of my earlier question Integer points of rational function in 2 variables .

No. The roots of $x^3 - x^2 - 2x + 1$ are $-(\zeta + \zeta^{-1})$ where $\zeta$ is a 7th root of unity; this soon implies [see below] that any prime factor is either $7$ or $\pm 1 \bmod 7$, and thus that all factors of $x^3 - x^2 - 2x + 1$ are congruent to $0$ or $\pm 1 \bmod 7$. In particular it is not possible for two factors to differ by $4$, so no number of the form $y^2 - 4 = (y-2) (y+2)$ can divide $x^3 - x^2 - 2x + 1$. added later: To show that any prime factor $p$ of $x^3 - x^2 - 2x + 1$ is either $7$ or $\pm 1 \bmod 7$, let $k$ be the finite field of order $p^2$, and $\zeta \in k$ a root of the quadratic equation $\zeta^2 + x\zeta + 1 = 0$ (any quadratic equation with coefficients in the $p$-element field has a root in $k$). Then $\zeta^7 = 1$, so either $\zeta = 1$ or the multiplicative group $k^\times$ of $k$ has a subgroup of size $7$. In the former case, $x = -2$, and then $x^3 - x^2 - 2x + 1 = -7$ so $p=7$. In the latter case, Lagrange's theorem gives $7 \mid \#k^\times = p^2-1$, so $p \equiv \pm 1 \bmod 7$. QED

Convergence rate of a sequence Suppose $x_0=x_1=1$, define $y_k=x_k+\frac{1}{2}(x_k-x_{k-1})$ and $x_{k+1}=y_k-\eta y_k^3$ where $\eta\in(0,1/8)$. If we know $x_k\to 0$ as $k\to\infty$. How to show that $x_k=\Theta(1/\sqrt{k})$? It suffices to show that $x_k^{-2}=\Theta(k)$. By Taylor expansion, we have $$x_{k+1}^{-2} = (y_k-\eta y_k^3)^{-2} = y_k^{-2}(1-\eta y_k^2)^{-2} = y_k^{-2}(1+2\eta y_k^2+o(y_k^2)) = y_k^{-2}+2\eta+o(1)$$ It seems that $x_{k+1}^{-2}$ and $y_k^{-2}$ form a linearly increases sequence. If here $y_k^{-2}$ is replaced by $x_k^{-2}$, then we obtain exactly $x_k^{-2}=\Theta(k)$. However, $y_k^{-2}\ne x_k^{-2}$, and I got stuck at this step. Can someone give a hint?

Suppose that we already know that $x_k>0\forall k$. Clearly $(x_k)$ and $(y_k)$ are decreasing sequences which converge to $0$. Then we can prove that $b_k:=\frac{x_{k}}{x_{k-1}}\to 1$. To do it note first that $b_k\in[0,1]\forall k$ and $$b_{k+1}=\frac{y_k-\eta y_k^3}{x_k}=\frac{x_k+\frac{1}{2}(x_k-x_{k-1})-\eta y_k^3}{b_kx_{k-1}}=\frac{3}{2}\frac{x_k}{b_kx_{k-1}}-\frac{1}{2b_k}-\eta\frac{y_k^3}{b_kx_{k-1}}=$$ $$=\frac{3}{2}-\frac{1}{2b_k}-\eta\frac{y_k^3}{b_kx_{k-1}}>\frac{3}{2}-\frac{1}{2b_k}-\eta\frac{x_k^3}{b_kx_{k-1}}=\frac{3}{2}-\frac{1}{2b_k}-\eta x_k^2.$$ So $b_{k+1}>\frac{3}{2}-\frac{1}{2b_k}-\eta x_k^2$. Now note that $x_k<1-\frac{3}{2}\eta$ for all $k\geq3$, so that $b_{k+1}>\frac{3}{2}-\frac{1}{2b_k}-\eta(1-\frac{3}{2}\eta)^2>\frac{3}{2}-\frac{1}{2b_k}-\frac{1}{8}(1-\frac{3}{16})^2=\frac{2903}{2048}-\frac{1}{2b_k}\sim1.407-\frac{1}{2b_k}$. Using this you can prove by induction that $b_k>0.7$ for all $k$ (as base case you need to check $b_2,b_3>0.7$). Using that fact and that: * *$b_{k+1}>\frac{3}{2}-\frac{1}{2b_k}-x_k^2$ *$x_k\to0$ it's not difficult to prove that $b_k\to 1$. Now let's prove that $x_k=\Theta(\frac{1}{\sqrt{k}})$. To do it we will compare it with the sequence $a_k=\frac{1}{10\sqrt{k}}$. Note that we have $a_{k+1}<a_k-10a_k^3$. So if we prove that for big enough $k$ we have $x_{k+1}>x_k-10x_k^3$, then our series will decrease slower than $a_k$ and we will be done. This is equivalent to proving $\frac{x_k-x_{k+1}}{x_k^3}<10$ for big enough $k$. So let's study $c_k:=\frac{x_k-x_{k+1}}{x_k^3}$. Of course $c_k$ is always positive, and we have $$c_{k}=\frac{x_k-x_{k+1}}{x_k^3}=\frac{x_k-y_k+\eta y_k^3}{b_k^3x_{k-1}^3}=\frac{\frac{1}{2}(x_{k-1}-x_k)+\eta y_k^3}{b_k^3x_{k-1}^3}=\frac{1}{2b_k^3}c_{k-1}+\frac{\eta}{b_k^3}\left(\frac{y_k}{x_{k-1}}\right)^3.$$ Now using that $b_k\to 1$ and $\frac{y_k}{x_{k-1}}\to1$, the inequality implies that for big $k$ we have $c_k<0.7c_{k-1}+1$. So for big enough $k$ we have $c_k<10$, as we wanted.

Equality of the sum of powers Hi everyone, I got a problem when proving lemmas for some combinatorial problems, and it is a question about integers. Let $\sum_{k=1}^m a_k^t = \sum_{k=1}^n b_k^t$ be an equation, where $m, n, t, a_i, b_i$ are positive integers, and $a_i \neq a_j$ for all $i, j$, $b_i \neq b_j$ for all $i, j$, $a_i \neq b_j$ for all $i, j$. Does the equality have no solutions? For $n \neq m$, it is easy to find solutions for $t=2$ by Pythagorean theorem, and even for $n = m$, we have solutions like $1^2 + 4^2 + 6^2 + 7^2 = 2^2 + 3^2 + 5^2 + 8^2$. For $t > 2$, similar equalities hold: $1^2 + 4^2 + 6^2 + 7^2 + 10^2 + 11^2 + 13^2 + 16^2 = 2^2 + 3^2 + 5^2 + 8^2 + 9^2 + 12^2 + 14^2 + 15^2$ and $1^3 + 4^3 + 6^3 + 7^3 + 10^3 + 11^3 + 13^3 + 16^3 = 2^3 + 3^3 + 5^3 + 8^3 + 9^3 + 12^3 + 14^3 + 15^3$, and we can extend this trick to all $t > 2$. The question is, if we introduce one more restriction, that is, $|a_i - a_j| \geq 2$ and $|b_i - b_j| \geq 2$ for all $i, j$, is it still possible to find solutions for the equation? For $t = 2$ we can combine two Pythagorean triples, say, $5^2 + 12^2 + 25^2 = 7^2 + 13^2 + 24^2$, but how about the cases for $t > 2$ and $n = m$?

One set of solutions for t = 3 is the class of numbers known as Taxicab Numbers, named after the number of a taxicab G. H. Hardy took, 1729, that Ramanujan mentioned was equal to 13 + 123 = 93 + 103. This particular example fails, as |10 - 9| = 1 < 2, but there are other Taxicab numbers, such as: 1673 + 4363 = 2283 + 4233 = 2553 + 4143. This might be a helpful site for your question. -Gabriel Benamy

'Sign matrices'-(-1,+1) square matrices My question arises from a discussion on an answer given by Maurizio Monge here.I do not know if there is a known terminology for such matrices. By "sign matrices," I mean square matrices whose entries are in ${-1,+1}$. For instance, $\begin{bmatrix} 1 &-1 \\ -1& -1 \end{bmatrix}$ , $\begin{bmatrix} -1&1&1 \\ 1&1&-1 \\ -1&-1&-1 \end{bmatrix}$ Clearly, there are $2^{n^2}$ sign matrices of size $n\times n$. So, we start their theory by enumerating them as follows. For a matrix of size $n\times n$ we consider a truth table of $n^2$ arguments and therefore $2^{n^2}$ rows. Each row corresponds to the entries in one matrix$(a_{11},a_{12},\dots,a_{1n},a_{21},a_{22},\dots,a_{nn})$. Let $M_{(n,k)}$ be the $n \times n$ sign matrix corresponding to the $k^th$ row of the truth table. Question: Does the following matrix product give the zero matrix for sign matrices of even size? $\prod_{k=1}^{2^{n^2}}M_{(n,k)}$ Thank you. As usual, I will be delighted if you point me to good references on this.

It is not part of the question but I consider it useful to give the result for the $M_3$ case( which I found using a $C$++ program). There are 512 rows and the matrix below is $M_3$. However, I am not completely certain since the compiler I used might have rounded off the numbers while multiplying the matrices though the data type used was long double,the most precise by far. \begin{matrix} -1 & 1 & -1 & 1 & -1 & -1 & -1 & -1 & 1\\ -1 & 1 & -1 & 1 & -1 & -1 & -1 & -1 & -1\\ -1 & 1 & -1 & -1 & 1 & 1 & 1 & 1 & 1\\ -1 & 1 & -1 & -1 & 1 & 1 & 1 & 1 & -1\\ -1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 & 1\\ -1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 & -1\\ \end{matrix} $$\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots$$ \begin{matrix} -1 & -1 & -1 & -1 & -1 & -1 & -1 & 1 & -1\\ -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & 1\\ -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \end{matrix} \begin{bmatrix} 1.50197\times10^{100} & 1.50197\times10^{100} & 1.50197\times10^{100}\\ 1.50197\times10^{100} & 1.50197\times10^{100} &1.50197\times10^{100} \\ 1.50197\times10^{100}& 1.50197\times10^{100} &1.50197\times10^{100} \end{bmatrix} It is sad that the "array size" blows out of the compiler's capacity when running my program for the $M_4$ case, the case that would either kill or save my question.

A Diophantine equation Let $a,b$ be integers, $a>b\ge 1$, $a^2(a+1)$ be divisible by $b$, and $3a^2$ be divisible by $b$. Let us consider the following expression: $\frac {1+3a+3a^2+a^2(a-b+1)/b} {1+\frac{3}{a}+\frac{3}{a^2}+\frac{b}{a^2(a-b+1)}}$. This fraction is always integer if $b=1$. For $b>1$, I know only one pair $a,b$ such that the fraction is integer, namely, $a=15,b=9$ (I checked all pairs with $a<10^4$). Can anyone prove that there are no other pairs $a,b$ such that the fraction is integer? This question is related to the algebraic graph theory. Thanks for any comments or hints!

If we substitute $f= \frac{a-b+1}{ab}$ and subtract $a^3$ the question becomes; when is this an integer: $$\frac{a^3 \times (f-1) \times (a^3 + \frac{1}{f})}{(a+1)^3 + \frac{1}{f} -1}$$ Note that $b=1$ implies that $f=1$ and this quotient is zero. Using Maple I found that $a=14$, $b=12$ is a second example which satisfies your conditions. (sorry, not a full answer, I'm new here and not sure how to post)

$ - \sum_{n=1}^\infty \frac{(-1)^n}{2^n-1} =? \sum_{n=1}^\infty \frac{1}{2^n+1}$ Numerical evidence suggests: $$ - \sum_{n=1}^\infty \frac{(-1)^n}{2^n-1} =? \sum_{n=1}^\infty \frac{1}{2^n+1} \approx 0.764499780348444 $$ Couldn't find cancellation via rearrangement. For the second series WA found closed form. Is the equality true?

This is just a formal proof using idea of Loïc Teyssier that appeared in the comments. We just rewrite both sides of the equation as double sums (using geometric series) and notice that they are equal after changing the order of summation. Since $$\frac{q}{1-q} = \sum_{k=1}^\infty q^k,$$ we have for the left hand side \begin{align} - \sum_{n=1}^\infty \frac{(-1)^n}{2^n-1} & = - \sum_{n=1}^\infty \frac{(-1)^n2^{-n}}{1-2^{-n}} \\ & = - \sum_{n=1}^\infty (-1)^n\sum_{k=1}^\infty (2^{-n})^k \\ & = \sum_{k,n = 1}^\infty (-1)^{n+1} 2^{-nk}. \end{align} Similarly, for the right hand side we use $$\frac{q}{1+q} = \sum_{k=1}^\infty (-1)^{k+1} q^k $$ to get \begin{align} \sum_{n=1}^\infty \frac{1}{1+2^n} &= \sum_{n=1}^\infty \frac{2^{-n}}{1+2^{-n}}\\ & = \sum_{n=1}^\infty \sum_{k=1}^\infty (-1)^{k+1}(2^{-n})^k \\ & = \sum_{k,n=1}^\infty (-1)^{k+1} 2^{-nk}. \end{align}

On class numbers $h(-d)$ and the diophantine equation $x^2+dy^2 = 2^{2+h(-d)}$ Given fundamental discriminant $d \equiv -1 \bmod 8$ such that the quadratic imaginary number field $\mathbb{Q}(\sqrt{-d})$ has odd class number $h(-d)$. Is it true that one can always solve the diophantine equation, $$x^2+dy^2 = 2^{2+h(-d)}\tag{1}$$ with odd integers $x,y$? For example, one can indeed solve, $$x^2+7y^2 = 2^3$$ $$x^2+23y^2 = 2^5$$ $$x^2+47y^2 = 2^7$$ $$x^2+71y^2 = 2^9$$ which have $d$ with $h(-d) = 1,3,5,7$, respectively, and it is solvable for all such $d$ with $h(-d) \leq 25$ in the link given above (I tested them all), but it would be nice to know if it is true in general.

Or, put the cheesy way: with $d = p \equiv 7 \pmod 8,$ there is a primitive positive quadratic form with coefficients $$ \langle 2, 1 , \frac{p+1}{8} \rangle. $$ As a result, with class number $h,$ by repeated Gauss composition we know the principal form $$ \langle 1, 1 , \frac{p+1}{4} \rangle $$ integrally represents $2^h.$ Note that, from Theorem 4.12 on page 64 of Buell, the composition algorithm of Shanks, as we continue with repeated Gauss composition, the middle coefficient remains $\equiv 1 \pmod 4,$ which means that we are showing that the principal form primitively represents $2^h.$ Next, we have the same class number in the genus of $$\langle 1, 0 , p \rangle, $$ although there is no longer a primitive form that represents $2$ itself. Page 118, Theorem 7.5 (a) in Buell, Binary Quadratic Forms. Finally, $$ 4 (x^2 + xy +\frac{p+1}{4} y^2 ) = 4 x^2 + 4 x y + (p+1) y^2 = (2x+y)^2 + p y^2, $$ so we can represent $4 \cdot 2^h$ as $u^2 + p v^2.$ EDIT, Tuesday: Note that $x^2 + xy +\frac{p+1}{4} y^2$ and $u^2 + p v^2$ represent exactly the same odd numbers, as $\frac{p+1}{4}$ is even, so to get an odd number we have $x^2 + xy = x(x+y)$ odd, so $x$ is odd and $y$ is even; take $u=x, v = y/2.$

Sets of squares representing all squares up to $n^2$ Let $S_n=\{1,2,\ldots,n\}$ be natural numbers up to $n$. Say that a subset $S \subseteq S_n$ square-represents $S_n^2$ if every square $1^2,2^2,\ldots,n^2$ can be represented by adding or subtracting at most one copy of squares of elements of $S$. Example. For $n=7$, the set $S$ of $|S|=5$ numbers $\{1, 2, 3, 5, 7\}$ square-represents $S_7^2$ because $4^2 = 5^2 - 3^2$ and $6^2 = 7^2 - 2^2 - 3^2$. The "at most one copy" condition means that one could represent each square missing from $S$ by a two-pan balance. My question is: Q. What is the minimum size $|S|$ to square-represent $S_n^2$ as $n$ gets large? In particular, is this size sublinear, $o(n)$? In some sense this asks for the frequency of occurrences of Pythagorean triples, quadruples, and other analogous solutions of equations of the form $a^2 = b^2 \pm c^2 \pm d^2 \pm \cdots$ with $a,b,c,d,\ldots$ distinct. This may be well-known to the experts, in which case I apologize for asking a naive question.

Here is a little example following Jeremy Rouse's construction. Let $A$ be the $a_i$ terms and $B$ the $b_i$ terms. $$A=\{1,2,3,4,5,6,8,11,16\}$$ $$B=\{0,0,0,0,42,78,142,263,519\}$$ For example $$a_9 = \lfloor \sqrt{b_8} \rfloor = \lfloor \sqrt{263} \rfloor = 16 \;,$$ and $$b_9=b_8+a^2_9 = 263+16^2 = 519 \;.$$ The missing squares can be achieved as follows: $$7^2 = 2^2 + 3^2 + 6^2$$ $$9^2 = 1^2 + 4^2 + 8^2$$ $$10^2 = 6^2 + 8^2$$ $$12^2 = 3^2 - 11^2 + 16^2$$ $$13^2 = -4^2+ 8^2+ 11^2$$ $$14^2 = 2^2 -8^2 + 16^2$$ $$15^2 = 1^2 + 2^2 - 6^2 + 16^2$$

Can a block matrix with at least 3 zero blocks of different size on the diagonal and 1's everywhere else have only integer eigenvalues? Let $M=\begin{pmatrix} \begin{array}{cccccccc} 0 & 0 & 1 & 1 & 1 & 1 & 1 &1\\ 0 & 0 & 1 & 1 & 1 & 1 & 1 &1\\ 1 & 1 & 0 & 0 & 0 & 1 & 1 &1\\ 1 & 1 & 0 & 0 & 0 & 1 & 1 &1\\ 1 & 1 & 0 & 0 & 0 & 1 & 1 &1\\ 1 & 1 & 1 & 1 & 1 & 0 & 0 &1\\ 1 & 1 & 1 & 1 & 1 & 0 & 0 &1\\ 1 & 1 & 1 & 1 & 1 & 1 & 1 &0\\ \end{array} \end{pmatrix}\implies\begin{pmatrix} \begin{array}{ccc} 0_{2\times2} & 1_{2\times3} & 1_{2\times2}&1_{2\times1}\\ 1_{3\times2} & 0_{3\times3} & 1_{3\times2}&1_{3\times1} \\ 1_{2\times2} & 1_{2\times3} & 0_{2\times2}&1_{2\times1} \\ 1_{1\times2} & 1_{1\times3} & 1_{1\times2}&0_{1\times1} \\ \end{array} \end{pmatrix}$ be a symmetric matrix whose entries are blocks(diagonal blocks entry must be zero and non-diagonal blocks entry must be one). How to prove/disprove that all the eigenvalues of $M$ are not integer if and only if the diagonal blocks are not of the same order.Note that the number of diagonal blocks $\geq 3$ and order of diagonal blocks$\geq 1$ .

You can have a look at Eigenvalues of complete multipartite graphs. One example with integer spectrum is $K_{2,2,2,8,8\ }.$

Integral quaternary forms and theta functions The following question arises when I attempt to understand the modular parameterization of the elliptic curve $$E:y^2-y=x^3-x$$ In Mazur-Swinnerton-Dyer and Zagier's construction, a theta function associated with a positive definite quadratic form is induced: $$\theta(q)=\sum_{x\in\mathbb{Z}^4}q^{\frac{1}{2}x^{T}Ax}$$ where $$A=\left(\begin{matrix}2 & 0 & 1 & 1\\ 0 & 4 & 1 &2\\ 1 & 1 & 10 & 1\\ 1 & 2 & 1 & 20 \end{matrix}\right)$$ $A$ is a positive definite matrix of determinant $37^2$, and we have $37A^{-1}=K^TAK$ where $K$ is an integral matrix of determinant $\pm 1$. Question: Suppose $A$ is a positive definite $4\times 4$ matrix with integral entries. All diagonal entries are even numbers. The determinant of $A$ is a square number $N^2$. Is it true that for every $N=p$ ($p>2$ is a prime number), there is at least one $A$ that $NA^{-1}=K^TAK$, where $K$ is an integral matrix of determinant $\pm 1$?

The answer is true, using the following construction. Let $B$ be the quaternion algebra of discriminant $p$ and let $O$ be a maximal order with an element $x$ satisfying $x^2 = -p$. The reduced norm is a quadratic form on $O$, with positive definite Gram matrix $A$ of determinant $p^2$. The matrix $A^{-1}$ then represents the reduced norm on the dual lattice $O^{\sharp}$. Recall that $\text{nrd}(\text{diff}(O)) = \text{discrd}(O) = p$ and since $O$ is maximal, $\text{diff}(O)$ is invertible and $O = \text{diff}(O) O^{\sharp}$ (see e.g. Voight, John, Quaternion algebras, ZBL07261776. Section 16.8), hence $$ p O^{\sharp} \subseteq \text{diff}(O) O^{\sharp} = O $$ which implies that $p \cdot (O^{\sharp} / O) = 0$ (therefore $O^{\sharp} / O$ is an abelian group of exponent $p$ and size $p^2$, so isomorphic to $(\mathbb{Z} / p \mathbb{Z})^2$). Next, we note that the matrix $A^{-1}$ is also the change of basis matrix between the chosen basis of $O$ and its dual, therefore $pA^{-1}$ is integral. This is not enough, but so far we have not used the element $x$. The matrix $pA^{-1}$ is the matrix representing the norm form on the ideal $xO^{\sharp}$, since $x^2 = -p$. But $nrd(xO^{\sharp})=p$ is a maximal order with an element $x$ such that $x^2 = -p$. By a theorem of Ibukiyama (see reference below), it is isomorphic (hence isometric as lattices) to $O$. This could also be seen directly (referring to some of the answers above - it is possible to do it using a finite number of cases) by explicitly writing down the maximal orders. I will list below explicit constructions, which are based on Ibukiyama, Tomoyoshi, On maximal orders of division quaternion algebras over the rational number field with certain optimal embeddings, Nagoya Math. J. 88, 181-195 (1982). ZBL0473.12012. - If $p \equiv 3 \bmod 4$, then $B = (-p,-1)$, and $O = \mathbb{Z}<(1+i)/2,j>$. In this case, we have $$ A = \left( \begin{array}{cccc} 2 & 1 & 0 & 0 \\ 1 & \frac{p+1}{2} & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 1 & \frac{p+1}{2} \end{array} \right) , pA^{-1} = \left( \begin{array}{cccc} \frac{p+1}{2} & -1 & 0 & 0 \\ -1 & 2 & 0 & 0 \\ 0 & 0 & \frac{p+1}{2} & -1 \\ 0 & 0 & -1 & 2 \end{array} \right), K = \left( \begin{array}{cccc} -1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right) $$ If $p \equiv 1 \bmod 4$, then choosing $q \equiv 3 \bmod 4$ such that $\left( \frac{q}{p} \right) = -1$, we can find $c$ such that $c^2 \equiv -p \bmod q$, and then $B = (-p,-q)$ and $$ O = \mathbb{Z} \oplus \mathbb{Z} \frac{1+j}{2} \oplus \mathbb{Z} \frac{i(1+j)}{2} \oplus \mathbb{Z} \frac{(c+i)j}{q} $$ In this case, we compute that $$ A = \left( \begin{array}{cccc} 2 & 1 & 0 & 0 \\ 1 & \frac{q+1}{2} & 0 & c \\ 0 & 0 & \frac{p(q+1)}{2} & p \\ 0 & c & p & \frac{2(p+c^2)}{q} \end{array} \right) , pA^{-1} = \left( \begin{array}{cccc} \frac{(q+1)(c^2+p)}{2q} & -c-\frac{c^2+p}{q} & -c & \frac{c(q+1)}{2} \\ -c-\frac{c^2+p}{q} & 2(c^2 + \frac{c^2+p}{q}) & 2c & -c(q+1) \\ -c & 2c & 2 & -q \\ \frac{c(q+1)}{2} & -c(q+1) & -q & \frac{q(q+1)}{2} \end{array} \right) $$ and if $\{e_1, e_2, e_3, e_4 \}$ is the above basis for $O$, then we see that $ \{e_2 e_4, ce_1 - e_4, e_1, j e_2 \} $ is a basis for the resulting module, which written in terms of the original basis yields the matrix $$ K = \left( \begin{array}{cccc} 0 & c & 1 & -\frac{q+1}{2} \\ -c & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ \frac{q+1}{2} & -1 & 0 & 0 \end{array} \right). $$ Referring to Will Jagy's wondering in the first answer, the reason that this does not work for most lattices is that they do not correspond to a maximal order (as in the answer by few_reps, the quotient of the lattices is actually cyclic) and there are only one or two (depending on $p \bmod 4$ isomorphism classes of maximal orders which contain a root of $-p$. For example, for $p = 37$, there are only two isomorphism classes of maximal orders in the quaternion algebra, and only one of them contains a root of $-p$.

Solution to a Diophantine equation Find all the non-trivial integer solutions to the equation $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4.$$

There is one idea. To search for the solution of the equation. $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=q$$ If we know any solution $(a,b,c)$ of this equation. Then it is possible to find another $(a_2, b_2, c_2)$. Make such a change. $$y=(a+b+2c)(q(a+b)-c)-(a+b)^2-(a+c)(b+c)$$ $$z=(a+2b+c)(2b-qa-(q-1)c)+(b+2a+c)(2a-qb-(q-1)c)$$ Then the following solution can be found by the formula. $$a_2=((5-4q)c-(q-2)(3b+a))y^3+((5-4q)b+4(1-q)c)zy^2+$$ $$+(3c+(q-1)(a-b))yz^2-az^3$$ $$b_2=((5-4q)c-(q-2)(3a+b))y^3+((5-4q)a+4(1-q)c)zy^2+$$ $$+(3c+(q-1)(b-a))yz^2-bz^3$$ $$c_2=2(q-2)cy^3+3(2-q)(a+b)zy^2+$$ $$+((5-4q)(a+b)+2(1-q)c)yz^2+(2c-(q-1)(a+b))z^3$$ I tried this formula to simplify, but nothing happens. Maybe someone will check in Maple?

How to deduce an equation from this 3 Diophantine equations with 5 variables? I have three equations: ${m \choose 2} + nk = {x \choose 2}$ ${n \choose 2} + mk = {y \choose 2}$ $x + y = m + n + k$ $m, n, k, x, y$ are natural numbers. I want to deduce from this 3 equations either $x = y$ or $m = n$. From where I got these equations makes me sure that this is only possible if $x = y$ and $m = n$. Just deducing either $x=y$ or $m=n$ is enough. I can show that if I show that $x + y$ is not divisible by 3. So it will be enough if we can show that $x + y$ is not divisible by 3.

The first two equalities imply $x>m$ and $y>n$ so one can substitute $x=m+X$, $y=n+Y$ and $k=X+Y$, with still $X,Y \in \mathbb N$: ${X \choose 2}=nX+nY-mX\tag{1}$ ${Y \choose 2}=mX+mY-nY\tag{2}$ From (1) follows: $\quad m=n+n\frac{Y}{X}-\frac{1}{X}{X \choose 2}$, then eliminate $m$ from (2): $\quad {Y \choose 2}+{X \choose 2}+\frac{Y}{X}{X \choose 2}=nX+nY+n\frac{Y^2}{X}$, and finally: $\quad 2n=X-\frac{X^2+2XY}{X^2+XY+Y^2}$. Clearly $X-2n=\frac{X^2+2XY+0Y^2}{X^2+XY+Y^2}\in(0,2)$, but since it is an integer it can only be $1$. This proves that $X=2n+1$ and by symmetry $Y=2m+1$. Substituting for $X$ and $Y$ in (1) or (2) finally yields $m=n$, so $X=Y$ and $x=y$.

Cycle index of $S_n \times S_m$ Given $n$, $m$ and $k$, I would like to evaluate cycle index of $S_n \times S_m$ for $c_1 = c_2 = ... = c_{nm} = k$. What is the fastest known algorithm to calculate it? For this particular case, is there any polynomial time algorithm? I know that $Z(S_n) = \frac{1}{n}\sum_{i=1}^na_iZ(S_{n-i})$. Maybe there is a similiar recursive formula for $Z(S_n \times S_m)$? Edit: For better explanation of the question here I give an example where $n = 2$, $m = 3$, $k = 2$. $Z(S_2) = \frac{1}{2}(a_1^2 + a_2)$ $Z(S_3) = \frac{1}{6}(b_1^3 + 3b_1b_2 + 2b_3)$ $Z(S_2 \times S_3) = \frac{1}{12} c_1^6 + \frac{1}{3} c_2^3 + \frac{1}{6} c_3^2 + \frac{1}{4} c_1^2 c_2^2 + \frac{1}{6} c_6$ And now substituting $c_1 = c_2 = c_3 = c_4 = c_5 = c_6 = k$ gives $\frac{1}{12} k^6 + \frac{1}{4} k^4 + \frac{1}{3} k^3 + \frac{1}{6} k^2 + \frac{1}{6} k$, so the answer I'm looking for is $13$. Note that I'm interested only in the value, not the polynomial itself.

$Z(S_n \times S_m)$ evaluated at $c_1 = c_2 = ... = c_{nm} = k$ equals $$\sum_{i_1+2i_2+\dots+ni_n=n}\frac{1}{1^{i_1} i_1!\cdots n^{i_n} i_n!} \sum_{j_1+2j_2+\dots+mj_m=m} \frac{1}{1^{j_1} j_1!\cdots m^{j_m} j_m!}\cdot k^{\sum_{p=1}^n\sum_{q=1}^m \gcd(p,q)i_p j_q}.$$ There are $p(n)\cdot p(m)$ terms in this double sum.

Minimal polynomial in $\mathbb Z[x]$ of seventh degree with given roots I am looking for a seventh degree polynomial with integer coefficients, which has the following roots. $$x_1=2\left(\cos\frac{2\pi}{43}+\cos\frac{12\pi}{43}+\cos\frac{14\pi}{43}\right),$$ $$x_2=2\left(\cos\frac{6\pi}{43}+\cos\frac{36\pi}{43}+\cos\frac{42\pi}{43}\right),$$ $$x_3=2\left(\cos\frac{18\pi}{43}+\cos\frac{22\pi}{43}+\cos\frac{40\pi}{43}\right)$$ $$x_4=2\left(\cos\frac{20\pi}{43}+\cos\frac{32\pi}{43}+\cos\frac{34\pi}{43}\right),$$ $$x_5=2\left(\cos\frac{10\pi}{43}+\cos\frac{16\pi}{43}+\cos\frac{26\pi}{43}\right),$$ $$x_6=2\left(\cos\frac{8\pi}{43}+\cos\frac{30\pi}{43}+\cos\frac{38\pi}{43}\right)$$ and $$x_7=2\left(\cos\frac{4\pi}{43}+\cos\frac{24\pi}{43}+\cos\frac{28\pi}{43}\right).$$ I see only that $\sum\limits_{k=1}^7x_k=-1$, but the computations for $\sum\limits_{1\leq i<j\leq7}x_ix_j$ and the similar are very complicated by hand and I have no any software besides WA, which does not help. Thank you for your help! Update. I got $$\sum\limits_{1\leq i<j\leq7}x_ix_j=-18.$$

By PARI / GP I get $x^7 + x^6 - 18*x^5 - 35*x^4 + 38*x^3 + 104*x^2 + 7*x - 49$ : K = nfinit (subst(polcyclo(43),x,y)) w = Mod(y,K.pol) f0(k) = (w^k + 1/w^k) f(k1,k2,k3) = f0(k1) + f0(k2) + f0(k3) v = [f(1,6,7),f(3,18,21),f(9,11,20),f(10,16,17),f(5,8,13),f(4,15,19),f(2,12,14)] /* = [x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7*x - 49, x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7*x - 49, x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7*x - 49, x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7*x - 49, x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7*x - 49, x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7*x - 49, x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7*x - 49] */ mps = [minpoly(w) | w<-v]

Can a product of Cohn matrices over the Eisenstein integers with non-zero, non-unit coefficients be a Cohn matrix? For $k > 1$, is it possible that $\begin{pmatrix} a_1 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} a_2 & 1 \\ -1 & 0 \end{pmatrix}\ldots \begin{pmatrix} a_k & 1 \\ -1 & 0 \end{pmatrix} = \pm \begin{pmatrix} b & 1 \\ -1 & 0 \end{pmatrix}$ if $a_1,a_2,\ldots a_k,b$ are Eisenstein integers and $|a_i| > 1$ for $i=1,2,\ldots k$? If the Eisenstein integers are replaced with the Gaussian integers, this is possible. $\begin{pmatrix} 3 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 - i & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 + i & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 - i & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} -2 & 1 \\ -1 & 0 \end{pmatrix} = -\begin{pmatrix} i & 1 \\ -1 & 0 \end{pmatrix}$ This problem came up in my research; I am primarily interested in the case where $b = (\pm 1 \pm \sqrt{-3})/2$, but I suspect that there might not be a solution for any choice of $b$. I originally asked this question on math.stackexchange (https://math.stackexchange.com/q/3903567/202799), but it seems to be more difficult than I first thought.

To my surprise, not only is there a solution for some $b$, there is actually a very simple infinite family of solutions for every $b$. Let $\omega = \frac{1 + \sqrt{-3}}{2}$. Then $\begin{pmatrix} a_0 + a_1\omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 -\omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 + \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 - \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 + \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 - \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} b_0 + b_1\omega & 1 \\ -1 & 0 \end{pmatrix} = -\begin{pmatrix} a_0 + b_0 - 1 + (a_1 + b_1 - 1)\omega & 1 \\ -1 & 0 \end{pmatrix}$ I wish I could say that there was some clever trick to finding this solution, but I cannot: it was found by a brute force search through $\approx 5$ million possibilities using Python after I realized that the bottom right coordinate of the product of $n$ Cohn matrices only depends on the inner $n - 2$ matrices. I do not know if there are shorter solutions. One can prove that there are no solutions with $k < 5$, but there could be solutions with $k = 5$ or $k = 6$. If there are, however, they have to involve some elements with squared norm at least $4$, as I checked everything smaller.

How to show a $3\times3$ matrix has three distinct eigenvalues? Here is a question I heared from others: Given four distinct positive real numbers $a_1,a_2,a_3,a_4$ and set $$a:=\sqrt{\sum_{1\leq i\leq 4}a_i^2}$$ $A=(x_{i,j})_{1\leq i\leq3,1\leq j\leq4}$ is a $3\times4$-matrix specified by $$ x_{i,j}=a_i\delta_{i,j}+a_j\delta_{4,j}-\frac{1}{a^2}(a_i^2+a_4^2)a_j $$ where $\delta_{i,j}$ is the Kronecker symbol or visually $$ A=\begin{pmatrix}a_1 &0&0&a_4\\ 0&a_2&0&a_4\\0&0&a_3&a_4\end{pmatrix}-\frac{1}{a^2} \begin{pmatrix} a_1(a_1^2+a_4^2) & a_2(a_1^2+a_4^2) & a_3(a_1^2+a_4^2) & a_4(a_1^2+a_4^2)\\ a_1(a_2^2+a_4^2) & a_2(a_2^2+a_4^2) & a_3(a_2^2+a_4^2) & a_4(a_2^2+a_4^2)\\ a_1(a_3^2+a_4^2) & a_2(a_3^2+a_4^2) & a_3(a_3^2+a_4^2) & a_4(a_3^2+a_4^2)\\ \end{pmatrix} $$ The question is to show that the $3\times3$-matrix $B=AA^T$ admits three distinct eigenvalues.($A^T$ is the transpose of $A$) What I am curious about is how many methods can be utilized to show a matrix has different eigenvalues? As for this question my idea is to calculate the characteristic polynomial $f$ of $B$ along with $f'$ which is a quadratic polynomial via Sagemath and show that neither of roots of $f'$ belongs to $f$. Or equivalently to calculate the resultant $R(f,f')$ of $f$ and $f'$ and show that $R(f,f')$ doesn't vanish for any distinct positive $a_i$'s. But the difficulties are both ways involve hideous calculation which I don't think I can write down by hand. So I'm wondering if there is a tricky way to get to that point? (e.g. an algebraic-geometry method?)

This question is from this year's Alibaba mathematics competition (qualifying round, which is finished 2 days ago), and here's my solution that could be wrong (I also participated in the competition and this is the solution I submitted). I tried to solve the problem geometrically to avoid tons of computations. First, we can deal with $A^{T}A$ instead of $AA^{T}$ since the first matrix's eigenvalues is same as the second eigenvalue's matrix with zero (consider SVD of $A$). The key point is that $A$ can be written as $A = BP$ where $$ B = \begin{pmatrix} a_1 & 0 & 0 & a_4 \\ 0 & a_2 & 0 & a_4 \\ 0 & 0 & a_3 & a_4 \end{pmatrix} $$ and $$ P = I_3 - \mathbf{v}\mathbf{v}^{T}, \mathbf{v} = \frac{1}{a}(a_1, a_2, a_3, a_4)^{T}. $$ Especially, the matrix $P$ is an orthogonal projection matrix that project a vector in $\mathbb{R}^{4}$ to the subspace of vectors that are perpendicular to $\mathbf{v}$. It satisfies $P^{T} = P^{2} = P$. To show that the eigenvalues are distinct, we will show that each eigenspace (for nonzero eigenvalues) has dimension 1. In other words, for a given eigenvalue, there exists a unique eigenvector (up to constant factor) corresponding to the eigenvalue. First, the above $\mathbf{v}$ is an eigenvector of $A^{T}A$ correspond to the eigenvalue 0 since $A\mathbf{v} = BP\mathbf{v} = \mathbf{0}$. Since the eigenvectors are orthogonal to each other, the other three eigenvectors are in the image of $P$ (the hyperplane perpendicular to $\mathbf{v}$). If we fix an (nonzero) eigenvalue $\lambda$ and a corresponding eigenvector $\mathbf{w}$, we have $P\mathbf{w} = \mathbf{w}$ and so $$ A^{T}A\mathbf{w} = PB^{T}BP\mathbf{w} = PB^{T}B\mathbf{w} = \lambda \mathbf{w}. $$ From this, the vector $B^{T}B\mathbf{w}$ should be written as $$ B^{T}B\mathbf{w} = \lambda \mathbf{w} + \beta \mathbf{v} $$ for some $\beta$. If we set $\mathbf{w} = (x_1, x_2, x_3, x_4)^{T}$, then expanding the above equation gives $$ \begin{pmatrix}a_1^2 & 0 & 0 & a_{1}a_{4} \\ 0 & a_{2}^{2} & 0 & a_{2}a_{4} \\ 0 & 0& a_{3}^{2} & a_{3}a_{4} \\ a_{1}a_{4} & a_{2}a_{4} & a_{3}a_{4} & 3a_{4}^{2} \end{pmatrix}\mathbf{w} = \begin{pmatrix} a_{1}^{2}x_{1} + a_{1}a_{4}x_{4} \\ a_{2}^{2}x_{2} + a_{2}a_{4}x_{4} \\ a_{3}^{2}x_{3} + a_{3}a_{4}x_{4} \\ a_{4}(a_{1}x_{1} + a_{2}x_{2} + a_{3}x_{3} + 3a_{4}x_{4})\end{pmatrix} = \begin{pmatrix} a_{1}^{2}x_{1} + a_{1}a_{4}x_{4} \\ a_{2}^{2}x_{2} + a_{2}a_{4}x_{4} \\ a_{3}^{2}x_{3} + a_{3}a_{4}x_{4} \\ 2a_{4}^{2}x_{4}\end{pmatrix} = \begin{pmatrix} \lambda x_1 + \beta' a_1 \\ \lambda x_2 + \beta' a_2 \\ \lambda x_3 + \beta' a_3 \\ \lambda x_4 + \beta' a_4\end{pmatrix}, \quad \beta' = \frac{\beta}{a} $$ Here we used $\langle \mathbf{v}, \mathbf{w} \rangle = a_{1}x_{1} + \cdots + a_{4}x_{4} = 0$ for the second equality. From this, we can show that $x_{4}$ should be nonzero (here's the point that distinctiveness of $a_i$'s are used), so we can assume that $x_4 = 1$ and the other components $x_1, x_2, x_3$ are uniquely determined. This proves our claim that each eigenspace has dimension 1, i.e. the eigenvalues are distinct.

Subtraction-free identities that hold for rings but not for semirings? Here is a concrete, if seemingly unmotivated, aspect of the question I am interested in: Question 1. Let $a$ and $b$ be two elements of a (noncommutative) semiring $R$ such that $1+a^3$ and $1+b^3$ and $\left(1+b\right)\left(1+a\right)$ are invertible. Does it follow that $1+a$ and $1+b$ are invertible as well? The answer to this question is * *"yes" if $ab=ba$ (because in this case, $1+a$ is a left and right divisor of the invertible element $\left(1+b\right)\left(1+a\right)$, and thus must itself be invertible; likewise for $1+b$). *"yes" if $R$ is a ring (because in this case, $1+a$ is a left and right divisor of the invertible element $1+a^3 = \left(1+a\right)\left(1-a+a^2\right) = \left(1-a+a^2\right)\left(1+a\right)$, and thus must itself be invertible; likewise for $1+b$). *"yes" if $1+a$ is right-cancellable (because in this case, we can cancel $1+a$ from $\left(1+a\right) \left(\left(1+b\right)\left(1+a\right)\right)^{-1} \left(\left(1+b\right)\left(1+a\right)\right) = 1+a$ to obtain $\left(1+a\right) \left(\left(1+b\right)\left(1+a\right)\right)^{-1} \left(1+b\right) = 1$, which shows that $1+a$ is invertible), and likewise if $1+b$ is left-cancellable. I am struggling to find semirings that are sufficiently perverse to satisfy none of these cases and yet have $\left(1+b\right)\left(1+a\right)$ invertible. (It is easy to find cases where $1+a^3$ is invertible but $1+a$ is not; e.g., take $a = \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}$ in the matrix semiring $\mathbb{N}^{2\times 2}$.) The real question I'm trying to answer is the following (some hopefully reasonably clear handwaving included): Question 2. Assume we are given an identity that involves only positive integers, addition, multiplication and taking reciprocals. For example, the identity can be $\left(a^{-1} + b^{-1}\right)^{-1} = a \left(a+b\right)^{-1} b$ or the positive Woodbury identity $\left(a+ucv\right)^{-1} + a^{-1}u \left(c^{-1} + va^{-1}u\right)^{-1} va^{-1} = a^{-1}$. Assume that this identity always holds when the variables are specialized to arbitrary elements of an arbitrary ring, assuming that all reciprocals appearing in it are well-defined. Is it then true that this identity also holds when the variables are specialized to arbitrary elements of an arbitrary semiring, assuming that all reciprocals appearing in it are well-defined? There is a natural case for "yes": After all, the same claim holds for commutative semirings, because in this case, it is possible to get rid of all reciprocals in the identity by bringing all fractions to a common denominator and then cross-multiplying with these denominators. However, this strategy doesn't work for noncommutative semirings (and even simple-looking equalities of the form $ab^{-1} = cd^{-1}$ cannot be brought to a reciprocal-free form, if I am not mistaken). Question 1 is the instance of Question 2 for the identity \begin{align} \left(1+a^3\right)^{-1} \left(1+b^3\right)^{-1} \left(1+a\right) \left(\left(1+b\right)\left(1+a\right)\right)^{-1} \left(1+b\right) = \left(1+a^3\right)^{-1} \left(1+b^3\right)^{-1} \end{align} (where, of course, the only purpose of the $\left(1+a^3\right)^{-1} \left(1+b^3\right)^{-1}$ factors is to require the invertibility of $1+a^3$ and $1+b^3$). Indeed, if $\alpha$ and $\beta$ are two elements of a monoid such that $\beta\alpha$ is invertible, then we have the chain of equivalences \begin{align} \left(\alpha\text{ is invertible} \right) \iff \left(\beta\text{ is invertible} \right) \iff \left( \alpha \left(\beta\alpha\right)^{-1} \beta = 1 \right) \end{align} (easy exercise).

The answer to your first question is yes (which was very surprising to me, to be honest). I have no idea whether the second question also has a positive answer. (By the way, don't let the work below fool you. This took me an entire week of serious computations to discover the main idea.) We will assume $(1+a^3)u=1$ and $d(1+a)=1$. We find that $$ d+au = d(1+(1+a)au) = d((1+a^3)u+(a+a^2)u)=d(1+a)(1+a^2)u=(1+a^2)u. $$ Thus, we compute $$ (1+a)d = d[1+(1+a)ad] = d[1+ad+a^2d] = d[u+a^3u+a^2d+ad] $$ $$ =d[a^2(d+au) + ad + u] = d[a^2(1+a^2)u+ad+u] = d[a(d+au)+a^4u+u] $$ $$ =d[a(1+a^2)u+(1+a^4)u] = d(1+a)(1+a^3)u=1. $$ Edited to add: A similar idea works for higher odd powers. The fifth power case is sufficient to give the main idea. Assume $(1+a^5)u=1=d(1+a)$. We find $$ d+(a+a^3)u = d[(1+a^5)u + (1+a)(a+a^3)u] = d(1+a)(1+a^2+a^4)u=(1+a^2+a^4)u. $$ Then we compute $$ (1+a)d=d^3[(1+a)^2+(1+a)^3ad] = d^3[(1+a)^2(1+a^5)u + (a+3a^2+3a^3+a^4)d] = d^3[u+2au+a^2u+2a^6u + (a+3a^2+3a^3)d + a^4[(a+a^3)u+d]] = \cdots $$ and you keep reducing monomials with $d$ to monomials involving only $a$ and $u$.

An infinite series that converges to $\frac{\sqrt{3}\pi}{24}$ Can you prove or disprove the following claim: Claim: $$\frac{\sqrt{3} \pi}{24}=\displaystyle\sum_{n=0}^{\infty}\frac{1}{(6n+1)(6n+5)}$$ The SageMath cell that demonstrates this claim can be found here.

Here is an elementary proof. We rewrite the series as $$\frac{1}{4}\int_0^1\frac{1-x^4}{1-x^6}\,dx=\frac{1}{8}\int_0^1\frac{dx}{1-x+x^2}+\frac{1}{8}\int_0^1\frac{dx}{1+x+x^2}.$$ It is straightforward to show that \begin{align*} \int_0^1\frac{dx}{1-x+x^2}&=\frac{2\pi}{3\sqrt{3}},\\ \int_0^1\frac{dx}{1+x+x^2}&=\frac{\pi}{3\sqrt{3}}, \end{align*} so we are done.

$2n \times 2n$ matrices with entries in $\{1, 0, -1\}$ with exactly $n$ zeroes in each row and each column with orthogonal rows and orthogonal columns I am interested in answering the following question: Question For a given $n$, does there exist a $2n \times 2n$ matrix with entries in $\{1, 0, -1\}$ having orthogonal rows and columns with exactly $n$ zeroes in each row and column? Conjectures * *For $n=2^k$, $k\ge0$ such a matrix always exists. *For $n=3$ such a matrix does not exist. *For $n=5$ such a matrix exists, for example: $$ \left( \begin{array}{cccccccccc} 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & -1 & 0 & -1 & -1 & -1 & 0 \\ 1 & 0 & 0 & -1 & 0 & 0 & -1 & 1 & 1 & 0 \\ 0 & -1 & -1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 \\ 1 & 0 & -1 & 0 & 0 & -1 & 1 & 0 & 0 & -1 \\ 1 & -1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & -1 & 1 & -1 & 0 & -1 & 0 & 1 \\ 0 & -1 & 1 & 0 & 0 & 0 & 0 & -1 & 1 & -1 \\ 0 & -1 & 1 & 0 & 0 & -1 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & -1 & 1 & 1 & 0 & 0 & -1 & -1 \\ \end{array} \right) $$ *For $n=7$ such a matrix does not exist.

Your first conjecture was proven by Nate in the comments. Your second conjecture is also true - there is no such matrix for $n=3$. If we just look at which entries are nonzero in each row, because any two rows are orthogonal, they must share an even number of nonzero entries, i.e. either share $0$ entries or $2$ entries. Only $3$ other rows can share $2$ nonzero entries with a given row, since otherwise there'd be more than $3$ nonzero entries in a column by the pidgeonhole principle, so $2$ rows must share $0$ nonzero entries, but then they share $3$ entries with each other, contradiction.

Equality of the sum of powers Hi everyone, I got a problem when proving lemmas for some combinatorial problems, and it is a question about integers. Let $\sum_{k=1}^m a_k^t = \sum_{k=1}^n b_k^t$ be an equation, where $m, n, t, a_i, b_i$ are positive integers, and $a_i \neq a_j$ for all $i, j$, $b_i \neq b_j$ for all $i, j$, $a_i \neq b_j$ for all $i, j$. Does the equality have no solutions? For $n \neq m$, it is easy to find solutions for $t=2$ by Pythagorean theorem, and even for $n = m$, we have solutions like $1^2 + 4^2 + 6^2 + 7^2 = 2^2 + 3^2 + 5^2 + 8^2$. For $t > 2$, similar equalities hold: $1^2 + 4^2 + 6^2 + 7^2 + 10^2 + 11^2 + 13^2 + 16^2 = 2^2 + 3^2 + 5^2 + 8^2 + 9^2 + 12^2 + 14^2 + 15^2$ and $1^3 + 4^3 + 6^3 + 7^3 + 10^3 + 11^3 + 13^3 + 16^3 = 2^3 + 3^3 + 5^3 + 8^3 + 9^3 + 12^3 + 14^3 + 15^3$, and we can extend this trick to all $t > 2$. The question is, if we introduce one more restriction, that is, $|a_i - a_j| \geq 2$ and $|b_i - b_j| \geq 2$ for all $i, j$, is it still possible to find solutions for the equation? For $t = 2$ we can combine two Pythagorean triples, say, $5^2 + 12^2 + 25^2 = 7^2 + 13^2 + 24^2$, but how about the cases for $t > 2$ and $n = m$?

For any $t$, if $m$ is sufficiently large relative to $t$, and $n$ is any positive integer (possibly equal to $m$), then the circle method proves that there exists an infinite sequence of increasingly large solutions such that the ratios between the $a_1,\ldots,b_n$ approach any real positive ratios you want (assuming that a real solution with those ratios exists for that $m,n,t$). This answers your question with much stronger inequalities than the ones you imposed. If you want, you can simultaneously specify the residues of the $a_i$ and $b_j$ modulo some fixed number $N$, provided that those residues are compatible with the equation. In fact, again when $m \gg t$, you can even fix the $b_j$ in advance: the solution to Waring's problem guarantees the existence of $a_1,\ldots,a_m$, and a slight strengthening lets you impose your inequalities too, if the $b_j$ are large enough. (Proof: the circle method again.)

Pythagorean 5-tuples What is the solution of the equation $x^2+y^2+z^2+t^2=w^2$ in polynomials over C ("Pythagorean 5-tuples")? There are simple formulas describing Pythagorean n-tuples for n=3,4,6: * *n=3. The formula for solutions of $x^2+y^2=z^2$ [4]: $x=d(p^2−q^2)$, $y=2dpq$, $z=d(p^2+q^2),$ where p,q,d are arbitrary polynomials. * *n=4. Similarly, all Pythagorean 4-tuples of polynomials are given by the identity [4, Theorem 2.2], [3, Theorem 7.1]: $(p^2+q^2-r^2-s^2)^2+(2pr+2qs)^2+(2ps-2qr)^2=(p^2+q^2+r^2+s^2)^2$ * *n=6. The following identity produces Pythagorean 6-tuples [3, Theorem 7.2]: $(m^2-n^2)^2+(2mn)^2+(2(n_0m_1-m_1n_0+m_3n_2-m_2n_3))^2+$ $(2(n_0m_2-m_2n_0+m_1n_3-m_3n_1))^2+$ $(2(n_0m_3-m_3n_0+m_2n_1-m_1n_2))^2=(m^2+n^2)^2$ where $m=(m_1,m_2,m_3,m_4)$, $n=(n_1,n_2,n_3,n_4)$, and $mn$ is the usual scalar product. These identities are somehow related to sl(2,R), sl(2,C), sl(2,H), but the case n=5 is missing in this description [3, Theorems 7.1 and 7.2]. The above formuli describe also Pythagorean n-tuples of integers. There are another descriptions of those; see [2] and [5, Chapter 5] for n<10, [6] for n<15, [1, Theorem 1 in Chapter 3] for arbitrary n, and also the answers below. There are reasons to believe that Pythagorean 5-tuples cannot be described by a single polynomial identity. Thus identities producing a "large" set of solutions are also of interest, like $(-p^2+q^2+r^2+s^2)^2+(2pq)^2+(2pr)^2+(2ps)^2=(p^2+q^2+r^2+s^2)^2$ This identity does not give all solutions because it produces only reducible polynomials y,z,t (once p,q,r,s are nonconstant). Examples of solutions which cannot be obtained by the approaches in the answers below are also interesting. Given a solution (x,y,z,t,w), methods to construct a new solution (x',y',z',t',w') are also of interest. For instance, $x'=w+y+z$, $y'=w+z+x$, $z'=w+x+y$, $t'=t$, $w'=2w+x+y+z$ [see the answer of Ken Fan below for generalizations to other n] or $x'+iy'+jz'+kt'=q(x+iy+jz+kt)q$, where $q$ is arbitrary polynomial with quaternionic coefficients [see the answer of Geoff Robinson below]. -- [1] L.J. Mordell, Diophantine Equations, Academic Press, London, 1969 [2] D. Cass, P.J. Arpaia, MATRIX GENERATION OF PYTHAGOREAN n-TUPLES, Proc. AMS 109:1 (1990), 1-7 [3] J. Kocik, Clifford Algebras and Euclid’s Parametrization of Pythagorean Triples, Adv. Appl. Clifford Alg. 17:1 (2007), 71-93 [4] R. Dietz, J. Hoschek and B. Juttler, An algebraic approach to curves and surfaces on the sphere and on other quadrics, Computer Aided Geom. Design 10 (1993) 211-229 [5] V. Kac, Infinite-Dimensional Lie Algebras (3rd edn. ed.), CUP, 1990 [6] E. Vinberg, The groups of units of certain quadratic forms (Russian), Mat. Sbornik (N.S.) 87(129) (1972), 18–36

In fact, the equation: $X^2+Y^2+Z^2+R^2=W^2$ Solutions look like this: $X=2psabk^2+a^2k^2s^2-ckabs^2-abk^2s^2$ $Y=2psabk^2+a^2k^2s^2-ckabs^2+2abk^2s^2$ $Z=2psabk^2+a^2k^2s^2+2ckabs^2-abk^2s^2$ $R=2p^2b^2k^2+c^2b^2s^2+b^2k^2s^2-ckb^2s^2-a^2k^2s^2+2psabk^2$ $W=2p^2b^2k^2+2psabk^2+c^2b^2s^2+b^2k^2s^2-ckb^2s^2+2a^2k^2s^2$ And the formula: $X=2psabk^2+abk^2s^2+ckabs^2-a^2k^2s^2$ $Y=a^2k^2s^2-ckabs^2+2abk^2s^2-2psabk^2$ $Z=a^2k^2s^2+2ckabs^2-abk^2s^2-2psabk^2$ $R=2p^2b^2k^2+c^2b^2s^2+b^2k^2s^2-ckb^2s^2-a^2k^2s^2-2psabk^2$ $W=2p^2b^2k^2+c^2b^2s^2+b^2k^2s^2-ckb^2s^2+2a^2k^2s^2-2psabk^2$ Where the numbers: $p,c,s,a,k,b$ integers and set us. Quite often, after a number of substitutions should be divided by the greatest common divisor.

Integral of the product of Normal density and cdf I am struggling with an integral pretty similar to one already resolved in MO (link: Integration of the product of pdf & cdf of normal distribution ). I will reproduce the calculus bellow for the sake of clarity, but I want to stress the fact that my computatons are essentially a reproduction of the discussion of the previous thread. In essence, I need to solve: $$\int_{-\infty}^\infty\Phi\left(\frac{f-\mathbb{A}}{\mathbb{B}}\right)\phi(f)\,df,$$ where $\Phi$ is cdf of a standard normal, and $\phi$ its density. $\mathbb{B}$ is a negative constant. As done in the aforementioned link, the idea here is to compute the derivative of the integral with respect to $\mathbb{A}$ (thanks to Dominated Convergence Theorem, integral and derivative can switch positions). With this, \begin{align*} \partial_A\left[\int_{-\infty}^\infty\Phi\left(\frac{f-A}{B}\right)\phi(f)\,df\right]&=\int_{-\infty}^\infty\partial_A\left[\Phi\left(\frac{f-A}{B}\right)\phi(f)\right]\,df=\int_{-\infty}^\infty-\frac{1}{B}\phi\left(\frac{f-A}{B}\right)\phi(f)\,df \end{align*} We note now that $$\phi\left(\frac{f-A}{B}\right)\phi(f)=\frac{1}{2\pi}\exp\left(-\frac{1}{2}\left[\frac{(f-A)^2}{B^2}+f^2\right]\right)=\exp\left(-\frac{1}{2B^2}\left[f^2(1+B^2)+A^2-2Af\right]\right)$$ $$=\frac{1}{2\pi}\exp\left(-\frac{1}{2B^2}\left[\left(f\sqrt{1+B^2}-\frac{A}{\sqrt{1+B^2}}\right)^2+\frac{B^2}{1+B^2}A^2\right]\right)$$ Finally, then, $$\partial_A\left[\int_{-\infty}^\infty\Phi\left(\frac{f-A}{B}\right)\phi(f)\,df\right]$$ $$\ \ \ \ \ =-\frac{1}{\sqrt{2\pi}B}\exp\left(-\frac{A^2}{2(1+B^2)}\right)\frac{1}{2\pi}\int_{-\infty}^\infty\exp\left(-\frac{1}{2B^2}\left[f\sqrt{1+B^2}-\frac{A}{\sqrt{1+B^2}}\right]^2\right)\,df$$ and with the change of variable \begin{align} \left[y\longmapsto f\frac{\sqrt{1+B^2}}{B}-\frac{A}{B\sqrt{1+B^2}}\Longrightarrow df=\frac{B}{\sqrt{1+B^2}}\,dy\right] \end{align} we get \begin{align} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\exp\left(-\frac{1}{2B^2}\left[f\sqrt{1+B^2}-\frac{A}{\sqrt{1+B^2}}\right]^2\right)\,df=\frac{B}{\sqrt{1+B^2}}\int_{-\infty}^{\infty}\phi(y)\,dy=\frac{B}{\sqrt{1+B^2}} \end{align} This means that \begin{align} \partial_A\left[\int_{-\infty}^\infty\Phi\left(\frac{f-A}{B}\right)\phi(f)\,df\right]&=-\frac{1}{\sqrt{2\pi}B}\exp\left(-\frac{A^2}{2(1+B^2)}\right)\frac{B}{\sqrt{1+B^2}}=-\frac{1}{\sqrt{1+B^2}}\phi\left(\frac{A}{\sqrt{1+B^2}}\right) \end{align} At this point, given that (as $\mathbb{B}$ is negative) $$\Phi\left(\frac{f-A}{\mathbb{B}}\right)\phi(f)=0$$ when $\mathbb{A}\rightarrow-\infty$, the integral we are looking for is equal to \begin{align} \int_{-\infty}^{\mathbb{A}}-\frac{1}{\sqrt{1+\mathbb{B}^2}}\phi\left(\frac{A}{\sqrt{1+\mathbb{B}^2}}\right)\,dA \end{align} Again with the obvious change of variables $$\left[y\longmapsto\frac{A}{\sqrt{1+\mathbb{B}^2}}\Longrightarrow\sqrt{1+\mathbb{B}^2}\,dy=dA\right]$$ one gets \begin{align} \int_{-\infty}^{\mathbb{A}}-\frac{1}{\sqrt{1+\mathbb{B}^2}}\phi\left(\frac{A}{\sqrt{1+\mathbb{B}^2}}\right)\,dA=-\frac{1}{\sqrt{1+\mathbb{B}^2}}\sqrt{1+\mathbb{B}^2}\int_{-\infty}^{\mathbb{A}/\sqrt{1+\mathbb{B}^2}}\phi(y)\,dy=-\Phi({\mathbb{A}/\sqrt{1+\mathbb{B}^2}}). \end{align} The problem here is that this number should obviously be positive, so at some point I am missing a signal. As the computations seem sound to me, I would like to see if anyone could help me to find my mistake. Many thanks to you all.

When you do the change of variable, you are doing: \begin{align} \left[y\longmapsto f\frac{\sqrt{1+B^2}}{B}-\frac{A}{B\sqrt{1+B^2}}\Longrightarrow df=\frac{B}{\sqrt{1+B^2}}\,dy\right] \end{align} .. which is correct. However, we need to apply the same change of variable to the bounds, ie to $-\infty$ and $+\infty$. Given that $\mathbb{B}$ is a negative constant, the bounds will change sign, becoming $+\infty$ and $-\infty$. So we will have: \begin{align} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\exp\left(-\frac{1}{2B^2}\left[f\sqrt{1+B^2}-\frac{A}{\sqrt{1+B^2}}\right]^2\right)\,df=\frac{B}{\sqrt{1+B^2}}\int_{\infty}^{-\infty}\phi(y)\,dy=-\frac{B}{\sqrt{1+B^2}} \end{align} This sign change will then propagate through to the final answer, reversing its sign too. Full working, based off your working above, but with some buggettes removed: Start with: $$ I = \def\A{\mathbb{A}} \def\B{\mathbb{B}} \int_{-\infty}^\infty \Phi\left( \frac{f - \A} {\B} \right)\phi(f)\, df $$ Take derivative wrt $\A$: $$ \partial_\A I = \int_{-\infty}^\infty \partial_\A \left( \Phi \left( \frac{f - \A}{\B} \right) \phi(f) \right) \, df $$ $$ =\int_{-\infty}^\infty \left( \frac{-1}{\B} \right) \phi\left( \frac{f - \A}{\B} \right) \phi(f) \, df $$ Looking at $E_1 = \phi\left(\frac{f - \A}{\B}\right) \phi(f)$: $$ E_1 = \frac{1}{2\pi} \exp \left( - \frac{1}{2} \left( \frac{f^2 - 2f\A + \A^2 + \B^2f^2} {B^2} \right) \right) $$ $$ =\frac{1}{2\pi} \exp\left( - \frac{1}{2\B^2} \left( \left( \sqrt{(1 + \B^2)}f - \A\frac{1}{\sqrt{1 + \B^2}} \right)^2 - \frac{\A^2} {1 + \B^2} + \A^2 \right) \right) $$ $$ - \frac{\A^2}{1+\B^2} + \A^2 $$ $$ = \frac{-\A^2 + \A^2 + \A^2\B^2} {1 + \B^2} $$ $$ = \frac{\A^2\B^2} {1 + \B^2} $$ Therefore $E_1$ is: $$ \frac{1}{2\pi} \exp \left( -\frac{1}{2\B^2} \frac{\A^2\B^2}{1 + \B^2} \right) \exp \left( - \frac{1}{2\B^2} \left( f\sqrt{1 + \B^2} - \frac{\A}{\sqrt{1 + \B^2}} \right)^2 \right) $$ $$ = \frac{1}{2\pi} \exp \left( -\frac{\A^2}{2(1 + \B^2)} \right) \exp \left( - \frac{1}{2\B^2} \left( f\sqrt{1 + \B^2} - \frac{\A}{\sqrt{1 + \B^2}} \right)^2 \right) $$ Make change of variable: $$ y = f\frac{\sqrt{1 + \B^2}}{\B} - \frac{A}{\B\sqrt{1 + \B^2}} $$ Therefore: $$ dy = \frac{\sqrt{1 + \B^2}}{\B}\,df $$ For the limits, we have $f_1 = -\infty$, and $f_2 = \infty$ $\sqrt{1 + \B^2}$ is always positive. $\B$ is always negative. Therefore: $$ y_1 = +\infty, y_2 = -\infty $$ Therefore: $$ \partial_\A I = \frac{-1}{\B}\int_{+\infty}^{-\infty} \frac{1}{2\pi} \exp \left( - \frac{\A^2} {2(1+\B^2)} \right) \exp \left( - \frac{1}{2} y^2 \right) \frac{\B}{\sqrt{1 + \B^2}} \, dy $$ $$ = \frac{1}{\sqrt{2\pi}} \frac{1}{\B} \frac{\B}{\sqrt{1 + \B^2}} \exp \left( - \frac{\A^2}{2(1 + \B^2)} \right) \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \exp\left( -\frac{1}{2} y^2 \right) \, dy $$ $$ = \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{1+\B^2}} \exp \left( - \frac{\A^2}{2(1 + \B^2)} \right) (1) $$ $$ = \frac{1}{\sqrt{1 + \B^2}} \phi\left( \frac{\A}{\sqrt{1 + \B^2}} \right) $$ Now we need to re-integrate back up again, since we currently have $\partial_\A I$, and we need $I$. Since we dont have limits, we'll need to find at least one known point. We have the following integral: $$ I = \frac{1}{\sqrt{1 + \B^2}} \int \phi\left( \frac{\A}{\sqrt{1 + \B^2}} \right) \,d\A $$ $$ = \frac{1}{\sqrt{1 + \B^2}} \sqrt{1 + \B^2} \Phi\left( \frac{\A} {\sqrt{1 + \B^2}} \right) + C $$ ... where $C$ is a constant of integration $$ = \Phi \left( \frac{\A}{\sqrt{1 + \B^2}} \right) + C $$ Looking at the original integral, we had/have: $$ I = \int_{-\infty}^\infty \Phi\left( \frac{f - \A}{\B} \right) \phi(f) \, df $$ We can see that, given that $\B$ is negative, as $\A \rightarrow \infty$, $\Phi\left(\frac{f-\A}{\B}\right) \rightarrow \Phi(\infty) = 1$. Therefore, as $\A \rightarrow \infty$, $\int_{-\infty}^\infty \Phi(\cdot)\phi(f)\,df \rightarrow 1$ Meanwhile, looking at the later expression for $I$, ie: $$ I= \Phi\left( \frac{\A}{\sqrt{1 + \B^2}} \right) + C $$ ... as $\A \rightarrow +\infty$, $\Phi\left( \frac{\A}{\sqrt{1 + \B^2}} \right) \rightarrow 1$ But we know that as $\A \rightarrow +\infty$, $I \rightarrow 1$. Therefore, $C = 0$ Therefore: $$ I = \Phi \left( \frac{\A} {\sqrt{1 + \B^2}} \right) $$

Solutions to $\binom{n}{5} = 2 \binom{m}{5}$ In Finite Mathematics by Lial et al. (10th ed.), problem 8.3.34 says: On National Public Radio, the Weekend Edition program posed the following probability problem: Given a certain number of balls, of which some are blue, pick 5 at random. The probability that all 5 are blue is 1/2. Determine the original number of balls and decide how many were blue. If there are $n$ balls, of which $m$ are blue, then the probability that 5 randomly chosen balls are all blue is $\binom{m}{5} / \binom{n}{5}$. We want this to be $1/2$, so $\binom{n}{5} = 2\binom{m}{5}$; equivalently, $n(n-1)(n-2)(n-3)(n-4) = 2 m(m-1)(m-2)(m-3)(m-4)$. I'll denote these quantities as $[n]_5$ and $2 [m]_5$ (this is a notation for the so-called "falling factorial.") A little fooling around will show that $[m+1]_5 = \frac{m+1}{m-4}[m]_5$. Solving $\frac{m+1}{m-4} = 2$ shows that the only solution with $n = m + 1$ has $m = 9$, $n = 10$. Is this the only solution? You can check that $n = m + 2$ doesn't yield any integer solutions, by using the quadratic formula to solve $(m + 2)(m +1) = 2(m - 3)(m - 4)$. For $n = m + 3$ or $n = m + 4$, I have done similar checks, and there are no integer solutions. For $n \geq m + 5$, solutions would satisfy a quintic equation, which of course has no general formula to find solutions. Note that, as $n$ gets bigger, the ratio of successive values of $\binom{n}{5}$ gets smaller; $\binom{n+1}{5} = \frac{n+1}{n-4}\binom{n}{5}$ and $\frac{n+1}{n-4}$ is less than 2—in fact, it approaches 1. So it seems possible that, for some $k$, $\binom{n+k}{5}$ could be $2 \binom{n}{5}$. This question was previously asked at Math StackExchange, without any answer, but some interesting comments were made. It was suggested that I ask here.

This isn’t a complete answer, but the problem “reduces” to finding the finitely many rational points on a certain genus 2 hyperelliptic curve. This is often possible by a technique involving a reduction to finding the rational points on a finite set of rank 0 elliptic curves—see for example “Towers of 2-covers of hyperelliptic curves” by Bruin and Flynn in Trans. Amer. Math. Soc. 357 (2005) #11 4329–4347. In your case, the curve is $u^2= 9t^6+16t^5-200t^3+256t+144$. There are the following 16 rational points: $(t,u)$ or $(t,-u)= (0,12)$, $(1,15)$, $(2,12)$, $(4,204)$, $(-1,9)$, $(-2,36)$, $(-4,180)$, and $(7/4,411/64)$. If these are the only rational points then the only non-trivial solution to your equation is $n=10$, $m=9$. To see this suppose that $$n(n-1)(n-2)(n-3)(n-4)=2m(m-1)(m-2)(m-3)(m-4).$$ Let $$y=(n-2)^2\text{ and }x=(m-2)^2.$$ Squaring both sides we find that $$y(y-1)(y-1)(y-4)(y-4)=4x(x-1)(x-1)(x-4)(x-4).$$ Suppose $y$ isn’t $0$. Then $4x/y$ is $t^2$ for some rational $t$ with $$(y-1)(y-4)=t(x-1)(x-4).$$ We replace $y$ by $4x/t^2$ in this equation and find that $$(t^5-16)x^2-(5t^5-20t^2)x+(4t^5-4t^4)=0.$$ So this last quadratic polynomial in $x$ has a rational root and its discriminant is a square. This gives the hyperelliptic curve above. Note that the case $n=10$, $m=9$ of your problem corresponds to the point $(7/4,411/64)$ on this curve. EDIT: More generally one can look for rational $m$ and $n$ with $[n]_5= 2[m]_5$. If $(t,u)$ is a rational point on the hyperelliptic curve with $t$ non-zero, set $x=(5t^5-20t^2+t^2u)/(2t^5-32)$ and $y=4x/t^2$. Then if $x$ is a square in $\mathbb Q$, one gets such an $m$ and $n$ with $m=2+\sqrt x$ and $n=2+\sqrt y$. Joro’s points lead in this way to the solutions $(n,m)=(10,9)$, $(10/3,5/3)$, and $(78/23,36/23)$, the last one being rather unexpected. (And as François notes, each $(n,m)$ gives a second solution $(4-n,4-m)$.) Perhaps these solutions and the trivial solutions with $m$ and $n \in \{0,1,2,3,4\}$ are the only rational solutions.

Symmetric sums and Representations of SO(3) I had tried to help someone on math.StackExchange to prove the identity: $$ (1-Tr(A))^2+\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2=4$$ I guess you could argue the left hand side is independent of basis. Then we can diagonalize. But I couldn't come up with an invariant way of expressing the 2nd term. Someone came up with $(1-\mathrm{Tr}\\,A)^2- \frac{1}{2}\mathrm{Tr}\left[(A-A^T)^2\right]=4$ and prove it. This look a bit Pythagoras' theorem since $1 - Tr A = 2 \cos \varphi$ so the other term must be $\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2 = 2 \sin \varphi$ which I also couldn't prove. Is this sum related to a direct sum of representations of SO(3)?

Using the parametrization $$ A= \begin{pmatrix} a^2+b^2-c^2-d^2&2bc-2ad &2bd+2ac \cr 2bc+2ad &a^2-b^2+c^2-d^2&2cd-2ab \cr 2bd-2ac &2cd+2ab &a^2-b^2-c^2+d^2\\ \end{pmatrix}, $$ which comes from quaternions of unit length we have $$ (1-tr (A))^2-\frac{1}{2} tr ((A-A^t)^2)-4= $$ $$ (9a^2 + b^2 + c^2 + d^2 + 3)(a^2 + b^2 + c^2 + d^2 - 1)=0, $$ since $a^2+b^2+c^2+d^2=1$.

Pythagorean triples related to non-isometric equidistant plane quadruples QUESTION   Do there exist integers   $u\ x\ A\ B$   such that   $x\ne 0$,   and the following two equalities hold: * *$ x^2 + (x-u)^2\ =\ A^2$ *$ x^2 + (x+u)^2\ =\ B^2$ ? REMARK   I have a family of pairs of quadruples   $S\ T\subseteq\mathbb Z^2$,   parametrized by   $(u\ x)$,   such that   $S\ T$ have the same six distances but are not isometric (with respect to the Euclidean distance). All six distances of such quadruples are integers   $\Leftrightarrow$   both integers   $x^2+(x-u)^2$   and   $x^2+(x+u)^2$   are full squares.

It's not even possible for the product of $x^2 + (x-u)^2$ and $x^2 + (x+u)^2$ to be a square unless $x=0$ or $u=0$: that product is $4x^4+u^4$, and the elliptic curve $4x^4+u^4 = y^2$ is isomorphic to $Y^2 = X^3 - X$, for which Fermat already proved that the obvious rational points are the only ones. For your application $x=0$ is forbidden and $u=0$ makes $A^2 = B^2 = 2x^2$ which would again imply $x=0$. To get from $4x^4+u^4=y^2$ to $Y^2 = X^3 - X$, let $y = 2x^2-m$ to get $4mx^2 = m^2-u^4$. If $x \neq 0$ this makes $m(m^2-u^4)$ a square. Then if $u\neq 0$ then write $m = Xu^2$ to get $u^6(X^3-X)$, etc.

Four-Square Theorem for Negative Coefficient What integers are not in the range of $a^2+b^2+c^2-x^2$ (for all integer combinations of a, b, c, and x)? This form is similar to that of Lagrange's Four-Square Theorem, for which the answer would be "none". The generalization by Ramanujan only seems to cover non-negative coefficients.

With indefinite forms, it is possible for ternary forms to be universal. Indeed, all are known. References are given in Modern Elementary Theory of Numbers by Leonard Eugene Dickson, (1939). With any integer $M$ and any odd $N,$ they are equivalent to (by an invertible linear change of variables) one of four: $$ xy - M z^2, $$ $$ 2xy - N z^2, $$ $$ 2 xy + y^2 - N z^2, $$ $$ 2 xy + y^2 - 2 N z^2. $$ If you allow $M=0$ you get a universal binary. There are universal quaternary forms without such a universal ternary "section," for example $$ w^2 - 2 x^2 + 3 y^2 - 6 z^2. $$ EEEEEDDDDIITTTTT: $2xy+ z^2$ is integrally equivalent to $a^2 + b^2 - c^2,$ by $$ \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & -1 & 1 \\ 1 & -1 & 1 \end{array} \right) \left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & -1 & -1 \\ 1 & 1 & 1 \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right). $$

How to prove that the odd continued fraction approximants of ln(1+X) are upper bounds? The odd order continued fraction approximants for $\ln(1+X)$ are $$X,\quad \frac{X^2+6X}{4X+6,}\quad \frac{X^3+21X^2+30X}{9X^2+36X+30,}\quad \dots.$$ In "Some bounds for the logarithmic function", Flemming Topsøe remarks that they are also $[n, n − 1]$-Padé approximants. Specific instances such as those shown above can trivially be seen to be upper bounds for $\ln(1+X)$ by taking the difference, differentiating, and observing that the result has the form $X^m/f(X)$ (where $m$ is odd and $f(X)$ is always positive). However, checking a few cases is not the same as proving the result in general. Topsøe presents closed forms for these approximants: rational functions of complicated summations involving multinomial coefficients. He also defines the numerators and denominators recursively. Nevertheless, proving the result by induction looks infeasible due to the sheer complexity of the formulas. Topsøe hints that these results are known to "experts", but his essay is maddeningly short on specifics. Am I overlooking something obvious? I need to formalise this proof by machine, so some sort of elementary argument would be preferable.

The main result of the paper you link is that $$\log(1+x) = \frac{x}{1+} \frac{x}{2+} \frac{x}{3+} \frac{2^2 x}{4+} \frac{2^2 x}{5+} \cdots$$ and what you want to know is that truncating this formula at an odd number of steps provides an upper bound. (I am using the standard shorthand $\frac{a}{b+} \frac{c}{d+} \cdots$ to denote $a/(b+c/(d+ \cdots ))$.) This is a general fact about $$\frac{a_1}{b_1+} \frac{a_2}{b_2+} \cdots$$ where $a_i$ and $b_i$ are any positive numbers. Define $c(m,n) = \frac{a_m}{b_m+} \frac{a_{m+1}}{b_{m+1}+} \cdots \frac{a_n}{b_n}$ for any positive $a_i$ and $b_i$. I claim that $$c(1,2) < c(1,4) < c(1,6) < \cdots < c(1,5) < c(1,3) < c(1,1). \quad (\ast)$$ We have $\frac{a_{2n-1}}{b_{2n-1}+} \frac{a_{2n}}{b_{2n}} >0$. Now, $\frac{a}{b+y}$ is a decreasing function of $y$, so that gives $$\begin{array}{rcr} \frac{a_{2n-2}}{b_{2n-2}+} \frac{a_{2n-1}}{b_{2n-1}+} \frac{a_{2n}}{b_{2n}} &<& \frac{a_{2n-2}}{b_{2n-2}} \\ \frac{a_{2n-3}}{b_{2n-3}+} \frac{a_{2n-2}}{b_{2n-2}+} \frac{a_{2n-1}}{b_{2n-1}+} \frac{a_{2n}}{b_{2n}} &>& \frac{a_{2n-3}}{b_{2n-3}+} \frac{a_{2n-2}}{b_{2n-2}} \\ \frac{a_{2n-4}}{b_{2n-4}+}\frac{a_{2n-3}}{b_{2n-3}+} \frac{a_{2n-2}}{b_{2n-2}+} \frac{a_{2n-1}}{b_{2n-1}+} \frac{a_{2n}}{b_{2n}} &<& \frac{a_{2n-4}}{b_{2n-4}+}\frac{a_{2n-3}}{b_{2n-3}+} \frac{a_{2n-2}}{b_{2n-2}} \\ \end{array}$$ etcetera. So $c(1,2n-2) < c(1,2n)$. A similar argument establishes that $c(1,2n-3) > c(1,2n-1)$ for all $n$, that $c(1,2n) < c(1,2n+1)$ and $c(1,2n+1) > c(1,2n)$. Putting all of that together, we have the inequalities of $(\ast)$. So all the odd convergents are upper bounds, and all the even convergents are lower bounds, for whatever the limit is. According to the paper you linked, the limit is $\log(1+x)$, so we win.

Hypergeometric sum specific value How to show? $${}_2F_1(1,1;\frac{1}{2}, \frac{1}{2}) = 2 + \frac{\pi}{2} $$ It numerically is very close, came up when evaluating: $$ \frac{1}{1} + \frac{1 \times 2}{1 \times 3} + \frac{1 \times 2 \times 3}{1 \times 3 \times 5} + \frac{1 \times 2 \times 3 \times 4}{1 \times 3 \times 5 \times 7} + \cdots = 1 + \frac{\pi}{2} ~ ?$$ I am missing something trivial, I am sure.

This is equivalent to showing that $\displaystyle\sum_{n=2}^\infty\frac{n!}{(2n-1)!!}=\frac\pi2$ , which, after multiplying both the numerator and the denominator with $(2n)!!=2^n\,n!$, and taking into account that $\dfrac{(2n)!}{n!^2}=$ $=\displaystyle{2n\choose n}$, can be rewritten as $\displaystyle\sum_{n=2}^\infty\frac{2^n}{\displaystyle{2n\choose n}}=\frac\pi2$ , which can ultimately be deduced from the wider identity $\displaystyle\sum_{n=1}^\infty\frac{(2x)^{2n}}{\displaystyle{2n\choose n}n^2}=2\arcsin^2x$, itself ultimately a consequence of integrating the Cauchy product between the binomial series expansion of $\arcsin'x=\dfrac1{\sqrt{1-x^2}}$ and that of its primitive. Hope this helps.

PDF of the product of normal and Cauchy distributions I am having trouble in finding out the resulting PDF of the product of normal and Cauchy distributions. It turns out that we have a general formula for calculating the PDF of product of two random distribution but the integral is not converging . Also, I tried using Mellin Transform method but it is getting too complicated. If anybody has any idea about how to approach the problem, please share it with me. Thanks

Assuming the normal is centered with variance $s$ and the Cauchy distribution has parameters $a, b$, combining this Wikipedia page and Mathematica gives $$ \text{ConditionalExpression}\left[\frac{i \left(\frac{e^{-\frac{z^2}{2 s^2 (a-i b)^2}} \left(2 \pi \text{erfi}\left(\frac{\left| z\right| }{\sqrt{2} (a s-i b s)}\right)+2 \text{Ei}\left(\frac{z^2}{2 (a-i b)^2 s^2}\right)+\log \left(\frac{s^2 (a-i b)^2}{z^2}\right)-\log \left(\frac{z^2}{s^2 (a-i b)^2}\right)-4 \log (-a+i b)-2 \log \left(\frac{s^2}{z^2}\right)\right)}{a-i b}-\frac{e^{-\frac{z^2}{2 s^2 (a+i b)^2}} \left(2 \pi \text{erfi}\left(\frac{\left| z\right| }{\sqrt{2} (a s+i b s)}\right)+2 \text{Ei}\left(\frac{z^2}{2 (a+i b)^2 s^2}\right)-\log \left(\frac{1}{(a+i b)^2}\right)+4 \log \left(-\frac{1}{a+i b}\right)+\log \left((a+i b)^2\right)\right)}{a+i b}\right)}{8 \sqrt{2} \pi ^{3/2} s}-\frac{i \left(\frac{e^{-\frac{z^2}{2 s^2 (a-i b)^2}} \left(2 \pi \text{erfi}\left(\frac{\left| z\right| }{\sqrt{2} (a s-i b s)}\right)-2 \text{Ei}\left(\frac{z^2}{2 (a-i b)^2 s^2}\right)+\log \left(\frac{1}{(a-i b)^2}\right)+4 \log (a-i b)-\log \left((a-i b)^2\right)\right)}{a-i b}+\frac{e^{-\frac{z^2}{2 s^2 (a+i b)^2}} \left(-2 \pi \text{erfi}\left(\frac{\left| z\right| }{\sqrt{2} (a s+i b s)}\right)+2 \text{Ei}\left(\frac{z^2}{2 (a+i b)^2 s^2}\right)-\log \left(\frac{1}{(a+i b)^2}\right)-4 \log (a+i b)+\log \left((a+i b)^2\right)\right)}{a+i b}\right)}{8 \sqrt{2} \pi ^{3/2} s},b\neq 0\land a=0\right] $$

Binomial coefficient identity It seems to be nontrivial (to me) to show that the following identity holds: $$ \binom {m+n}{n} \sum_{k=0}^m \binom {m}{k} \frac {n(-1)^k}{n+k} = 1. $$ This quantity is related to the volume of the certain polytope.

This is a specialization at $x = n$ of the rational function identity $$ \frac{1}{x(x+1)\cdots(x+m)} = \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{x+k}, $$ and this identity could be proved either by partial fractions with unknown coefficients and limits to find the coefficients (see my comment on Todd Trimble's answer) or by induction on $m$. For the inductive step, divide both sides of the above identity by $x+m+1$: $$ \frac{1}{x(x+1)\cdots(x+m)(x+m+1)} = \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{(x+k)(x+m+1)}. $$ Split up the right side by partial fractions to make it $$ \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{m+1-k}\left(\frac{1}{x+k} - \frac{1}{x+m+1}\right). $$ Break this into a sum of two terms: $$ \frac{1}{m!}\sum_{k=0}^m \binom{m}{k}\frac{1}{m+1-k}\frac{(-1)^k}{x+k} - \frac{1}{m!}\left(\sum_{k=0}^m\binom{m}{k}\frac{(-1)^k}{m+1-k}\right)\frac{1}{x+m+1}. $$ Since $\binom{m}{k}/(m+1-k) = \binom{m+1}{k}/(m+1)$, the difference is $$ \frac{1}{(m+1)!}\sum_{k=0}^m \binom{m+1}{k}\frac{(-1)^k}{x+k} - \frac{1}{(m+1)!}\left(\sum_{k=0}^m\binom{m+1}{k}(-1)^k\right)\frac{1}{x+m+1}. $$ The sum inside parentheses is $(1-1)^{m+1} - (-1)^{m+1} = (-1)^m$, so finally we have $$ \frac{1}{x(x+1)\cdots(x+m)(x+m+1)} = \frac{1}{(m+1)!}\sum_{k=0}^m \binom{m+1}{k}\frac{(-1)^k}{x+k} - \frac{1}{(m+1)!}\frac{(-1)^m}{x+m+1}. $$ Since $-(-1)^m = (-1)^{m+1}$, we're done.

Negative coefficient in an almost cyclotomic polynomial Let $a,b,c,d$ be four prime numbers. We set the polynomial : $$P(X)=\frac{(1-X^{abc})(1-X^{abd})(1-X^{acd})(1-X^{bcd})(1-X^a)(1-X^b)(1-X^c)(1-X^d)}{(1-X)^2(1-X^{ab})(1-X^{ac})(1-X^{ad})(1-X^{bc})(1-X^{bd})(1-X^{cd})}$$ By numerical tests, i see that $P(X)$ always has at least one negative coefficient, how can i prove it?

Suppose that $a<b<c<d$. We show that the coefficient of $X^c$ or of $X^{b+c-1}$ of $P(X)$ is negative. In order to do so, it suffices to work in the power series ring $\mathbb Q[[X]]$ modulo $X^{b+c}$. Note that $ac>b+c$ and so on, hence \begin{equation} P(X)\equiv\frac{(1-X^a)(1-X^b)(1-X^c)(1-X^d)}{(1-X)^2(1-X^{ab})}\pmod{X^{b+c}}. \end{equation} Set \begin{equation} F(X)=\frac{1-X^a}{1-X}\cdot\frac{1-X^b}{1-X}\cdot\frac{1}{1-X^{ab}}=(1+\dots+X^{a-1})(1+\dots+X^{b-1})(1+X^{ab}+\dots), \end{equation} so $P(X)\equiv F(X)(1-X^c-X^d)$ (recall that $c+d>b+c$). For a power series $G$ let $G[k]$ be the coefficient of $X^k$. So $P[k]=F[k]-F[k-c]-F[k-d]$. The coefficients of $F$ lie between $0$ and $a$, and $F[k]=a$ if an only if the remainder of $k\pmod{ab}$ lies between $a-1$ and $b-1$. Furthermore, $F[k]=0$ if the remainder of $k\pmod{ab}$ lies between $a+b-1$ and $ab-1$. Now assume that $P[b+c-1]\ge0$. From $P[b+c-1]=F[b+c-1]-F[b-1]-F[b+c-1-d]$ and $F[b-1]=a$ we infer that $F[b+c-1]=a$. Thus the remainder of $(b+c-1)\pmod{ab}$ lies between $a-1$ and $b-1$. This implies that the remainder of $c\pmod{ab}$, call it $r$, fulfills $ab+a-b\le r\le ab-1$ or $r=0$. The latter cannot happen, because then $a$ would divide the prime $c$. Thus the former holds. But $a+b-1\le ab+a-b$, hence $F[c]=0$. But then $P[c]=F[c]-F[0]=0-1=-1$, and we are done.

Integer solution For every prime $p$, does there exists integers $x_1$, $x_2$ and $x_3$ ($0\leq x_1, x_2, x_3 \leq \lfloor cp^{1/3}\rfloor$ and $c$ is some large constant) such that $\frac{p-1}{2}-\lfloor 2cp^{1/3} \rfloor \leq f(x_1,x_2,x_3) \leq \frac{p-1}{2}$, where, $f(x_1,x_2,x_3)=x_1+x_2+x_3+2(x_1x_2+x_2x_3+x_3x_1)+4x_1x_2x_3$.

I doubt that the lower bound $\frac{p-1}{2} - 2cp^{1/3}$ holds for all $p$. Here is a proof for the weaker bound $\frac{p-1}{2} - cp^{1/2}$. First of all, the inequality $\frac{p-1}{2}-cp^{1/2} \leq f(x_1,x_2,x_3) \leq \frac{p-1}{2}$ is essentially equivalent to $$p- O(p^{1/2}) \leq (2x_1+1)(2x_2+1)(2x_3+1) \leq p.$$ Let us take $y=\left\lceil\frac{p^{1/3}-1}{2}\right\rceil$. Then $$p \leq (2y+1)^3 < (p^{1/3}+2)^3 < p + O(p^{2/3}).$$ Define $$z = \left( \frac{(2y+1)^3 - p}{4(2y+1)} \right)^{1/2} .$$ Clearly, we have $z=O(p^{1/6})$. Now, let us take $x_1 = y$, $x_2 = y - \lceil z\rceil$, $x_3 = y + \lceil z\rceil$ so that $$(2x_1+1)(2x_2+1)(2x_3+1) = (2y+1)^3 - 4\lceil z\rceil^2(2y+1)$$ $$ < (2y+1)^3 - 4z^2(2y+1) = p.$$ On the other hand, we have $$(2x_1+1)(2x_2+1)(2x_3+1) = (2y+1)^3 - 4\lceil z\rceil^2(2y+1) $$ $$ > (2y+1)^3 - 4(z+1)^2(2y+1) = p - O(p^{1/2}).$$

Can a block matrix with at least 3 zero blocks of different size on the diagonal and 1's everywhere else have only integer eigenvalues? Let $M=\begin{pmatrix} \begin{array}{cccccccc} 0 & 0 & 1 & 1 & 1 & 1 & 1 &1\\ 0 & 0 & 1 & 1 & 1 & 1 & 1 &1\\ 1 & 1 & 0 & 0 & 0 & 1 & 1 &1\\ 1 & 1 & 0 & 0 & 0 & 1 & 1 &1\\ 1 & 1 & 0 & 0 & 0 & 1 & 1 &1\\ 1 & 1 & 1 & 1 & 1 & 0 & 0 &1\\ 1 & 1 & 1 & 1 & 1 & 0 & 0 &1\\ 1 & 1 & 1 & 1 & 1 & 1 & 1 &0\\ \end{array} \end{pmatrix}\implies\begin{pmatrix} \begin{array}{ccc} 0_{2\times2} & 1_{2\times3} & 1_{2\times2}&1_{2\times1}\\ 1_{3\times2} & 0_{3\times3} & 1_{3\times2}&1_{3\times1} \\ 1_{2\times2} & 1_{2\times3} & 0_{2\times2}&1_{2\times1} \\ 1_{1\times2} & 1_{1\times3} & 1_{1\times2}&0_{1\times1} \\ \end{array} \end{pmatrix}$ be a symmetric matrix whose entries are blocks(diagonal blocks entry must be zero and non-diagonal blocks entry must be one). How to prove/disprove that all the eigenvalues of $M$ are not integer if and only if the diagonal blocks are not of the same order.Note that the number of diagonal blocks $\geq 3$ and order of diagonal blocks$\geq 1$ .

It is not a complete solution, but can be such of them by some calculations. You can assign a graph to the matrice $M$ in each case and analyze them. If all-zero blocks have equal size, their size $t$ must be the divisors of $8$. So, all possible cases are $1$, $2$, $4$ and $8$. If $t=1$ then you have the complete graph $K_8$ and all its eigenvalues are integral. If $t=2$ then you have $4$ zero eigenvalues and the rest are twice of the eigenvalues of $K_4$ (By the rank of resulted matrix and constructing quotient matrix). If $t=4$, you have the graph $K_{4,4}$ which has integral eigenvalues. If $t=8$, the case is obvious. So, when the zero blocks have equal sizes, the eigenvalues are integral. For the cases when the zero blocks have not equal sizes, you obtain at least $4$ zero eigenvalues and the remaining eigenvalues can be obtain by quotient matrix of original matrix. Also, in each of those cases, you have at least some special induced subgraphs (such as $P_3$) which by interlacing theorem, you can analyze the behavior of eigenvalues of the original matrix.

An irresistible inequality The following occurred while working on some research project. Since the methods of proof I used were lengthy, I wish to see a skillful or insightful approach (perhaps even conceptual). Anyhow, here it is. Let $$f(x)=\left(\frac{x}{e^x-1}\right)^2 + \left(\frac{x+1}{e^{x+1}+1}\right)^2.$$ Can one give a short and elegant proof of these statements? (1) $f(x)$ is a strictly decreasing function of $x$ over $\mathbb{R}$. (2) In fact, the statement holds true if $e$ is replaced by any real number $t>1$.

I do not know wether this helps you or not, but you may do as follows. Denote $f(x)=(e^x-1)/x$. Note that $f'(x)=\frac{fe^x(x-1+e^{-x})}{x(e^x-1)}>0$, so $f$ is a positive increasing function. Lemma. The function $1/f=x/(e^x-1)$ is a (positive decreasing) convex function. Proof. $$ (1/f)''=\frac{e^x(2+x-(2-x)e^x)}{(e^x-1)^3}, $$ we have to check that it is non-negative. If $x\geqslant 2$, this is clear. If $0<x<2$, this reduces to $$ e^x\leqslant \frac{2+x}{2-x}=1+x+x^2/2+x^3/2^2+x^4/2^3+\dots,\,\,(1) $$ that holds coefficient-wise: $n!\geqslant 2^{n-1}$ for $n\geqslant 1$. Finally, the case $x<0$ reduces to $x>0$, since $2+x-(2-x)e^x$ and $2-x-(2+x)e^{-x}$ always have opposite signs. Corollary. $g:=1/f^2=x^2/(e^x-1)^2$ is convex. Proof. $-(1/f^2)'=2(-1/f)'(1/f)$, both multiples are positive decreasing functions, thus their product also decreases. Now we claim that $g(x-a)+x^2/(e^x+1)^2$ decreases for each $a\geqslant 0$, for $a=1$ we get your statement (and for other $a$ something equivalent to your remark). Since $g'$ increases, we have $g'(x-a)\leqslant g'(x)$, so it suffices to check this for $a=0$. We have $g(x)+x^2/(e^x+1)^2=2x^2(e^{2x}+1)/(e^{2x}-1)^2$. Denote $2x=y$, we need to check that $y^2(e^y+1)/(e^y-1)^2$ decreases. Taking logarithmic derivative, this is equivalent to $$\frac{2}x+\frac{e^x}{e^x+1}-\frac{2e^x}{e^x-1}\leqslant 0.$$ For $x=-y>0$ we have $$\frac2{y}-\frac2{e^y-1}\geqslant \frac2y-\frac2{y+y^2/2}=\frac1{1+y/2}>\frac1{1+e^y}$$ as desired. For $x>0$ we have $$ \frac{2}x+\frac{e^x}{e^x+1}-\frac{2e^x}{e^x-1}\leqslant \frac{2+x}x-\frac{2e^x}{e^x-1}= \frac{(2-x)e^x-(2+x)}{x(e^x-1)}\leqslant 0 $$ by (1).

Is there always one integer between these two rational numbers? It appears that for each integer $k\geq2$, there is always one integer $c$ that satisfies the inequalities below. Can this be proved? $$\frac{3^k-2^k}{2^k-1}<c\leq \frac{3^k-1}{2^k}.$$ Note that for $k\geq2$ the lower bound is always a proper fraction and will never match an integer. Edit 28/1/2018 I have a short proof here on Overleaf. <--- It's done! Edit #28 is perfect! edit It looks like Waring's problem already has a solution. The sequence: https://oeis.org/A060692 "Corresponds to the only solution of the Diophantine equation 3^n = x*2^n + y*1^n with constraint 0 <= y < 2^n." This is equivalent to our $a+c$ for the Waring if statement. Oops, there is no proof that the diophantine equation is always $\leq 2^n.$ Here is a proof sketch of Waring's Here is a brute force function: aplusc[k_] := Module[{c}, c = 1; While[0 < 3^k - 2^k (++c)]; 3^k - 2^k (--c) + c] where we increment $c$ until the calculation becomes negative, then we decrement by one to get $c$. We recalculate using that $c$ and add them together. A060692(k) equals this value. Closed form: here. The brute force function illustrates the sawtooth pattern. We can also create a(k), b(k), and c(k) using the same module. And we can use b(k) for the proof. Trivially, we can show b$(k) + 1 \leq 2^k.$ edit. Some more information and references: https://oeis.org/A002379 PM

We show that $\left\lfloor\frac{3^n-1}{2^n-1}\right\rfloor$ and $\left\lfloor \left(\frac{3}{2} \right)^n\right\rfloor$ have a common floor when $\frac{3^n-1}{2^n-1}-\left(\frac{3}{2} \right)^n =$ $ \left\{\frac{3^n-1}{2^n-1}\right\}-\left\{\left(\frac{3}{2} \right)^n\right\} = \delta(n),$ where $\left\{\cdot\right\}$ are the fractional parts and $\delta(n)$ is the difference function. We verify that the order of the fractions, $\frac{3^n-1}{2^n-1}>\left(\frac{3}{2} \right)^n,$ is true for $n\not=0$ and set $n\geqslant1$ to constrain to the positive numbers. Notice that the denominators are consecutive integers and therefore relatively prime to each other. The lower denominator, $2^n-1,$ must be greater than one, so we set $n\geqslant2$ as our final constraint. By the ratio test, it is easy to show $\frac{3^n-1}{2^n-1}$ and $\left(\frac{3}{2} \right)^n$ are always proper fractions for $n\geqslant2$ and therefore they always have fractional parts. We define $\delta(n):=\frac{3^n-2^n}{4^n-2^n}$ and its one's complement, $1-\delta(n),$ as $\frac{4^n-3^n}{4^n-2^n}.$ We define the interval which contains the common floor, $x,$ as lower bound, $\frac{3^n-1}{2^n-1}-1$ and upper bound, $\left(\frac{3}{2} \right)^n.$ From this we can show that the upper bound minus the lower bound is equal to $\frac{4^n-3^n}{4^n-2^n}.$ The RHS below is $\left\{\left(\frac{3}{2} \right)^n\right\}+\left(1- \left\{\frac{3^n-1}{2^n-1}\right\}\right).$ We assume a common floor: Let $(n|x)\in \mathbb{Z},\ x\geqslant 1,\text{ and } n\geqslant 2,$ $\frac{4^n-3^n}{4^n-2^n}=\frac{3^n-2^n x}{2^n}+ \frac{\left(2^n-1\right) x+2^n-3^n}{2^n-1},\ \ \text{expand; subtract LHS from both sides}\Rightarrow$ $\left(\frac{3}{2}\right)^n-\frac{3^n}{2^n-1}+\frac{1}{2^n-1}+1=\frac{2^n x}{2^n-1}-\frac{x}{2^n-1}+\left(\frac{3}{2}\right)^n-\frac{3^n}{2^n-1}+\frac{1}{2^n-1}-x+1;\ \ \Rightarrow$ $0=\frac{2^n x}{2^n-1}-\frac{x}{2^n-1}-x,\ \ \text{simplify}\Rightarrow\text{True.}$ We assume different floors: Let $(n|w|x)\in \mathbb{Z},\ w\geqslant 1,\ x\geqslant 1,\text{ and } n\geqslant 2,$ $\frac{4^n-3^n}{4^n-2^n}=\frac{3^n-2^n x}{2^n}+ \frac{\left(2^n-1\right) w+2^n-3^n}{2^n-1},\ \ \text{expand; subtract LHS from both sides}\Rightarrow$ $\left(\frac{3}{2}\right)^n-\frac{3^n}{2^n-1}+\frac{1}{2^n-1}+1=\frac{2^n w}{2^n-1}-\frac{w}{2^n-1}+\left(\frac{3}{2}\right)^n-\frac{3^n}{2^n-1}+\frac{1}{2^n-1}-x+1;\ \ \Rightarrow$ $0=\frac{2^n w}{2^n-1}-\frac{w}{2^n-1}-x,\ \ \text{simplify}\Rightarrow w=x.$ Next, we craft two unit-intervals within which the bounds reside: \begin{align} &\left(\frac{3^n-1}{2^n-1}-1-\left\{\frac{3^n-1}{2^n-1}\right\},\ \frac{3^n-1}{2^n-1}-\left\{\frac{3^n-1}{2^n-1}\right\}\right),\ \ (1)\\ &\left(\left(\frac{3}{2} \right)^n-\left\{\left(\frac{3}{2} \right)^n\right\},\ \left(\frac{3}{2} \right)^n+1-\left\{\left(\frac{3}{2} \right)^n\right\}\right).\ \ (2) \end{align} Since lower bound of (1) $<\left(\frac{3}{2} \right)^n-1$, upper bound of (2) $>\frac{3^n-1}{2^n-1},$ and the remaining bounds equal $x,\text{ we have a common floor because }x$ is the greatest integer less than the fraction, which is the floor definition.$\ \ \ \square$ Edit 6/1 I forgot to show that $x$ was the floor.

Primality test for specific class of Proth numbers Can you provide a proof or a counterexample for the following claim : Let $P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)$ Let $N=k\cdot 2^n+1$ such that $n>2$ , $0< k <2^n$ and $\begin{cases} k \equiv 1,7 \pmod{30} \text{ with } n \equiv 0 \pmod{4} \\ k \equiv 11,23 \pmod{30} \text{ with } n \equiv 1 \pmod{4} \\ k \equiv 13,19 \pmod{30} \text{ with } n \equiv 2 \pmod{4} \\ k \equiv 17,29 \pmod{30} \text{ with } n \equiv 3 \pmod{4} \end{cases}$ Let $S_i=S_{i-1}^2-2$ with $S_0=P_k(8)$ , then $N$ is prime iff $S_{n-2} \equiv 0 \pmod N$ . You can run this test here . A list of Proth primes sorted by coefficient $k$ can be found here . I have tested this claim for many random values of $k$ and $n$ and there were no counterexamples . Note that for $k=1$ we have Inkeri's primality test for Fermat numbers . Reference : Tests for primality, Ann. Acad. Sci. Fenn. Ser. A I 279 (1960), 1-19.

Your criterion is equivalent to: $$(4+\sqrt{15})^{k2^{n-1}}\equiv (4+\sqrt{15})^{\frac{N-1}{2}}\equiv -1 (\bmod N \mathbb{Z}[\sqrt{15}]).$$ The point is that $P_m(8)=(4+\sqrt{15})^m+(4-\sqrt{15})^m$. Moreover, $S_i=(4+\sqrt{15})^{k2^i}+(4-\sqrt{15})^{k2^i}$, which one may prove by induction, using the fact that $x\mapsto x^2-2$ is semi-conjugate to $z\mapsto z^2$ under the relation $x=z+1/z$. Then the condition $S_{n-2} = (4+\sqrt{15})^{k2^{n-2}}+(4-\sqrt{15})^{k2^{n-2}} \equiv 0 (\bmod N)$ is equivalent to $(4+\sqrt{15})^{k2^{n-1}} \equiv -1 (\bmod N)$. By Fermat's little theorem (see Theorem 1 of this paper) in the quadratic number field $\mathbb{Q}[\sqrt{15}]$, for any odd prime $p$, one has $(4+\sqrt{15})^{p-\left(\frac{15}{p}\right)}\equiv 1 (\bmod p)$, where $\left(\frac{15}{p}\right)$ is the Legendre symbol (note that $(4+\sqrt{15})(4-\sqrt{15})=1$, so $4+\sqrt{15}$ is a unit in the ring of integers $\mathbb{Z}[\sqrt{15}]$). One also has by Theorem 2 of this paper that if $l$ is the smallest integer with $(4+\sqrt{15})^l \equiv -1 (\bmod p)$, and $(4+\sqrt{15})^K\equiv -1(\bmod p)$, then $K=lu$, where $u$ is odd. Also, the smallest integer with $(4+\sqrt{15})^e \equiv 1 (\bmod p)$ is $e=2l$. Now, let's show that your criterion implies that $N$ is prime. For contradiciton, let $p < N$ be a prime factor of $N$, then by your criterion, $(4+\sqrt{15})^{k2^{n-1}}\equiv -1 (\bmod p)$. Then $k2^{n-1} = l u$, where $l$ is the smallest exponent $>0$ with $(4+\sqrt{15})^k\equiv -1(\bmod p)$, and $u$ is an odd integer by Theorem 2. So $l=2^{n-1}\delta$, with $\delta | k$, and hence $e \geq 2^n$. Theorem 1 implies that $(4+\sqrt{15})^{p\pm 1}\equiv 1(\bmod p)$. Then $p\pm 1\geq 2^n$, for every prime $p | N$. $N$ is not a square from your congruence conditions on $k$ and $n$ (check $(\bmod 15)$), so it has a factorization $pq$ with $p\geq 2^n-1$, $q\geq p+2$. But then $N = p\cdot q \geq p(p+2)\geq (2^n-1)(2^n+1)=2^n\cdot 2^n-1 > k 2^n +1 =N$ since $k < 2^n$, a contradiction. Going in the reverse direction (following the proof of Theorem 1 of this paper), your congruence conditions imply that $N\equiv 2,8 (\bmod 15)$ and $N\equiv 1(\bmod 8)$, hence $\left(\frac{15}{N}\right)=1, \left(\frac{2}{N}\right)=1,\left(\frac{5}{N}\right)=-1$. We have $(4+\sqrt{15})=\frac{5+\sqrt{15}}{5-\sqrt{15}}$. Hence $$(4+\sqrt{15})^{\frac{N-1}{2}}\equiv -1 (\bmod N)$$ if and only if $$(5+\sqrt{15})^{\frac{N-1}{2}}\equiv - (5-\sqrt{15})^{\frac{N-1}{2}} (\bmod N)$$ if and only if $$(5+\sqrt{15})^{N-1} \equiv - 10^{\frac{N-1}{2}} \equiv - 2^{\frac{N-1}{2}} 5^{\frac{N-1}{2}} \equiv - \left(\frac{2}{N}\right) \left(\frac{5}{N}\right ) \equiv 1 (\bmod N),$$ which holds by Theorem 1 and the properties of Legendre symbols.

Oddity of generalized Catalan numbers: Part I The famous (classical) Catalan numbers $C_{1,n}=\frac1{n+1}\binom{2n}n$ satisfy the following well-known arithmetic property: $$\text{$C_{1,n}$ is odd iff $n=2^j-1$ for some $j$}.\tag1$$ Consider the "second generation" of Catalan numbers $C_{2,n}$ which can be found on OEIS with A236339. One possible expression is given in the manner $$C_{2,n}=\frac1{n+1}\sum_{k=0}^{\lfloor n/3\rfloor}(-1)^k\binom{2n-2k}{n-2k}\binom{n-2k}k2^{n-3k}.$$ Working in the spirit of (1), I was curious to find any perpetual behavior. Here is my observation: QUESTION 1. Is this true? If so, how does the proof go? $$\text{$C_{2,n}$ is odd iff $n=2^{2j}-1$ for some $j$}.$$ The $2$-adic valuation of $x\in\mathbb{N}$ is the highest power $2$ dividing $x$, denoted by $\nu(x)$. Let $s(x)$ stand for the sum of the binary digits of $x$. We may now state a stronger claim: QUESTION 2. Is this true? $$\nu(C_{2,n})= \begin{cases} s(3m+1)-1 \qquad \, \text{if $n=3m$}, \\ s(3m-1)+2 \qquad \,\text{if $n=3m-1$}, \\ s(3m-1) \qquad \qquad \text{if $n=3m-2$}. \end{cases}$$ NOTE. Evidently, Question 2 implies Question 1. A related fact: $\nu(C_{1,n})=s(n+1)-1$.

The answer to QUESTION 1 is Yes. The following proof simplifies my suggestion in the comments. According to the OEIS entry, the shifted generating function $$ g(x) = \sum_{n=0}^\infty C_{2,n} x^{n+1} = x + 2x^2 + 8x^3 + 39x^4 + \cdots $$ satisfies $g^4 - 2g^2 + g = x$. Reducing $\bmod 2$ gives $g^4 + g = x.$ Therefore, modulo 2 we have $$\begin{eqnarray} g &=& x + g^4 = x + (x+g^4)^4 \cr &=& x + x^4 + g^{16} = x + x^4 + (x+g^4)^{16} \cr &=& x + x^4 + x^{16} + (x+g^4)^{64} = \cdots, \end{eqnarray}$$ whence $$ g = x + x^4 + x^{16} + x^{64} + \cdots = \sum_{j=0}^\infty x^{2^{2j}}. $$ Therefore $C_{2,n}$ is odd if and only if $n+1 = 2^{2j}$ for some $j=0,1,2,3,\ldots$, QED Remark. A similar proof can be given for the characterization of odd Catalan numbers, starting from the equation $c-c^2=x$ satisfied by the shifted generating function $c = \sum_{n=0}^\infty C_{1,n} x^{n+1}$. The coefficients of the solution $$x - x^4 + 4 x^7 - 22 x^{10} + 140 x^{13} - 969 t^{16} + - \cdots$$ of $g^4+g = x$ can also be given in closed form $-$ the $x^{3m+1}$ coefficient is $$\frac{(-1)^m}{3m+1} {4m\choose m}$$ found at OEIS A002293; see also the link here $-$ but getting the parity from that formula is not as easy as using $g^4+g=x$ directly.

Bounds on the mills ratio How do I show the following bounds on the mills ratio : $\frac{1}{x}- \frac{1}{x^3} < \frac{1-\Phi(x)}{\phi(x)} < \frac{1}{x}- \frac{1}{x^3} +\frac{3}{x^5} \ \ \ \ \ \ \ $ for $ \ \ \ x>0$ where $\Phi()$ is the CDF of the Normal distribution , and $\phi()$ is the density function of the Normal distribution ? Also , is there a similar bound when $x < 0$ ? I am aware of the proof of the fact that the mills ratio is bounded below by $\frac{x}{1+x^2}$ and above by $\frac{1}{x}$ , but I am unable to prove this inequality .

Here's a sketch and a link for how I prove it. Let $$ f(x) = - \left( \frac{1}{x} - \frac{1}{x^3} + \frac{3}{x^5}\right) \phi(x) .$$ Now show (lemma): $\frac{df}{dx} = \left(1 + \frac{15}{x^6}\right)\phi(x)$. (To prove this, use that $\frac{d\phi}{dx} = - x \phi(x)$, the product rule, and some cancellation.) Now suppose $x > 0$: \begin{align*} 1 - \Phi(x) &= \int_{t=x}^{\infty} \phi(x) dx \\ &\leq \int_{t=x}^{\infty} \left(1 + \frac{15}{x^6}\right) \phi(x) dx \\ &= \lim_{t\to\infty} f(t) - f(x) \\ &= - f(x) \\ &= \left(\frac{1}{x} - \frac{1}{x^3} + \frac{3}{x^5} \right) \phi(x) . \end{align*} Using the next term in the series gives $f(x) = -\left(\frac{1}{x} - \frac{1}{x^3} + \frac{3}{x^5} - \frac{15}{x^7}\right)\phi(x)$ and $\frac{df}{dx} = \left(1 - \frac{105}{x^8}\right)\phi(x)$. Notice because of the alternating positive/negative terms, $\frac{df}{dx}$ is now $\phi$ times something less than one, so plugging it into the same proof gives a lower bound on $1 - \Phi(x)$. I have a blog post on the general form of this (sorry to not point you to a more formal reference).

Two versions of Sylvester identity MathWorld presents the following two versions of Sylvester's determinant identity, relating to an $n\times n$ matrix $\mathbb{A}$: First: $$ |\mathbb{A}||A_{r\,s,p\,q}| = |A_{r,p}||A_{s,q}| - |A_{r,q}| |A_{s,p}| $$ where $r$ and $s$ ($p$ and $q$) are sets that indicate which rows (columns) of $\mathbb{A}$ are to be deleted (correcting MathWorld's typo). Second: $$ |\mathbb{A}|\left[ a_{k\,k}^{(k-1)}\right]^{n-k-1} = \left| \begin{matrix} a_{k+1\, k+1}^{(k)} & \cdots & a_{k+1\, n}^{(k)} \\ \vdots & \ddots & \vdots \\ a_{n\, k+1}^{(k)} & \cdots & a_{n\, n}^{(k)} \\ \end{matrix}\right| $$ where $$ a_{i\, j}^{(k)} = \left| \begin{matrix} a_{11} & \cdots & a_{1\,k} & a_{1 \, j} \\ \vdots & \ddots & \vdots& \vdots \\ a_{k\ 1} & \cdots & a_{k\,k}& a_{k\, j} \\ a_{i\ 1} & \cdots & a_{i\, k} & a_{i\, j} \\ \end{matrix} \right| $$ for $k<i$, $j \leq n$. Would anyone help me prove that these two versions are indeed equivalent. Note: As has been pointed our below, MathWorld's claim is obviously (using a counterexample) incorrect; the second version implies only a special case (when r,s,p,q are single numbers) of the first 'version'.

For the special case that $r,s,p,q$ are single elements, it is shown in these notes (page 7) how the first identity (known as the Desnanot-Jacobi identity) follows from the second identity. Apply the second identity to the matrix we thus arrive at the first identity, illustrated graphically as source Update: Since I could not find the first identity in the literature, for the more general case when $r,s,p,q$ each contain more than a single element, I tried to check it for an example. I took $n=6$, $r=1,2$, $s=5,6$, $p=1,2$, $q=5,6$. For the $6\times 6$ matrix $A$ I took $$A=\left( \begin{array}{cccccc} 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 2 & 0 & 2 & 2 & 0 & 0 \\ 2 & 1 & 0 & 0 & 2 & 1 \\ 2 & 0 & 2 & 1 & 0 & 2 \\ 1 & 0 & 0 & 0 & 2 & 2 \\ \end{array} \right)$$ The left-hand-side of the first identity is 0, $\det A \det \left( \begin{array}{cc} 2 & 2 \\ 0 & 0 \\ \end{array} \right) = 24\cdot 0 = 0$, but for the right-hand-side I find a nonzero answer: $$ \det \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 2 & 0 & 2 & 2 \\ 2 & 1 & 0 & 0 \\ \end{array} \right) \det \left( \begin{array}{cccc} 2 & 2 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 2 & 1 & 0 & 2 \\ 0 & 0 & 2 & 2 \\ \end{array} \right) - \det \left( \begin{array}{cccc} 2 & 0 & 2 & 2 \\ 2 & 1 & 0 & 0 \\ 2 & 0 & 2 & 1 \\ 1 & 0 & 0 & 0 \\ \end{array} \right) \det \left( \begin{array}{cccc} 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 \\ 2 & 2 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ \end{array} \right)=$$ $$\qquad\qquad= (-4) \cdot 4 - (-2)\cdot (-2)=-20\neq 0$$ Incidentally, I did find a determinantal identity of a somewhat similar form in Tao's blog (last equation, Karlin's identity). But it is not quite of the form of the first identity in the OP. So unless I have made a mistake, my conclusion is that the first identity in the OP only holds when $r,s,p,q$ are single elements, but not more generally.

Minimal polynomial in $\mathbb Z[x]$ of seventh degree with given roots I am looking for a seventh degree polynomial with integer coefficients, which has the following roots. $$x_1=2\left(\cos\frac{2\pi}{43}+\cos\frac{12\pi}{43}+\cos\frac{14\pi}{43}\right),$$ $$x_2=2\left(\cos\frac{6\pi}{43}+\cos\frac{36\pi}{43}+\cos\frac{42\pi}{43}\right),$$ $$x_3=2\left(\cos\frac{18\pi}{43}+\cos\frac{22\pi}{43}+\cos\frac{40\pi}{43}\right)$$ $$x_4=2\left(\cos\frac{20\pi}{43}+\cos\frac{32\pi}{43}+\cos\frac{34\pi}{43}\right),$$ $$x_5=2\left(\cos\frac{10\pi}{43}+\cos\frac{16\pi}{43}+\cos\frac{26\pi}{43}\right),$$ $$x_6=2\left(\cos\frac{8\pi}{43}+\cos\frac{30\pi}{43}+\cos\frac{38\pi}{43}\right)$$ and $$x_7=2\left(\cos\frac{4\pi}{43}+\cos\frac{24\pi}{43}+\cos\frac{28\pi}{43}\right).$$ I see only that $\sum\limits_{k=1}^7x_k=-1$, but the computations for $\sum\limits_{1\leq i<j\leq7}x_ix_j$ and the similar are very complicated by hand and I have no any software besides WA, which does not help. Thank you for your help! Update. I got $$\sum\limits_{1\leq i<j\leq7}x_ix_j=-18.$$

I also used PARI/GP with the following program: z1 = Mod(z, (z^43-1)/(z-1)); e(n) = lift(Mod(3,43)^n); c(n) = z1^n + z1^-n; r(n) = c(1*n) + c(6*n) + c(7*n); p = prod(n=1,7, x - r(e(n))); lift(p) with the resulting output z^7+z^6-18*z^5-35*z^4+38*z^3+104*z^2+7*z-49 A simpler program with complex numbers is z1=exp(2*Pi*I/43); z2=z1^6; z3=z1^7; bestappr(prod(n=1,7, m=lift(Mod(3,43)^n);\ x - 2*real(z1^m + z2^m + z3^m)), 10^9)

Evaluations of three series involving binomial coefficients Question. How to prove the following three identities? \begin{align}\sum_{k=1}^\infty\frac1{k(-2)^k\binom{2k}k}\left(\frac1{k+1}+\ldots+\frac1{2k}\right)=\frac{\log^22}3-\frac{\pi^2}{36},\tag{1} \end{align} \begin{align}\sum_{k=1}^\infty\frac1{k2^k\binom{3k}k}\left(\frac1{k+1}+\ldots+\frac1{2k}\right) =\frac{3}{10}\log^22+\frac{\pi}{20}\log2-\frac{\pi^2}{60},\tag{2} \end{align}\begin{align}\sum_{k=1}^\infty\frac1{k^22^k\binom{3k}k}\left(\frac1{k+1}+\ldots+\frac1{2k}\right) =-\frac{\pi G}2+\frac{33}{32}\zeta(3)+\frac{\pi^2}{24}\log2,\tag{3} \end{align} where $G$ denotes the Catalan constant $\sum_{k=0}^\infty(-1)^k/{(2k+1)^2}$. Remark. Motivated by my study of congruences, in 2014 I tried to evaluate the three series in $(1)$-$(3)$, and this led me to discover $(1)$-$(3)$ which can be easily checked numerically via Mathematica. But I'm unable to prove the above three identities. Also, Mathematica could not evaluate the three series. For more backgrounds of this topic, you may visit http://maths.nju.edu.cn/~zwsun/165s.pdf. Your comments are welcome!

Here is a proof of the first identity. The others can probably be done similarly. We have $$\frac1{k\binom{2k}{k}}=\frac12 B(k,k)=\frac12 \int_0^1 t^{k-1}(1-t)^{k-1}\,dt,$$ $$\frac{1}{k+1}+\cdots+\frac{1}{2k} = \int_0^1 \frac{1-x^{2k}}{1+x}\, dx$$ and $$\sum_{k=1}^\infty \frac{1}{(-2)^k}t^{k-1}(1-t)^{k-1}(1-x^{2k}) = -\frac{1}{2+t(1-t)}+\frac{x^2}{2+t(1-t)x^2}.$$ Combining all these together, we get $$\sum_{k=1}^\infty\frac1{k(-2)^k\binom{2k}k}\left(\frac1{k+1}+\ldots+\frac1{2k}\right) = I_1 + I_2,$$ where $$I_1 := -\int_0^1\frac{dx}{2(1+x)}\int_0^1 dt\,\frac{1}{2+t(1-t)} = -\frac13 \log(2)^2,$$ $$I_2 := \int_0^1 dx \frac{x^2}{2(1+x)} \int_0^1 dt\, \frac{1}{2+t(1-t)x^2} = \frac23\log(2)^2 - \frac{\pi^2}{36}.$$ So, $$I_1 + I_2 = \frac13\log(2)^2 - \frac{\pi^2}{36}.$$

Finding all proper divisors of $a_3z^3 +a_2z^2 +a_1z+1$ of the form $xz+1$ Let $n=a_3z^3+a_2z^2+a_1z+1$ where $a_1<z, \ a_2<z, \ 1 \le a_3<z, z>1$ are non negative integers. To obtain proper divisors of $n$ of the form $xz+1$, one may perform trial divisions $xz+1 \ | \ n$, for all $xz+1 \le \sqrt n$. Trial division however is inefficient as $z$ becomes large. The method below is much more efficient. Since $xz+1 \ | \ n$ , we may write $(xz+1)(yz+1)=n$. Assume $y \le x$. We consider two cases; **Case 1: ** $1 \le x <z$ and $1 \le y<z$ $xyz+x+y=a_3z^2+a_2z+a_1$. Since every positive integer has a unique base $z$ representation we have; \begin{equation} x+y=C\cdot z+a_1 , C=0 \ \text{or} \ 1 \ \end{equation} \begin{equation} xy+C = a_3z+a_2 \end{equation} Solving the two equations for the cases $C=0$ and $C=1$, $x$ and $y$ can be determined. **Case 2: ** $z\le x<z^2$ and $1 \le y<z$ Let $x=x_1z+x_0$ and $y = y_0$, $1\le x_1<z, \ 0 \le x_0<z$, $ 1 \le y_0 <z$. So \begin{equation} x_1y_0z^2+(x_0y_0+x_1)z+x_0+y_0=a_3z^2+a_2z+a_1 \end{equation} Comparing coefficients of powers of $z$; \begin{equation} x_0+y_0=C_1 \cdot z+a_1, \ C_1 = 0 \ \text{or} \ 1 \end{equation} \begin{equation} x_0y_0+x_1+C_1=C_2 \cdot z+a_2 \end{equation} \begin{equation} x_1y_0+C_2=a_3 \end{equation} From the last equation, either $y_0 \le \sqrt a_3$ or $x_1 \le \sqrt a_3$. So we do trial divisions $yz+1 \ | \ n$ for all $y = y_0 \le \sqrt a_3$. Also for each value of $x_1 \le \sqrt a_3 $, we solve the three equations simultaneously to find $x_0, y_0$ and $C_2$ when $C_1=0$ and $C_1 = 1$. Cases 1 and 2 exhaust all the possible cases. In this example, we had $n < z^4$. How can this method be modified to efficiently find all proper divisors of the form $xz+1$ of an arbitrary positive integer $n$, $n \equiv \ 1 ($ mod $ z) $ or at least when $n<z^5 $?

Such an extension is highly unlikely to exist. Already in the simple case of $z=2$, it's equivalent to just factoring a given odd integer $n$, which is a famous hard problem.

Divisibility of (finite) power sum of integers Consider the power sum $$S_a(b)=1^{2b}+2^{2b}+\cdots+(3a-2)^{2b}.$$ Let $\nu_3(x)$ denote the $3$-adic valuation of $x$. QUESTION 1. (milder) Is this true? $$\nu_3\left(\frac{S_a(b)}{S_a(1)}\right)=0.$$ QUESTION 2. Is this true? $\nu_3(S_a(b))=\nu_3(2a-1)$.

Notice that $$2S_a(b) \equiv 1^{2b} + 2^{2b} + \dots + (6a-4)^{2b} \pmod{6a-3}.$$ From Faulhaber's formula, we have $$1^{2b} + 2^{2b} + \dots + (6a-4)^{2b} \equiv B_{2b} (6a-3) \pmod{6a-3}.$$ It follows that $$S_a(b) \equiv \frac32B_{2b} (2a-1)\pmod{3(2a-1)},$$ where $\nu_3(\frac32 B_{2b})=0$ by von Staudt–Clausen theorem. Hence, $\nu_3(S_a(b))=\nu_3(2a-1)$. ADDED. Explanation of the Faulhaber's formula implication: Lemma. Let $n,m$ be positive integers, $m$ is even. Then $$1^{m} + 2^{m} + \dots + (n-1)^{m} \equiv B_{m} n \pmod{n}.$$ Proof. For $m=2$, the statement can be verified directly: $$1^2 + 2^2 + \dots + (n-1)^2 = \frac{(n-1)n(2n-1)}{6} \equiv B_2 n\pmod{n}.$$ For the rest assume $m\geq 4$. Using Faulhaber's formula and taking into account that $B_{m-1}=0$, we have \begin{split} & 1^{m} + 2^{m} + \dots + (n-1)^{m} \equiv 1^{m} + 2^{m} + \dots + n^{m} \\ &\equiv B_{m} n + \frac{1}{m+1} \sum_{k=3}^{m+1} \binom{m+1}{k} B_{m+1-k} n^k \pmod{n}. \end{split} It remains to show that $\frac{1}{m+1} \binom{m+1}{k} B_{m+1-k} n^k \equiv 0\pmod{n}$ for any integer $k\in [3,m+1]$. Consider any prime $p\mid n$, and let $t:=\nu_p(n)\geq 1$. Our goal is to show that $\nu_p\big(\frac{1}{m+1} \binom{m+1}{k} B_{m+1-k}\big) + t(k-1) \geq 0$. It is enough to focus on the case $t=1$, from which the case $t>1$ follows instantly. Since by von Staudt–Clausen theorem the denominators of Bernoulli numbers are square-free, we need to show that $$\nu_p\big(\frac{1}{m+1} \binom{m+1}{k}\big) + k - 2 \geq 0.$$ Noticing that $\nu_p\big(\frac{1}{m+1} \binom{m+1}{k}\big) = \nu_p\big(\frac1k\binom{m}{k-1}\big)\geq -\nu_p(k)$, our goal reduces to showing that $$k-2-\nu_p(k)\geq 0.$$ For $k=3$ and $k=4$, this inequality is trivial, while for $k\geq 5$ it follows from the bound $\nu_p(k)\leq \log_2(k)$. QED

Primality test for $\frac{(10 \cdot 2^n)^m-1}{10 \cdot 2^n-1} - 2$ and $\frac{(10 \cdot 2^n)^m+1}{10 \cdot 2^n+1} - 2$ Here is what I observed: For $\frac{(10 \cdot 2^n)^m-1}{10 \cdot 2^n-1} - 2$ : Let $N$ = $\frac{(10 \cdot 2^n)^m-1}{10 \cdot 2^n-1} - 2$ : when $m$ is a number $m \ge 3$ and $n \ge 0$. Let the sequence $S_i=2 \cdot T_{10 \cdot 2^n}(S_{i-1}/2)$ where $T_{n}(x)$ is the Chebyshev's polynomial of the first kind with $S_0=L_{10 \cdot 2^n}$ where $L_{n}$ is the $n_{th}$ Lucas number. Then $N$ is prime if and only if $S_{m-1} \equiv L_{30 \cdot 2^n-2}\pmod{N}$. You can run this test here. For $\frac{(10 \cdot 2^n)^m+1}{10 \cdot 2^n+1} - 2$ : Let $N$ = $\frac{(10 \cdot 2^n)^m+1}{10 \cdot 2^n+1} - 2$ : when $m$ is a odd number $m \ge 3$ and $n \ge 0$. Let the sequence $S_i=2 \cdot T_{10 \cdot 2^n}(S_{i-1}/2)$ where $T_{n}(x)$ is the Chebyshev's polynomial of the first kind with $S_0=L_{10 \cdot 2^n}$ where $L_{n}$ is the $n_{th}$ Lucas number. Then $N$ is prime if and only if $S_{m-1} \equiv L_{30 \cdot 2^n+2}\pmod{N}$. You can run this test here. Is there a way to explain this? I don't know how to start for proving it. If you found a counterexample please tell me.

It's often the case with such tests that the "only if" part is more or less easy to prove, while the "if" part is inaccessible for proving or disproving. Below I prove the "only if" part, ie. assuming that $N$ is prime. First notice that Chebyshev polynomials appear here just for efficient computation of Lucas number $$S_{m-1} = L_{(10\cdot 2^n)^m}.$$ Since $N\equiv -1\pmod{10}$, we have that the period of $L_k$ modulo $N$ divides $N-1$. It remains to notice that $$(10\cdot 2^n)^m\ =\ (N+2)(10\cdot 2^n\mp 1)\pm 1\ \equiv\ 30\cdot 2^n\mp 2\pmod{N-1},$$ implying that $S_{m-1} = L_{(10\cdot 2^n)^m} \equiv L_{30\cdot 2^n\mp 2}\pmod{N}$. PS. As for the "if" part - since its conclusion is based on a single congruence, we get an analog of Lucas preudoprimality test, which is known to have false positives. However, here we are restricted to exponentially growing numbers, where finding false positives becomes much harder if feasible.

Surface fitting with convexity requirement Hi all, Consider a cloud of points in 3D space (x,y,z). The data is well-behaved, once plotted the surface looks like some sort of spheroid. I assumed a form for the fitting function f(x,y,z) = c1 x^2 + c2 y^2 + c3 z^2 + c4 x^2y^2 + ..etc The coefficients were obtained using a least squares approximation. My only problem is that the surface has some concave portions. Does anyone know how to express a constraint that would generate the coefficients (c1,c2, ...) but ensure convexity at all points? Thanks a lot! J

For your apparent purpose, in dimension $n$ it is convenient to begin with a homogeneous polynomial of total degree $2n$ with all individual exponents even. Then, for translates and rotations, all sorts of lower degree and odd exponent terms may show up. In $\mathbb R^2,$ a rounded version of an ordinary square is $$ A(x^4 + y^4) + B x^2 y^2 = 1.$$ The ordinary unit circle is $A=1, B = 2.$ Disjoint hyperbolas are $A=1, B=-2.$ A somewhat squared shape, indeed the $L^4$ "unit circle," is $A=1, B=0.$ An alternative to Piet Hein's "superellipse" is $A=1, B=1.$ Finally, a real analytic curve that passes through all 8 lattice points with $ |x| \leq 1,\; |y| \leq 1$ other than the origin itself is $A=1, B = -1,$ or $$ x^4 - x^2 y^2 + y^4 = 1.$$ At some point I wanted a smooth version of an ordinary cube in $\mathbb R^3,$ meaning that it passed through all 26 integer lattice points with $ |x| \leq 1,\; |y| \leq 1,\; |z| \leq 1$ other than the origin itself. I wrote $$ A( x^6 + y^6 + z^6) + B (y^4 z^2 + z^4 x^2 + x^4 y^2 + y^2 z^4 + z^2 x^4 + x^2 y^4) + C x^2 y^2 z^2 = 1.$$ To find $A,B,C$ it is only necessary to check the $(x,y,z)$ triples $(0,0,1),(0,1,1),(1,1,1),$ and evidently $A=1, B=-\frac{1}{2}, C=1$ works. so $$ ( x^6 + y^6 + z^6) - \frac{1}{2}(y^4 z^2 + z^4 x^2 + x^4 y^2 + y^2 z^4 + z^2 x^4 + x^2 y^4) + x^2 y^2 z^2 = 1$$ is a rounded cube. I recall graphing this in spherical coordinates with $\rho$ a function of $\theta, \phi.$ The trouble was that it is very flat near the axes, so without spherical coordinates many different patches were necessary. It is obvious that this is a star-shaped body around the origin, it needs just a little more work to confirm that it is compact.

Hypergeometric sum specific value How to show? $${}_2F_1(1,1;\frac{1}{2}, \frac{1}{2}) = 2 + \frac{\pi}{2} $$ It numerically is very close, came up when evaluating: $$ \frac{1}{1} + \frac{1 \times 2}{1 \times 3} + \frac{1 \times 2 \times 3}{1 \times 3 \times 5} + \frac{1 \times 2 \times 3 \times 4}{1 \times 3 \times 5 \times 7} + \cdots = 1 + \frac{\pi}{2} ~ ?$$ I am missing something trivial, I am sure.

You may use the general formula from Brychkov, Prudnikov, Marichev, Integral and Series, Vol.3: $$ _{2}F_{1}(1,1;\frac{1}{2};x)=(1-x)^{-1}\left(1+\frac{\sqrt{x}\arcsin{\sqrt{x}}}{\sqrt{1-x}}\right) $$ and put in it $x=\frac{1}{2}$.

When is $(q^k-1)/(q-1)$ a perfect square? Let $q$ be a prime power and $k>1$ a positive integer. For what values of $k$ and $q$ is the number $(q^k-1)/(q-1)$ a perfect square, that is the square of another integer? Is the number of such perfect squares finite? Note that $(q^k-1)/(q-1)$ is the number of points in a finite projective space of dimension $k-1$. The above question is related to the following one: How many non-isomorphic finite projective spaces are there whose numbers of points are perfect squares. Preliminary calculation shows that $(q^k-1)/(q-1)$ is a perfect square when $(k,q)$ takes on one of the values $(2,3)$, $(5,3)$, $(4,7)$, $(2,8)$, in which cases it is equal to $2^2$, $11^2$, $20^2$, $3^2$, respectively. When $k=2$, $(q^k-1)/(q-1)$ is a perfect square if and only if $q=3$ or 8. When $k=3$, $(q^k-1)/(q-1)$ cannot be a perfect square. When $q=2$, $(q^k-1)/(q-1)$ cannot be a perfect square.

The equation $$ \frac{x^k-1}{x-1}=y^m$$ is known as the Nagell-Ljunggren equation. It is conjectured that for $x\geq 2$, $y\geq 2$, $k\geq 3$, $m\geq 2$, the only solutions are $$ \frac{3^5-1}{3-1}=11^2,\qquad \frac{7^4-1}{7-1}=20^2,\qquad \frac{18^3-1}{18-1}=7^3.$$ For $m=2$, the equation was solved by Ljunggren (Norsk. Mat. Tidsskr. 25 (1943), 17-20). Hence for your problem (where $m=2$ and $x$ is a prime power) we can assume that $k=1$ or $k=2$. In the first case, $y=1$ and $x$ is arbitrary, which we can regard as trivial solutions. In the second case, $x=y^2-1$ is a prime power, whence it is easy to see that either $x=8$ and $y=3$, or $x=3$ and $y=2$. To summarize, your equation only has the four solutions listed in your post, besides the trivial solutions ($k=1$ and $y=1$).

Infinite limit of ratio of nth degree polynomials The Problem I have two recursively defined polynomials (skip to the bottom for background and motivation if you care about that) that represent the numerator and denominator of a factor and I want to find the limit of that factor as n goes to infinity. $$n_0 = d_0 = 1$$ $$n_n = d_{n-1}x - n_{n-1}$$ $$d_n = d_{n-1}(x-1)-n_{n-1}$$ I was able represent this recursive relationship as a matrix and use eigenvalue matrix decomposition to find a closed form for $d_n$ and $n_n$. They are not very pretty: $$d_n = \frac{2^{-n-1}}{\sqrt{x-4}} \left(\left(\sqrt{x-4}+\sqrt{x}\right) \left(x+\sqrt{x-4} \sqrt{x}-2\right)^n+\left(\sqrt{x-4}-\sqrt{x}\right) \left(x-\sqrt{x-4} \sqrt{x}-2\right)^n+\frac{2 \left(\left(x-\sqrt{x-4} \sqrt{x}-2\right)^n-\left(x+\sqrt{x-4} \sqrt{x}-2\right)^n\right)}{\sqrt{x}}\right)$$ $$ n_n = 2^{-n-1} \left(\left(x+\sqrt{x-4} \sqrt{x}-2\right)^n+\left(x-\sqrt{x-4} \sqrt{x}-2\right)^n+\frac{\left(x+\sqrt{x-4} \sqrt{x}-2\right)^n-\left(x-\sqrt{x-4} \sqrt{x}-2\right)^n}{\sqrt{\frac{x-4}{x}}}\right)$$ The first few terms of the ratio ($r_n = n_n/d_n$) are: $$r_1 = \frac{x-1}{x-2}$$ $$r_2 = \frac{x^2-3 x+1}{x^2-4 x+3}$$ $$r_3 = \frac{x^3-5 x^2+6 x-1}{x^3-6 x^2+10 x-4}$$ $$r_4 = \frac{x^4-7 x^3+15 x^2-10 x+1}{x^4-8 x^3+21 x^2-20 x+5}$$ And so on. All of these equations seem to have poles located solely at $0\le x\le 4$, moreover the denominator and numerator seem to never have imaginary roots, but I haven't proven that. $x$ is assumed to be a positive real number, and because there are no poles $x > 4$ the limit seems to be well behaved and equal to 1(?) outside of this range. But within this range, I am extremely curious about what the limit is, if it even exists or is possible to evaluate. Can someone let me know if it's possible to evaluate this limit in this range? And if so can you let me know how to go about it? Background and Motivation An ideal transmission line can be modeled as an inductor and capacitor, the inductor is in series with the load and the capacitor is in parallel. The impedance of an inductor is given by $Z_L = i \omega L$ and the impedance of a capacitor is $Z_C = \frac{1}{i \omega C}$. If we string several transmission lines together and end in an open circuit then we can start evaluating from the end using the impedance addition laws. We first add $Z_L + Z_C$ and then to combine with the second to last capacitor we must $\frac{1}{\frac{1}{Z_L+Z_C}+\frac{1}{Z_C}}$. When we do that we find, interestingly that the ratio between $Z_C$ and this value is our real valued ratio $r_1$. We can then solve for $r_n$ by breaking up the ratio into a numerator and denominator and starting from the (n-1)th iteration, and find the relationship given at the top of the question, with $x = \omega ^2 L C$. Because a series of short transmission lines strung together is the same thing as a long transmission line, I expected the limit of this ratio to be equal to 1. However, when attempting to evaluate this limit it seems that it is not that simple.

The explanation for your observations is that you are dealing with a self-adjoint difference equation in disguise. Let me make a transformation along these line, though I won't analyze it through to the end. If we let $Y_n=(n_n,d_n)^t$, then this satisfies $$ Y_n = \begin{pmatrix} -1 & x \\ -1 & x-1 \end{pmatrix} Y_{n-1} . $$ Now write $Y_n=A^n W_n$, with $A=\begin{pmatrix} -1 & 0\\ -1&-1\end{pmatrix}$, so $$ A^n = (-1)^n \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} $$ (this is variation of constants, relative to $x=0$). By a calculation, $$ J(W_n-W_{n-1}) = -xH_n W_n, \quad\quad H_n=-JA^{-n}\begin{pmatrix} 0&1\\ 0&1\end{pmatrix}A^{n-1},\quad J=\begin{pmatrix} 0&-1\\ 1&0\end{pmatrix}. $$ An equation of this form is called a (discrete) canonical system, and it corresponds to a symmetric difference expression in the space $\ell^2_H$ if $H_n$ is a positive definite matrix. Our luck holds here because a calculation gives that $$ H_n = \begin{pmatrix} (n-1)^2 & n-1 \\ n-1 & 1 \end{pmatrix} \ge 0 , $$ as required. This explains why the zeros are real, though it doesn't show why they are in $[0,4]$. For this, one would have to analyze the spectrum of this system more carefully (you could rewrite it as a Jacobi difference equation).

Quasi-concavity of $f(x)=\frac{1}{x+1} \int_0^x \log \left(1+\frac{1}{x+1+t} \right)~dt$ I want to prove that function \begin{equation} f(x)=\frac{1}{x+1} \int\limits_0^x \log \left(1+\frac{1}{x+1+t} \right)~dt \end{equation} is quasi-concave. One approach is to obtain the closed form of the integral (provided below) and then prove that the result is quasi-concave. I tried this but it seems to be difficult. Do you have any idea how to prove the quasi-concavity? \begin{align} f(x)&=\frac{1}{x+1} \int\limits_0^x \log \left(1+\frac{1}{x+1+t} \right)~dt\\ &= \frac{1}{x+1} \left[ \log(2x + 2) + (2x+1)\log \left(1+\frac{1}{2x+1} \right) -\log(x+2) - (x+1)\log \left( 1+\frac{1}{x+1} \right) \right] \end{align}

The derivative is: ${d f(x) \over d x} = \frac{-\log (x+1)+\log (x+2)+\log \left(\frac{x+1}{2 x+1}\right)}{(x+1)^2}$, which is: and hence is both positive and negative, as Mark L. Stone wrote.

Want to prove an inequality I want to show that $9*\left[\frac{xy}{x+y}-q(1-q)\right]-12*[xy-q(1-q)]+(1-q-x)^{3}+(x+y)^{3}+(q-y)^{3}-1\geq0$ where $0<q<1$ $0<x<1-q$ $0<y<q$ $(x+y)\left[1+max\{\frac{1-q}{y},\frac{q}{x}\}\right]\leq3$ I play with it numerically. It is right. But don't know how to prove it analytically. Anybody can help? Thanks a lot.

Remark: Actually, the inequality is quite easy. The trick is to take $x, y$ as parameters. Denote the expression by $f(q)$. It suffices to prove that $f(q) \ge 0$ provided that \begin{align*} &x, y > 0,\\ &x + y < 1,\\ &3y > (x + y)^2,\\ &3x > (x + y)^2, \\ &y < q < 1 - x, \\ &1 + y - \frac{3y}{x + y} \le q \le \frac{3x}{x + y} - x. \end{align*} (Note: Condition $(x+y)\left[1+max\{\frac{1-q}{y},\frac{q}{x}\}\right]\leq3$ is equivalent to $(x + y)[1 + (1-q)/y] \le 3$ and $(x + y)(1 + q/x) \le 3$ which are equivalent to $1 + y - \frac{3y}{x + y} \le q \le \frac{3x}{x + y} - x$. Also, from $y < \frac{3x}{x + y} - x$, we have $3x > (x + y)^2$, and from $1 + y - \frac{3y}{x + y} < 1 - x $, we have $3y > (x + y)^2$.) Note that $f(q)$ is quadratic and concave (the coefficient of $q^2$ is $-3x - 3y$). Also, we have \begin{align*} f(y) &= \frac{3x(1 - x - y)(2y - x)}{x + y},\\ f(1 - x) &= \frac{3y(1 - x - y)(2x - y)}{x + y},\\ f\left(1 + y - \frac{3y}{x + y}\right) &= \frac{3y(2 - x - y)(x - 2y)}{x + y}, \\ f\left(\frac{3x}{x + y} - x\right) &= \frac{3x(2 - x - y)(y - 2x)}{x + y}. \end{align*} We split into three cases: (1) If $x > 2y$, we have $1 + y - \frac{3y}{x + y} > y$ and $1 - x \le \frac{3x}{x + y} - x$, and $f(1 + y - \frac{3y}{x + y}) \ge 0$ and $f(1 - x) \ge 0$. Thus, $f(q) \ge 0$. (2) If $y/2 \le x \le 2y$, we have $1 + y - \frac{3y}{x + y} \le y$ and $1 - x \le \frac{3x}{x + y} - x$, and $f(y) \ge 0$ and $f(1 - x) \ge 0$. Thus, we have $f(q) \ge 0$. (3) If $x < y/2$, we have $1 + y - \frac{3y}{x + y} \le y$ and $1 - x > \frac{3x}{x + y} - x$, and $f(y) \ge 0$ and $f(\frac{3x}{x + y} - x) \ge 0$. Thus, $f(q) \ge 0$. We are done.

Estimation of Hypergeometric function ${_3F_2}$ Is there any way to estimate the following function, which is a result of sum of ratios of Gamma functions? $$ {_3F_2}\begingroup \renewcommand*{\arraystretch} % your pmatrix expression \left[ \begin{array}{c@{}c} \begin{array}{c} -q, \frac{M}{2}, \frac{1}{2}+\frac{M}{2}\\ \frac{1}{2}, -q-\frac{n-M}{2}+1 \end{array} ;& 1 \end{array}\right]\endgroup, $$ where $q, M, n \in N, n\geq M$

Here are explicit forms of $$f(q,n,m)={_3F_2}\begingroup \renewcommand*{\arraystretch} % your pmatrix expression \left[ \begin{array}{c@{}c} \begin{array}{c} -q, \frac{m}{2}, \frac{1}{2}+\frac{m}{2}\\ \frac{1}{2}, -q-\frac{n-m}{2}+1 \end{array} ;& 1 \end{array}\right]\endgroup,$$ for small values of $q$: $$f(1,n,m)=-\frac{m^2+n}{m-n},$$ $$f(2,n,m)=\frac{m^4+m^2 (6 n+8)+3 n (n+2)}{3 (m-n-2) (m-n)},$$ $$f(3,n,m)=-\frac{m^6+5 m^4 (3 n+8)+m^2 (45 n^2 +210n+184)+15 n (n+2) (n+4)}{15 (m-n-4) (m-n-2) (m-n)}.$$ For large $m$ or $n$ we have the limits $$\lim_{m\rightarrow\infty}m^q f(q,n,m)=(-1)^q\frac{1}{(2q-1)!!}, $$ $$\lim_{n\rightarrow\infty}n\bigl(f(q,n,m)-1\bigr)=q(m+m^2).$$

The exact constant in a bound on ratios of Gamma functions The answer to another question (Upper bound of the fraction of Gamma functions) gave an asymptotic upper bound for an expression with Gamma functions: $$\left(\frac{\Gamma(a+b)}{a\Gamma(a)\Gamma(b)}\right)^{1/a}\!\leq \,C\,\frac{a+b}a, \forall a,b\geq\frac12$$ What is the best possible value for the constant $C$ in that statement?

The optimal $C$ is $\mathrm{e}$. Proof: We have $$\ln C \ge \ln a - \ln(a + b) + \frac{\ln \Gamma(a + b) -\ln a - \ln\Gamma(a) - \ln\Gamma(b)}{a}.$$ Let $$F(a, b) := \ln a - \ln(a + b) + \frac{\ln \Gamma(a + b) -\ln a - \ln\Gamma(a) - \ln\Gamma(b)}{a}.$$ We have $$\frac{\partial F}{\partial b} = - \frac{1}{a+b} + \frac{\psi(a + b) - \psi(b)}{a} \ge 0 \tag{1}$$ where $\psi(\cdot)$ is the digamma function defined by $\psi(u) = \frac{\mathrm{d} \ln \Gamma(u)}{\mathrm{d} u} = \frac{\Gamma'(u)}{\Gamma(u)}$. The proof of (1) is given at the end. Fixed $a\ge 1/2$, we have $$G(a) := \lim_{b\to \infty} F(a, b) = \ln a + \frac{ -\ln a - \ln\Gamma(a)}{a} $$ where we have used $$\lim_{b\to \infty} -\ln(a + b) + \frac{\ln \Gamma(a + b) - \ln\Gamma(b)}{a} = 0.$$ (Note: Use $\sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x} \le \Gamma(x) \le \sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x}\mathrm{e}^{\frac{1}{12x}}$ for all $x > 0$.) We have $$ a^2 G'(a) = a - 1 - a\psi(a) + \ln a + \ln \Gamma(a) \ge 0.\tag{2} $$ The proof of (2) is given at the end. We have $$\lim_{a\to \infty} G(a) = 1.$$ (Note: Use $\sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x} \le \Gamma(x) \le \sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x}\mathrm{e}^{\frac{1}{12x}}$ for all $x > 0$.) Thus, the optimal $C$ is $\mathrm{e}$. Proof of (1): Using Theorem 5 in [1]: for all $u > 0$, $$\ln u - \frac{1}{2u} - \frac{1}{12u^2} < \psi(u) < \ln u - \frac{1}{2u} - \frac{1}{12(u+1/14)^2},$$ we have \begin{align*} &- \frac{1}{a+b} + \frac{\psi(a + b) - \psi(b)}{a}\\ \ge{}&- \frac{1}{a+b} + \frac{1}{a} \left(\ln (a+b) - \frac{1}{2(a+b)} - \frac{1}{12(a+b)^2}\right)\\ &\qquad - \frac{1}{a}\left(\ln b - \frac{1}{2b} - \frac{1}{12(b+1/14)^2}\right)\\ ={}& \frac{1}{a}\ln(1 + a/b) - \frac{1}{a+b} - \frac{1}{2a(a+b)} - \frac{1}{12a(a+b)^2} + \frac{1}{2ab} + \frac{1}{12a(b+1/14)^2}\\ \ge{}& \frac{1}{a}\left(\ln(1 + a/b) - \frac{a/b}{1 + a/b}\right) + \frac{1}{2ab} - \frac{1}{2a(a+b)} + \frac{1}{12a(b+1/14)^2} - \frac{1}{12a(a+b)^2}\\ \ge{}&0 \end{align*} where we use $\ln(1+x) \ge \frac{x}{1+x}$ for all $x \ge 0$. We are done. $\phantom{2}$ Proof of (2): Using $\Gamma(x) \ge \sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x}$ and $\psi(u) < \ln u - \frac{1}{2u} - \frac{1}{12(u+1/14)^2}$ for all $u > 0$ (Theorem 5 in [1]), we have \begin{align*} &a - 1 - a\psi(a) + \ln a + \ln \Gamma(a)\\ \ge{}& a - 1 - a \left(\ln a - \frac{1}{2a} - \frac{1}{12(a+1/14)^2}\right) + \ln a + \frac12\ln(2\pi) + (a-1/2)\ln a - a\\ ={}& \frac12\ln(2\pi a) - \frac12 + \frac{a}{12(a+1/14)^2}\\ \ge{}& 0. \end{align*} We are done. Reference [1] L. Gordon, “A stochastic approach to the gamma function”, Amer. Math. Monthly, 9(101), 1994, 858-865.

A question on an identity relating certain sums of Harmonic numbers In the description of this question, it was established that \begin{align} \sum_{n=2}^{\infty} (\zeta(n)^{2}-1) &= \frac{7}{4} - \zeta(2) + 2 \sum_{m=1}^{\infty} \frac{H_{m-1- \frac{1}{m}} - H_{- \frac{1}{m}} - H_{m-1} }{m} \qquad(1) \end{align} In the answer to that very same question, Fedor Petrov showed that an integral representation can be found for this sum. If we proceed from his derivation and use this calculation, we obtain \begin{align} \sum_{n=2}^{\infty}(\zeta(n)^{2}-1) &= 1 + \int_0^1 \sum_{k=0}^\infty\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 + \sum_{k=0}^{\infty} \int_{0}^{1}\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 - \sum_{k=0}^{\infty} \frac{H_{- \frac{1}{k+2}}}{k+2} \\ &= 1- \sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m}.\qquad (2) \end{align} Now, we can equate the two expressions. Note that not all terms in the first sum can be separated, as only $\sum \frac{H_{-\frac{1}{m}}}{m}$ converges as a standalone series. Grouping like terms together, we find $$\sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m} = \frac{3}{4} - \zeta(2) + 2 \sum_{m=2}^{\infty} \frac{H_{m - 1- \frac{1}{m}} - H_{m-1} }{m}. \qquad \qquad (*)$$ What I find interesting here, is that (series involving) Harmonic numbers with both negative and positive fractional arguments can be related to one another. Questions: * *Do identities like the $(*)$-marked equation appear in the literature? *Can it be shown that one side of the equation amounts to the other side, only by means of algebraic manipulations and without invoking the aforementioned integral representation? *I seem to have made some calculation error, because the sums don't appear to add up to the same number. Can this error be identified? Answer: this has been answered by Carlo Beenakker. The correct identity has now been established. The first two questions remain open.

Q3: The first identity (1) is not correct, it should read \begin{align} \sum_{n=2}^{\infty}(\zeta(n)^{2} -1) &= \frac{7}{4} - \zeta(2) + 2\sum_{m=2}^{\infty} \frac{H_{m-1-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m-1}}{m} . \end{align} The final identity (*) then becomes $$ \sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m} = \frac{3}{4} - \zeta(2) + 2 \sum_{m=2}^{\infty} \frac{H_{m -1- \frac{1}{m}} - H_{m-1} }{m}. $$ Left-hand-side and right-hand-side both evaluate to $-1.4653$.

Divisibility condition implies $a_1=\dotsb=a_k$? Let $a_1, a_2, \dotsc, a_k$ be $k$ positive integers, $k\ge2$. For all $n\ge n_0$ there is a positive integer $f(n)$ such that $n$ and $f(n)$ are relatively prime and $a_{1}^{f(n)}+\dotsb+a_{k}^{f(n)}$ is a multiple of $a_{1}^n+\dotsb+a_{k}^n$. Is it true that $a_1=a_2=\dotsb=a_k$ or is it possible to construct a counterexample?

Here's a a tweak of Seva's idea that gives a counterexample. Note that if $r$ is odd, then $2^{n}+1$ divides $2^{rn} + 1$. Let $k = 6$, $a_{1} = 1$, $a_{2} = a_{3} = a_{4} = 2$, $a_{5} = a_{6} = 4$. Then $a_{1}^{n} + \cdots + a_{6}^{n} = 1 + 3 \cdot 2^{n} + 2 \cdot 4^{n} = (1+2^{n})(1+2^{n+1})$. If $n$ is any positive integer, let $f(n) = 2n^{2} + n - 1$. We see that $f(n) \equiv -1 \pmod{n}$ and so $\gcd(n,f(n)) = 1$. Now, $$a_{1}^{f(n)} + \cdots + a_{6}^{f(n)} = (1+2^{2n^{2} + n - 1})(1+2^{2n^{2} + n}).$$ The second factor is $2^{n(2n+1)}+1$ and so is a multiple of $2^{n} + 1$. On the other hand, the first factor is $2^{(n+1)(2n-1)}+1$ and so it is a multiple of $2^{n+1} + 1$. Thus, $a_{1}^{f(n)} + \cdots + a_{6}^{f(n)}$ is a multiple of $a_{1}^{n} + \cdots + a_{6}^{n}$.

Finding all proper divisors of $a_3z^3 +a_2z^2 +a_1z+1$ of the form $xz+1$ Let $n=a_3z^3+a_2z^2+a_1z+1$ where $a_1<z, \ a_2<z, \ 1 \le a_3<z, z>1$ are non negative integers. To obtain proper divisors of $n$ of the form $xz+1$, one may perform trial divisions $xz+1 \ | \ n$, for all $xz+1 \le \sqrt n$. Trial division however is inefficient as $z$ becomes large. The method below is much more efficient. Since $xz+1 \ | \ n$ , we may write $(xz+1)(yz+1)=n$. Assume $y \le x$. We consider two cases; **Case 1: ** $1 \le x <z$ and $1 \le y<z$ $xyz+x+y=a_3z^2+a_2z+a_1$. Since every positive integer has a unique base $z$ representation we have; \begin{equation} x+y=C\cdot z+a_1 , C=0 \ \text{or} \ 1 \ \end{equation} \begin{equation} xy+C = a_3z+a_2 \end{equation} Solving the two equations for the cases $C=0$ and $C=1$, $x$ and $y$ can be determined. **Case 2: ** $z\le x<z^2$ and $1 \le y<z$ Let $x=x_1z+x_0$ and $y = y_0$, $1\le x_1<z, \ 0 \le x_0<z$, $ 1 \le y_0 <z$. So \begin{equation} x_1y_0z^2+(x_0y_0+x_1)z+x_0+y_0=a_3z^2+a_2z+a_1 \end{equation} Comparing coefficients of powers of $z$; \begin{equation} x_0+y_0=C_1 \cdot z+a_1, \ C_1 = 0 \ \text{or} \ 1 \end{equation} \begin{equation} x_0y_0+x_1+C_1=C_2 \cdot z+a_2 \end{equation} \begin{equation} x_1y_0+C_2=a_3 \end{equation} From the last equation, either $y_0 \le \sqrt a_3$ or $x_1 \le \sqrt a_3$. So we do trial divisions $yz+1 \ | \ n$ for all $y = y_0 \le \sqrt a_3$. Also for each value of $x_1 \le \sqrt a_3 $, we solve the three equations simultaneously to find $x_0, y_0$ and $C_2$ when $C_1=0$ and $C_1 = 1$. Cases 1 and 2 exhaust all the possible cases. In this example, we had $n < z^4$. How can this method be modified to efficiently find all proper divisors of the form $xz+1$ of an arbitrary positive integer $n$, $n \equiv \ 1 ($ mod $ z) $ or at least when $n<z^5 $?

An extension to the case when $n<z^5$ with some restrictions on $x$ and $y$: Let $n=a_4z^4+a_3z^3+a_2z^2+a_1z+1$, $a_i < z $, $a_4>0, z>1$. We are looking for positive integes $x$ and $y$ such that $(xz+1)(yz+1)=n$. We add a restriction on $x$ and $y$; If $x=x_k \cdot z^k + \cdots + x_0$ and $y=y_t \cdot z^t + \cdots + y_0$ are the base $z$ represenations of $x$ and $y$, then $x_i \le x_k$ for all $i<k$ and $y_i \le y_t$ for all $i<t$. Case 1: $x=x_1\cdot z+x_0$ and $y=y_0$ We get \begin{equation} x_1y_0z^3+(x_0y_0+x_1)z^2+(x_0+y_0)z+1=a_4z^4+a_3z^3+a_2z^2+a_1z+1 \end{equation} Comparing coefficients of powers of $z$; \begin{equation} x_0+y_0=C_1\cdot z+a_1 , C_1=0 \ \text{or} \ 1 \end{equation} \begin{equation} x_0y_0+x_1+C_1=C_2\cdot z+a_2 \end{equation} \begin{equation} x_1y_0+C_2=a_4\cdot z+a_3 \end{equation} From the last equation, $x_1y_0 < (a_4+1) \cdot z$ and since it's assumed that $x_0 \le x_1$, we have $x_0y_0+x_1+C_1<(a_4+2)\cdot z$ , therefore $C_2 \le a_4+1$. By solving the three equations for $C_2=0, 1, \cdots , a_4+1$, all solutions with $x_0 \le x_1$ are found. Case 2: $x=x_1\cdot z+x_0$ and $y=y_1\cdot z+y_0$ We get $x_1y_1z^4 + (x_0y_1 + x_1y_0)z^3 + (x_0y_0+x_1+y_1)z^2+(x_0+y_0)z+1=a_4z^4+a_3z^3+a_2z^2+a_1z+1$ Comparing coefficients of powers of $z$; \begin{equation} x_0+y_0=C_1\cdot z +a_1 \ \ C_1 = 0 \ \text{or} \ 1 \end{equation} \begin{equation} x_0y_0+x_1+y_1+C_1=C_2\cdot z +a_2 \end{equation} \begin{equation} x_0y_1+x_1y_0+C_2=C_3\cdot z +a_3 \end{equation} \begin{equation} x_1y_1+C_3=a_4 \end{equation} From the last equation, we have $x_1y_1 \le a_4 < z$. Assuming $x_0 \le x_1$ and $y_0 \le y_1$ we have, \begin{equation} x_0y_0+x_1+y_1+C_1 < 3z \end{equation} Therefore $C_2 \le 2$. And from the second last equation we have, \begin{equation} x_0y_1+x_1y_0+C_2 \le 2z \end{equation} Therefore $C_3 \le 2$. Solving the four equations for all the 18 combinations of $C_1 \le 1$, $C_2 \le 2$, $C_3 \le 2$, all solutions $x_1,x_0,y_1,y_0$ are found. The last case when $x$ and $y$ have $3$ base $z$ digits and $1$ base $z$ digits respectively can be tackled in the same way.

EM-wave equation in matter from Lagrangian Note I am not sure if this post is of relevance for this platform, but I already asked the question in Physics Stack Exchange and in Mathematics Stack Exchange without success. Setup Let's suppose a homogeneous dielectric with a (spatially) local dielectric response function $\underline{\underline{\epsilon}} (\omega)$ (in general a tensor), such that we have the linear response relation $$\mathbf{D}(\mathbf{x}, \omega) = \underline{\underline{\epsilon}} \left(\omega \right) \mathbf{E}(\mathbf{x}, \omega) \, ,$$ for the displacement field $\mathbf{D}$ in the dielectric. We can now write down a Lagrangian in Fourier space, describing the EM-field coupling to the dielectric body $$\mathcal{L}=\frac{1}{2}\left[\mathbf{E}^{*}\left(x,\omega\right) \cdot (\underline{\underline{\epsilon}} \left(\omega \right)-1) \mathbf{E}\left(x,\omega\right)+|\mathbf{E}|^{2}\left(x,\omega\right)-|\mathbf{B}|^{2}\left(x,\omega\right) \right] \, .$$ If we choose a gauge \begin{align} \mathbf{E} &= \frac{i \omega}{c} \mathbf{A} \\ \mathbf{B} &= \nabla \times \mathbf{A} \, , \end{align} such that we can write the Lagrangian (suppressing arguments) in terms of the vector potential $\mathbf{A}$ as $$\mathcal{L} =\frac{1}{2}\left[\frac{\omega^2}{c^2} \mathbf{A}^{*} \cdot (\underline{\underline{\epsilon}}- \mathbb{1}) \mathbf{A}+ \frac{\omega^2}{c^2} |\mathbf{A}|^{2}-|\nabla \times \mathbf{A}|^{2}\right] \, . $$ And consequently we have the physical action $$ S[\mathbf{A}] = \int d \mathbf{x} \int \frac{d \omega}{2 \pi} \; \mathcal{L} \left(\mathbf{A}\right) \, . $$ Goal My goal is to derive the EM-wave equation for the electric field in the dielectric media. Idea So my ansatz is the following: If we use Hamilton's principle, we want the first variation of the action to be zero \begin{align} 0 = \delta S[\mathbf{A}] &= \left.\frac{\mathrm{d}}{\mathrm{d} \varepsilon} S[\mathbf{A} + \varepsilon \mathbf{h}] \right|_{\varepsilon=0} \\ &= \left.\frac{\mathrm{d}}{\mathrm{d} \varepsilon} \int d \mathbf{x} \int \frac{d \omega}{2 \pi} \; \mathcal{L} (\mathbf{A} + \varepsilon \mathbf{h}) \right|_{\varepsilon=0} \\ &= \int d \mathbf{x} \int \frac{d \omega}{2 \pi} \; \frac{1}{2} \Bigg( \frac{\omega^2}{c^2} \mathbf{A}^* \cdot ({\underline{\underline{\epsilon}}}-\mathbb{1}) \mathbf{h} + \frac{\omega^2}{c^2} \mathbf{h}^* \cdot ({\underline{\underline{\epsilon}}}- \mathbb{1}) \mathbf{A} + \frac{\omega^2}{c^2} \mathbf{A}^* \cdot \mathbf{h} + \frac{\omega^2}{c^2} \mathbf{h}^* \cdot \mathbf{A} \\ &\quad \quad \quad \quad \quad \quad \quad \quad- (\nabla \times \mathbf{A}^* ) \cdot ( \nabla \times \mathbf{h}) - (\nabla \times \mathbf{h}^* ) \cdot ( \nabla \times \mathbf{A}) \Bigg) \\ &= \int d \mathbf{x} \int \frac{d \omega}{2 \pi} \; \frac{1}{2} \Bigg( \frac{\omega^2}{c^2} \left[ ({\underline{\underline{\epsilon}}}^{\dagger}-\mathbb{1}) \mathbf{A} \right]^* \cdot \mathbf{h} + \frac{\omega^2}{c^2} \left[({\underline{\underline{\epsilon}}}- \mathbb{1}) \mathbf{A} \right] \cdot \mathbf{h}^* + \frac{\omega^2}{c^2} \mathbf{A}^* \cdot \mathbf{h} + \frac{\omega^2}{c^2} \mathbf{A} \cdot \mathbf{h}^* \\ &\quad \quad \quad \quad \quad \quad \quad \quad- (\nabla \times \nabla \times \mathbf{A}^* ) \cdot \mathbf{h} - (\nabla \times \nabla \times \mathbf{A} ) \cdot \mathbf{h}^* \Bigg) \\ &= \int d \mathbf{x} \int \frac{d \omega}{2 \pi} \; \frac{1}{2} \Bigg( \underbrace{\left[ \frac{\omega^2}{c^2} \left[ ({\underline{\underline{\epsilon}}}^{\dagger}-\mathbb{1}) \mathbf{A} \right]^* + \frac{\omega^2}{c^2} \mathbf{A}^* - \nabla \times \nabla \times \mathbf{A}^* \right]}_{\stackrel{!}{=} 0} \cdot \mathbf{h} \\ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + \underbrace{\left[ \frac{\omega^2}{c^2} \left[({\underline{\underline{\epsilon}}}- \mathbb{1}) \mathbf{A} \right] + \frac{\omega^2}{c^2} \mathbf{A} - \nabla \times \nabla \times \mathbf{A} \right]}_{\stackrel{!}{=} 0} \cdot \mathbf{h}^* \Bigg) \, , \end{align} for all $\mathbf{h}(\mathbf{x}, \omega)$. And consequently we get the equations \begin{align} \frac{\omega^2}{c^2} \left[ ({\underline{\underline{\epsilon}}}^{\dagger}-\mathbb{1}) \mathbf{A} \right]^* + \frac{\omega^2}{c^2} \mathbf{A}^* - \nabla \times \nabla \times \mathbf{A}^* &= 0 \\ \frac{\omega^2}{c^2} \left[({\underline{\underline{\epsilon}}}- \mathbb{1}) \mathbf{A} \right] + \frac{\omega^2}{c^2} \mathbf{A} - \nabla \times \nabla \times \mathbf{A} &= 0 \, . \end{align} If we suppose a lossy dielectric body, such that $\underline{\underline{\epsilon}}^{\dagger} \neq \underline{\underline{\epsilon}}$, the equations are in contradiction. Time-domain An analogues derivation in the time-domain (which I can post here on request), yields the wave equation (in Fourier space) $$\frac{\omega^2}{2 c^2} (\underline{\underline{\epsilon}}-1) \mathbf{A} + \frac{\omega^2}{2 c^2} (\underline{\underline{\epsilon}}^{\dagger} -1) \mathbf{A} + \frac{\omega^2}{c^2} \mathbf{A} - \left( \nabla \times \nabla \times \mathbf{A} \right) = 0 \, .$$ This result is also not resembling the expected result, for $\underline{\underline{\epsilon}}^{\dagger} \neq \underline{\underline{\epsilon}}$. Question What went wrong in the calculation?

Expanding on Carlo Beenakker's comment, one can't just expect to substitute in a complex dielectric function to properly describe absorption. Rather, the relevant question is how to structure a Lagrangean such as to generate the desired damping term in the equation of motion. For example, to generate linear damping in a harmonic oscillator, one can use $$ L=e^{\gamma t} \left( \frac{m\dot{q}^2 }{2} -\frac{kq^2 }{2} \right) . $$ Since the electromagnetic field is a collection of oscillators, one might be able to use a generalization of this.

What are the properties of this set of infinite matrices and operations on them? Consider infinite matrices of the form $$\left( \begin{array}{ccccc} a_0 & a_1 & a_2 & a_3 & . \\ 0 & a_0 & a_1 & a_2 & . \\ 0 & 0 & a_0 & a_1 & . \\ 0 & 0 & 0 & a_0 & . \\ . & . & . & . & . \\ \end{array} \right)$$ The elements on each diagonal coincide. My questions are: * *Do they form a commutative ring? *Can they be extended to form a field? Now, let define an operation $\operatorname{reg} A=\sum_{k=0}^\infty B_k a_k,$ where $B_k$ are Bernoulli numbers. What are the properties of this operation? Let's define another operation $\det' A=\exp(\Re \operatorname{reg} \log A)$. What are the properties of this operation? Motivation part. This is meant to be a matrix representation of divergent integrals and series. For instance, $\sum_{k=1}^\infty 1= \left( \begin{array}{ccccc} 0 & 1 & 0 & 0 & . \\ 0 & 0 & 1 & 0 & . \\ 0 & 0 & 0 & 1 & . \\ 0 & 0 & 0 & 0 & . \\ . & . & . & . & . \\ \end{array} \right)$ $\sum_{k=0}^\infty 1= \left( \begin{array}{ccccc} 1 & 1 & 0 & 0 & . \\ 0 & 1 & 1 & 0 & . \\ 0 & 0 & 1 & 1 & . \\ 0 & 0 & 0 & 1 & . \\ . & . & . & . & . \\ \end{array} \right)$ $\sum_{k=0}^\infty k= \left( \begin{array}{ccccc} 1/12 & 1/2 & 1/2 & 0 & . \\ 0 & 1/12 & 1/2 & 1/2 & . \\ 0 & 0 & 1/12 & 1/2 & . \\ 0 & 0 & 0 & 1/12 & . \\ . & . & . & . & . \\ \end{array} \right)$ $\int_0^\infty x dx=\int_0^\infty \frac 2{x^3}=\left( \begin{array}{ccccc} 1/6 & 1/2 & 1/2 & 0 & . \\ 0 & 1/6 & 1/2 & 1/2 & . \\ 0 & 0 & 1/6 & 1/2 & . \\ 0 & 0 & 0 & 1/6 & . \\ . & . & . & . & . \\ \end{array} \right)$ There are also some expressions that include divergent integrals that can be represented this way: $(-1)^{\int_0^\infty dx}=\left( \begin{array}{ccccccc} i & -\pi & -\frac{i \pi ^2}{2} & \frac{\pi ^3}{6} & \frac{i \pi ^4}{24} & -\frac{\pi ^5}{120} & . \\ 0 & i & -\pi & -\frac{i \pi ^2}{2} & \frac{\pi ^3}{6} & \frac{i \pi ^4}{24} & . \\ 0 & 0 & i & -\pi & -\frac{i \pi ^2}{2} & \frac{\pi ^3}{6} & . \\ 0 & 0 & 0 & i & -\pi & -\frac{i \pi ^2}{2} & . \\ 0 & 0 & 0 & 0 & i & -\pi & . \\ 0 & 0 & 0 & 0 & 0 & i & . \\ . & . & . & . & . & . & . \\ \end{array} \right)$ The $\operatorname{reg}$ operation gives the regularized value of the integral or series.

If the matrices have entries from a (unital) ring $R$ then the set of such matrices is isomorphic to $R[[x]]$, the ring of formal power series over $R$. To see this, observe that the map sending the infinite matrix with $a_0 = 0$, $a_1 = 1$ and $a_k = 0$ for $k \ge 2$ to $x$ is a ring isomorphism. This also answers the second question: if $R$ is an integral domain then set of matrices embeds canonically in the field of fractions of $R[[x]]$ and this is the smallest field containing $R[[x]]$. In particular, if $R$ is a field then this field is $\{ \sum_{k=-m}^\infty a_k x^k : a_k \in R, m \in \mathbb{N}_0 \}$. I'm uncertain how $\mathrm{reg}$ is (well)-defined, but certainly one can take $R$ to be the polynomial ring $\mathbb{C}[z]$ and then something like $\sum_{k=0}^\infty B_k(z) x^k$ is a well-defined element of $R[[x]] = \mathbb{C}[z][[x]]$. If, as in the correction then one wants Bernoulli numbers rather than the polynomials, just specialize to $\mathbb{C}[[x]]$ by evaluating at $z=0$.

Are there finitely many primes $x$ such that for a fixed odd prime $p$, $n=x^{p-1}+x^{p-2}+\dotsb + x+1$ is composite and $x \mid \phi(n)$? Let \begin{equation} n =x^{p-1}+x^{p-2}+\dotsb + x+1 \end{equation} where $x$ and $p$ are odd primes. If $p$ is set to $5$, it appears $x=5$ is the only prime $x$ such that $n$ is composite and $x \mid \phi(n) $ (verified up to $x \le 2\cdot 10^6$). Setting $p$ to $7$, we get only two values of $x$; $x=7$ and $x = 281$. If $x$ is allowed to take composite values, it appears there are infinitely many $x$ such that $x \mid \phi(n)$. Therefore, the following Conjecture is reasonable. Conjecture 1 Let \begin{equation} n =x^{p-1}+x^{p-2}+\dotsb + x+1 \end{equation} where $x$ and $p$ are odd primes. For a fixed odd prime $p$, there are finitely many primes $x$ such that $x \mid \phi(n) $ with $n$ composite. If Conjecture 1 is true, then, for a fixed prime $p$, there exists an upper bound $x_\text{max}$ such that $x \nmid \phi(n) $ for all $x>x_\text{max} $ with $n$ composite. If Conjecture 1 is true, Theorem 1 gives a fast primality test for integers $x^{p-1}+x^{p-2}+\dotsb + x+1$ with $x > x_\text{max}$. Theorem 1 Assuming Conjecture 1. Let $n =x^{p-1}+x^{p-2}+\dotsb + x+1$ where $x$ and $p$ are odd primes with $x>x_\text{max}$. If there exists a positive integer $b$ such that $b^{n-1} \equiv 1 \pmod n$ and $b^{(n-1)/x} \not\equiv 1 \pmod n$ then $n$ is prime. Proof. Assuming Conjecture 1, we have $ x \mid \phi(n) $ if and only if $n$ is prime. Assume $n$ is composite. Using the properties of order of an integer, one can deduce that $ord_nb \mid (n-1) /x$. It follows that if $n$ is composite then $b^{(n-1)/x} \equiv 1 \pmod n$, contradicting our hypothesis. Therefore $n$ must be prime. Note: As $x$ is prime then $x \mid \phi(n) $ implies $n = (ux+1)(vx+1)$ for some non negative integers $u$, $v$ with $ux+1$ prime. It can also be shown that $n = (sp+1)(tp+1)$ for some non negative integers $s$, $t$ with $sp+1$ prime. And from this post Positive divisors of $P(x,n)=1+x+x^2+ \cdots + x^n$ that are congruent to $1$ modulo $x$, if $ux+1 \mid n$ then $ux+1 \mid \frac{ u^{p}+1} {u+1}$. Perhaps these observations might be useful in proving Conjecture 1 and establishing $x_\text{max}$.

Here's a proof of Conjecture 1 for the case $p=5$. The proof depends on the truth of the following overwhelmingly true unproven result : Let \begin{equation} P(x) = x^4 +x^3 +x^2 +x+1. \end{equation} Then all positive integers $x$ such that $P(x) $ has a proper divisor congruent to 1 modulo $x$ are given as follows : Let $r(m)=m^2 +m-1$ and $q(m) = (r(m) +2)^2 - 2$, where $m$ is a positive integer. For a particular positive integer $m$, define sequences $A_n, B_n$ as follows: $A_1 = m^3+2m^2 +2m$ , $A_2 = q(m)A_1+r(m)$, $A_n = q(m)A_{n-1}-A_{n-2}+r(m)$ $B_1 = m^5+2m^4 +3m^3 + 3m^2 +m$ , $B_2 = q(m)B_1+r(m)-m$, $B_n = q(m)B_{n-1}-B_{n-2}+r(m)$ Then all positive integers $x$ such that $P(x) $ has a proper divisor congruent to 1 modulo $x$ are given by $x=A_n$ and $x=B_n$, $n\ge 1$, $m \ge 1$. It can be shown by induction that $A_n$ and $B_n$ are always composite except when $m=1$ and $n=1$ in which case $x=A_1=5$ is prime. If the unproven result here is true (very likely the case), then Conjecture 1 is settled for the case $p=5$. (Having no proper divisor congruent to 1 modulo $x$ for all primes $x>5 $ implies $x \nmid \phi(P(x)) $ for all composites $P(x) $, $x >5 $ prime) There's no reason to believe that the case $p=5$ is special. Conjecture 1 is most likely true for all odd primes. ADDED (Compositeness of $A_n$ and $B_n$) For $n \equiv 0, 1 \ (\mathrm{mod} \ 3)$, it can be shown by induction that $m$ divides $A_n$ and $B_n$. And when $n \equiv 2 \ (\mathrm{mod \ 3})$, one can prove that $m+1$ divides $A_n$ and $B_n$. Therefore it's clear that $A_n$ and $B_n$ are composite if $m>1$. The remaining case $m=1$ is interesting. When $m=1$, $A_n$ and $B_n$ are a product of two sequences i.e $A_n = f_n\cdot g_n$ where $f_1=1 , f_2 = 3, f_n = 3f_{n-1}-f_{n-2}$ and $g_1=5 , g_2 = 12, g_n = 3g_{n-1}-g_{n-2}$. Similarly $B_n = f_n\cdot g_n$ but with initial values changed; $f_1 = 2, f_2= 5$ and $g_1 = 5, g_2 = 14$ So $x = A_1 = 5$ is the only prime $x$ such that $P(x) =x^4 +x^3 +x^2 +x+1 $ contains a proper divisor congruent to $1$ modulo $x$

$2n \times 2n$ matrices with entries in $\{1, 0, -1\}$ with exactly $n$ zeroes in each row and each column with orthogonal rows and orthogonal columns I am interested in answering the following question: Question For a given $n$, does there exist a $2n \times 2n$ matrix with entries in $\{1, 0, -1\}$ having orthogonal rows and columns with exactly $n$ zeroes in each row and column? Conjectures * *For $n=2^k$, $k\ge0$ such a matrix always exists. *For $n=3$ such a matrix does not exist. *For $n=5$ such a matrix exists, for example: $$ \left( \begin{array}{cccccccccc} 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & -1 & 0 & -1 & -1 & -1 & 0 \\ 1 & 0 & 0 & -1 & 0 & 0 & -1 & 1 & 1 & 0 \\ 0 & -1 & -1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 \\ 1 & 0 & -1 & 0 & 0 & -1 & 1 & 0 & 0 & -1 \\ 1 & -1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & -1 & 1 & -1 & 0 & -1 & 0 & 1 \\ 0 & -1 & 1 & 0 & 0 & 0 & 0 & -1 & 1 & -1 \\ 0 & -1 & 1 & 0 & 0 & -1 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & -1 & 1 & 1 & 0 & 0 & -1 & -1 \\ \end{array} \right) $$ *For $n=7$ such a matrix does not exist.

There is no such matrix if $n\equiv 3\pmod 4$. Suppose otherwise. Each column represents a vector of length $\sqrt n$. Since those vectors are pairwise orthogonal, their sum is a vector whose scalar square is $2n^2$. On the other hand, the sum of all columns has odd entries, so their squares are all congruent to $1$ modulo $8$. Hence the sum of those squares is congruent to $2n$ modulo $8$. Hence we should have $2n^2\equiv 2n\pmod 8$, or $8\mid 2n(n-1)$ which does not hold.