Finding $(a, b, c)$ with $ab-c$, $bc-a$, and $ca-b$ being powers of $2$ This is a 2015 IMO problem.  It seems difficult to solve.


  Find all triples of positive integers $(a, b, c)$ such that each of the numbers $ab-c$, $bc-a$, and $ca-b$ is a power of $2$.


Four such triples are $(a,b,c)=(2,2,2),(2,3,2),(3,5,7),(2,6,11)$.
Suppose $a,b,c\in\mathbb{N}$ and $\alpha,\beta,\gamma\in\mathbb{N}_0$ are such that $bc-a=2^\alpha$, $ca-b=2^\beta$, and $ab-c=2^\gamma$. It is easily seen that either all of $a,b,c$ are odd, or at least two of them are even.  We claim that all solutions $(a,b,c)$ are permutations of $(2,2,2)$, $(2,2,3)$, $(2,6,11)$, and $(3,5,7)$.

If two of $a$, $b$, and $c$ are equal, say $a=b$, then we have $2^\alpha=bc-a=a(c-1)$, or $c-1=2^\nu$ and $a=2^{\alpha-\nu}$ for some $\nu\in\mathbb{N}_0$ such that $\nu\leq\alpha$.  Hence, $2^{2\alpha-2\nu}=a^2=ab=2^\gamma+c=2^\gamma+2^\nu+1$, which means $a=2$ and $\{\gamma,\nu\}=\{0,1\}$.  If $\nu=0$, then $(a,b,c)=(2,2,2)$; if $\nu=1$, then $(a,b,c)=(2,2,3)$.  From now on, we assume that $a$, $b$, and $c$ are pairwise distinct.  Without loss of generality, assume that $a&lt;b&lt;c$.  It follows immediately that $\alpha&gt;\beta&gt;\gamma$.

If $\gamma=0$, we have $ab-c=2^\gamma=1$, or $c=ab-1$.  Plugging this in, we have $$2^\alpha=bc-a=b(ab-1)-a=ab^2-(a+b)$$ and $$2^\beta=ca-b=(ab-1)a-b=a^2b-(a+b)\,.$$  If both $a$ and $b$ are odd, then $c$ is even, and we have a contradiction.  If $a$ or $b$ is even, then both are even.  Suppose that $k$ and $l$ are the largest positive integers such that $2^k\mid a$ and $2^l\mid b$.  We claim that $k=l=1$.  If $k\neq l$, then from $2^\alpha-2^\beta=ab(a-b)$, we obtain $\beta=k+l+\min\{k,l\}$.  Nonetheless, from $2^\alpha+2^\beta=(a+b)(ab-2)$, we get $\beta=1+\min\{k,l\}&lt;k+l+\min\{k,l\}$, which is a contradiction.  Thus, $k=l$.  Write $a=2^ks$ and $b=2^kt$, where $s$ and $t$ are odd integers.  We see that $$2^\alpha+2^\beta=(a+b)(ab-2)=2^{k+1}(s+t)\left(2^{2k-1}st-1\right)$$ and $$2^\alpha-2^\beta=2^{3k+1}st\left(\frac{s-t}{2}\right)\,.$$  This means $\beta\geq 2^{3k+1}$.  That is, $2^{2k}\mid s+t$, whence $4\mid s+t$.  This would then mean that $\frac{s-t}{2}$ is odd.  Hence, $\beta=3k+1$.  Now, we have $$2^{3k+1}=2^\beta=a^2b-(a+b)=2^{3k}s^2t-2^k(s+t)\geq2^k\left(2^{2k}-2\right)s^2t\,.$$  If $k&gt;1$, then $s^2t\leq 2$, implying that $s=t=1$, but this violates the assumption that $a&lt;b$.  Consequently, $k=1$, so $s^2t\leq 4$, which gives $s=1$ and $t=3$.  Thence, $(a,b,c)=(2,6,11)$.

From now on, we assume that $\gamma&gt;0$.  If $a$ is even, then $$2^\beta-2^\gamma=(ca-b)-(ab-c)=(c-b)(a+1)$$ leads to $2^\gamma\mid c-b$, and $$2^\beta+2^\gamma=(ca-b)+(ab-c)=(c+b)(a-1)$$ leads to $2^\gamma \mid c+b$.  Ergo, $2^\gamma\mid (c+b)-(c-b)=2b$ and $2^\gamma\mid (c+b)+(c-b)=2c$, or $2^{\gamma-1}\mid b$ and $2^{\gamma-1}\mid c$.  Now, as $2^\gamma=ab+c$, $2\mid a$, and $2^{\gamma-1}\mid b$, we conclude that $2^\gamma\mid c$.  Therefore, $2^\gamma\mid b$ as well, whence $c=2^\gamma w$ for some odd $w\in\mathbb{N}$.  Similarly, $2^\beta\mid b$ and $2^\beta\mid a$ (by considering $2^\alpha+2^\beta$ and $2^\alpha-2^\beta$), with $b=2^\beta v$ and $a=2^\delta u$ for some odd $v,u\in\mathbb{N}$ and for some $\delta\in\mathbb{N}$ such that $\delta&gt;\beta$.  If $b$ or $c$ is even, we can use a similar method to achieve the same result.

Consequently, in the case where $\gamma&gt;0$, if any of $a$, $b$, and $c$ is even, then $a=2^\delta u$, $b=2^\beta v$, and $c=2^\gamma w$ for some odd $u,v,w\in\mathbb{N}$ and $\delta \in \mathbb{N}$ with $\delta&gt;\beta$.  From $2^\beta=ca-b$ and $2^\gamma=ab-c$, we get $$2^\beta=\left(ab-2^\gamma\right)a-b=a^2b-2^\gamma a-b=2^{2\delta+\beta}u^2v-2^{\gamma+\delta} u - 2^\beta v\geq \left(2^{2\delta+\beta}-2^{\gamma+\delta}-2^\beta\right)u^2v\,.$$
Therefore, $2\leq 2^{2\delta-1}u^2v\leq \left(2^{2\delta}-2^{\gamma+\delta-\beta}-1\right)u^2v \leq 1$, which is absurd.  That is, $a$, $b$, and $c$ must all be odd.

Now, $\gamma&gt;0$ and all $a$, $b$, and $c$ are odd.  From the equality $$2^{2\beta}-2^{2\gamma}=\left(2^\beta-2^\gamma\right)\left(2^\beta+2^\gamma\right)=\big((c-b)(a+1)\big)\big((c+b)(a-1)\big)=\left(c^2-b^2\right)\left(a^2-1\right)\,,$$
we get $2^6\mid 2^{2\beta}-2^{2\gamma}$, making $\gamma \geq 3$.  Plugging $c=ab-2^\gamma$ into $bc-a=2^\alpha$ and $ca-b=2^\beta$ to get $\left(\frac{b^2-1}{2^\gamma}\right)a-b=2^{\alpha-\gamma}$ and $\left(\frac{a^2-1}{2^\gamma}\right)b-a=2^{\beta-\gamma}$.  Since $\alpha&gt;\beta&gt;\gamma$, we deduce that $\frac{b^2-1}{2^\gamma}$ and $\frac{a^2-1}{2^\gamma}$ are odd integers.  Hence, there exist $i,j\in\{-1,+1\}$ and odd $u,v\in\mathbb{N}$ such that $a=2^{\gamma-1}u+i$ and $b=2^{\gamma-1}v+j$.  As $c=ab-2^\gamma$, $c=2^\gamma w+k$ for some $w\in\mathbb{N}$ and $k:=ij$.  From $$2^{\gamma}w\left(2^{\gamma-1}(iu+jv)+2\right)=(c-k)(ia+jb)=(ca-b)i+(bc-a)j=2^\beta i+2^\alpha j\,,$$ 
we conclude that $2^{\beta-\gamma-1}\mid w$ but $2^{\beta-\gamma}\nmid w$, or $c=2^{\beta-1}z+k$ for some odd $z\in\mathbb{N}$.  Moreover, $$2^\beta=ca-b&gt; ca-c=c(a-1)=\left(2^{\beta-1}z+k\right)(a-1) \geq \left(2^{\beta-1}-1\right)(a-1)$$ yields $a-1\leq 2$.  Therefore, $a=3$, and this forces $z=1$ and $k=-1$.  That is, $c=2^{\beta-1}-1$; i.e.,
$$2^\beta=ca-b=3\left(2^{\beta-1}-1\right)-b\,,$$
or $b=2^{\beta-1}-3$. Since $2^{\gamma-1}u+i=a$, we must have $\gamma=3$, $u=1$, and $i=-1$.  Finally, $2^\gamma=ab-c$ implies that $8=3\left(2^{\beta-1}-3\right)-\left(2^{\beta-1}-1\right)=2^\beta-8$.  That is, $\beta=4$, so $(a,b,c)=(3,5,7)$.
Without loss of generality: replace $b$ with $a+k1$ and $c$ with $a+ k2$

$(a, a+k1, a +k2) $

$ab- c = a2 + k1a -a -k2 $

$bc- a = a2 + k1k2 + (k1+k2 -1)a$

$ca- b = a2+ (k2-1)a - k1$

All these must by $2N$ for some $n$ in $I &gt; 0$