An asymptotic from below to number of solutions to $xyz + x + y = n$ Let $n$ be a positive integer. One can show (not that easy, but still via elementary methods) that the number of triples $(x,y,z)$ of positive integers satisfying $xyz + x + y = n$ is $O(n^{\frac{1}{3}+\varepsilon})$ for any $\varepsilon &gt; 0$. (Use that $xz+1$ divides $n-x$, take without loss of generality $x&lt; n^{\frac{1}{3}}$ or $z&lt;n^{\frac{1}{3}}$, etc.)

Hence I was wondering - is it also true that this number is at least $Cn^{\frac{1}{3}}$ for some constant $C&gt;0$?
No. I'll show there are arbitrarily large $n$ with at most $c(\log n)^2$ solutions.

Let $s(n)$ be the number of solutions $(x, y, z)$ of $xyz + x + y = n$, and $t(n)$ the number of solutions of $xyz \leq n$, so we have $t(n) \geq \sum_{k=1}^n s(k)$, since $xyz + x + y \leq n$ implies $xyz \leq n$. Note that for given $x, y$, the number of $z$ with $xyz \leq n$ is $\lfloor n/xy \rfloor$, hence
$$t(n) = \sum_{1 \leq x, y \leq n} \left \lfloor \frac{n}{xy} \right\rfloor \leq \sum_{1 \leq x, y \leq n} \frac{n}{xy} = nH_n^2 \leq 2n(\log n)^2$$
holds for sufficiently large $n$. If we also had $s(n) &gt; 16(\log n)^2$ for large $n$, this would mean 
$$t(n) \geq \sum_{k=\lceil n/2 \rceil}^n s(k) &gt; \sum_{k=\lceil n/2 \rceil}^n 16(\log k)^2 \geq (n/2) 16(\log (n/2))^2 \geq 2n(\log n)^2$$
for sufficiently large $n$, a contradiction, so there are arbitrarily large $n$ with $s(n) \leq 16(\log n)^2$.
xyz+x+y=n.  ----- (1)

Solution is,

(x,y,z,n)=((p-38),(p-11),(1),(p^2-47p+369))  

for  p>38

example, p=40, (x,y,z,n)=(2,29,1,89)

since 'p' can take many value's, equation (1) has infinite number of solutions.