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 $\varepsilon > 0$. (Use that $xz+1$ divides $n-x$, take without loss of generality $x< n^{\frac{1}{3}}$ or $z<n^{\frac{1}{3}}$, etc.)
Hence I was wondering - is it also true that this number is at least $Cn^{\frac{1}{3}}$ for some constant $C>0$?
 A: 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 $s(n) > 16(\log n)^2$ for large $n$, this would mean 
$$t(n) \geq \sum_{k=\lceil n/2 \rceil}^n s(k) > \sum_{k=\lceil n/2 \rceil}^n 16(\log k)^2 \geq (n/2) 16(\log (n/2))^2 \geq 2n(\log n)^2$$
for sufficiently large $n$, a contradiction, so there are arbitrarily large $n$ with $s(n) \leq 16(\log n)^2$.
A: The equation is
$$
x+y+z x y=n\tag 1
$$
I will find the number of solutions of (1) given a positive integer $n$, when $x,y,z$ are positive integers. For this done we assume $z$ is a parameter and rewrite (1) in the form 
$$
nz+1=(xz+1)(yz+1)\tag 2
$$
Set $N=nz+1$ and $AB=N$. Then the number of solutions of (2) when $x,y\geq 0$, $z>1$ is 
$$
r^{*}(z,n)=\sum_{
\begin{array}{cc}
 A,B>0\\
 AB=nz+1\\
 A\equiv 1(z)\\
 B\equiv 1(z)
\end{array}
}1.\tag 3
$$ 
We assume $z\geq 2$. The case $z=1$ is easy (I will leave it). Hence equation $(1)$ have solutions when $x,y\geq1$ and $z\geq2$:
$$
r(n)=-2(n-1)+\sum^{n}_{k=2}r^{*}(k,n)=-2(n-1)+\sum^{n}_{k=2}\sum_{
\begin{array}{cc}
 0<d|(nk+1)\\
 d\equiv 1(k)
\end{array}
}1
$$ 
The term $-2(n-1)$ in $r(n)$ is for removing the zero solutions $x=0$ or $y=0$. Hence the number of solutions of (1) is
$$
r(n)=-2n+d(n+1)+\sum^{n}_{k=2}\sum_{
\begin{array}{cc}
 0<d|(nk+1)\\
 d\equiv 1(k)
\end{array}
}1
$$
where $d_a(n)=\sum_{d|n,d\equiv1(a)}1$.
A: 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.
