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 $C>0$?
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 $s(n) \leq 16(\log n)^2$.
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 $d_a(n)=\sum_{d|n,d\equiv1(a)}1$.
$xyz+x+y=n \overset{x\to u+v\\y\to u-v}{\implies} (u z + 1)^2 - (vz)^2 = nz+1$
gp-code for test first small $n,z$:
xyzn()=
{
for(n=1, 100,
k= 0;
for(z=1, 1000,
T= thue('x^2-1, n*z+1);
for(i=1, #T,
X= T[i][1]; Y= T[i][2];
if(X!=0, if(Y!=0,
v= Y/z;
if(v==floor(v),
u= (X-1)/z;
if(u==floor(u),
x= u+v; y= u-v;
if(x>0 && y>0,
\\ print("n = "n"; (x,y,z) = ("x","y","z")");
k++
)
)
)
))
)
);
if(k, print("n = "n"; #sol = "k));
)
};
Output:
n = 7; #sol = 2
n = 10; #sol = 2
n = 11; #sol = 2
n = 13; #sol = 2
n = 14; #sol = 2
n = 15; #sol = 2
n = 16; #sol = 4
n = 19; #sol = 4
n = 20; #sol = 2
n = 21; #sol = 2
n = 22; #sol = 6
n = 23; #sol = 4
n = 25; #sol = 2
n = 26; #sol = 4
n = 27; #sol = 2
n = 28; #sol = 4
n = 29; #sol = 2
n = 30; #sol = 2
n = 31; #sol = 8
n = 32; #sol = 4
n = 34; #sol = 6
n = 35; #sol = 2
n = 36; #sol = 4
n = 37; #sol = 4
n = 38; #sol = 6
n = 39; #sol = 4
n = 40; #sol = 4
n = 41; #sol = 2
n = 42; #sol = 2
n = 43; #sol = 6
n = 44; #sol = 6
n = 45; #sol = 2
n = 46; #sol = 10
n = 47; #sol = 6
n = 49; #sol = 2
n = 50; #sol = 4
n = 51; #sol = 4
n = 52; #sol = 8
n = 53; #sol = 4
n = 54; #sol = 4
n = 55; #sol = 8
n = 56; #sol = 8
n = 57; #sol = 2
n = 58; #sol = 8
n = 59; #sol = 4
n = 61; #sol = 6
n = 62; #sol = 8
n = 63; #sol = 4
n = 64; #sol = 10
n = 66; #sol = 6
n = 67; #sol = 6
n = 68; #sol = 6
n = 69; #sol = 2
n = 70; #sol = 6
n = 71; #sol = 10
n = 72; #sol = 4
n = 73; #sol = 6
n = 74; #sol = 6
n = 75; #sol = 2
n = 76; #sol = 12
n = 77; #sol = 2
n = 78; #sol = 6
n = 79; #sol = 10
n = 80; #sol = 6
n = 81; #sol = 2
n = 82; #sol = 12
n = 83; #sol = 6
n = 84; #sol = 2
n = 85; #sol = 6
n = 86; #sol = 10
n = 87; #sol = 4
n = 88; #sol = 4
n = 89; #sol = 2
n = 90; #sol = 4
n = 91; #sol = 10
n = 92; #sol = 12
n = 93; #sol = 4
n = 94; #sol = 14
n = 95; #sol = 8
n = 96; #sol = 4
n = 97; #sol = 2
n = 98; #sol = 6
n = 99; #sol = 4
n = 100; #sol = 10
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.
