An asymptotic from below to number of solutions to $xyz + x + y = n$

Question

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 $C>0$?

Answer 1

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 $s(n) \leq 16(\log n)^2$.

Answer 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.

Answer 3

$xyz+x+y=n \overset{x\to u+v\\y\to u-v}{\implies} (u z + 1)^2 - (vz)^2 = nz+1$

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

Answer 4

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 $d_a(n)=\sum_{d|n,d\equiv1(a)}1$.