<s>[INST] Prove that the diophantine equation $(xz+1)(yz+1)=az^{3} +1$ has no solutions in positive integers $x, y, z$ with $z>a^{2} +2a$. Let $a$ be a positive integer that is not a perfect cube. From experimental data, it appears all solutions to $(xz+1)(yz+1)=az^{3} +1$ in positive integers $x, y, z$ occur when $z \le a^{2} +2a$ i.e it appears there are no solutions in $x, y,z$ with $z> a^{2} +2a$. Can this observation be proved?
To motivate the question, we shall prove that on the contrary if $a$ is a perfect cube, there are infinitely many positive integer solutions in $x, y, z$.
Proof.
Let $a=m^{3} $ for some integer $m$. Using the identity $n^{3} +1 =(n+1)(n^{2}-n+1)$, we see that $az^{3} +1=(mz)^{3} +1= (mz+1)((mz)^{2}-mz+1) $.
A family of solutions is then given by $x=m$, $y=m^{2}z - m$  where $z$ takes on any positive integer.
How do I go about proving the striking observation: There are no positive integer solutions  $x, y, z$ with $z>a^{2} +2a$ when the integer $a$ is not a perfect cube? Is there any counterexample?
 [/INST] A concise Proof based on an earlier proof:
The given equation $(xz+1)(yz+1)=az^3+1$ can be rewritten as $az^2-xyz-(x+y)=0$. We shall show that for any solution $(x,y,z)$, we have $z \le a^2+2a. \ $
Note that $z \ | \ x+y$, therefore $z \le x+y. \ $  Treating $x, y$ as constants, the only positive solution for $z \ $ is   \begin{equation} z = \frac{xy+\sqrt{x^2y^2+4a(x+y)}} {2a}
\end{equation} In order for $z$ to be rational, the discriminant must be a perfect square. Therefore $w^2 = x^2y^2+4a(x+y)$. We see that $w > xy$ and $w \equiv xy \ ( $mod$ \ 2)$. We can write $w = xy + 2t$, $t > 0$.
Substituting $w$ above,  $(xy+2t)^2 = x^2y^2+4a(x+y)$. Expanding and simplifying, $txy - ax - ay +t^2 = 0$. Multiplying through by $t$ and factoring, $(tx-a)(ty-a)=a^2 - t^3$. We must have $t \le a-1$ otherwise $RHS<0$ and $LHS \ge 0$. Because $a$ is not a perfect cube, $a^2 - t^3 \not = 0$. The remainder of the proof utilizes the result: If $ab = c \  $ where $a,b, c \not = 0$ are integers then $a+b \le c+1$ if $c>0$ and $a+b \le -(c+1)$ if $c < 0$.
We now consider two cases:
Case $1: \ $ $ a^2 - t^3 >0 \ ;$
Using the result above on the factored equation, we have $(tx-a)+(ty-a) \le a^2 - t^3+1 \le a^2$.
Hence, $z \le x+y \le (a^2+2a)/t \le a^2+2a \\$.
Case $2: \ $ $ a^2 - t^3 < 0 \ ;$
As in case $1$, we have $(tx-a)+(ty-a) \le t^3 - a^2-1 \ $,  $x+y \le t^2 - (a^2-2a+1)/t < t^2$.  Hence , $z \le x+y < t^2 \le (a-1)^2 < a^2 +2a$
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