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?
 A: Similar to finishing
Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers?
where I had an acceptable bound but needed help from Gerry Myerson to improve to the sharp bound.
We have $$ (xz+1)(yz+1) = a z^3 + 1  $$
This becomes
$$ a z^3 - xyz^2 - (x+y)z=0$$
or
$$ a z^2 - xyz - (x+y) = 0 $$
We get
$$    z = \frac{ xy + \sqrt{ x^2 y^2 + 4a(x+y) } }{2a}   $$
Let me also record
$$\color{fuchsia}{ z(az-xy) = x+y }$$
which follows directly from $ a z^2 - xyz - (x+y) = 0 $
Note also the simple
$$\color{fuchsia}{ z \leq x+y }$$
It is necessary to have square discriminant to get a rational value for $z,$   take
$$  w^2 =  x^2 y^2 + 4a(x+y) $$
We have $$ w > xy  $$  and
$$  w \equiv xy \pmod 2. $$
Therefore we can define an integer $t,$   when it all works, with
$$ w = xy+2t $$
Now $$ z = \frac{xy+w}{2a} = \frac{xy+xy+2t}{2a} = \frac{2xy+2t}{2a} = \frac{xy+t}{a}  $$
$$ z =  \frac{xy+t}{a}  $$
There are always three flavors for any $a$
$$ t=a-1 \; , \; y = 1 \; , \; x = a^2 - 3a +1 \; , \; z = a-2 $$
$$ t=1 \; , \; y = 2a-1 \; , \; x = 2a +1 \; , \; z = 4a $$
$$ t=1 \; , \; y = a+1 \; , \; x = a^2 +a -1 \; , \; z = a^2+2a $$
From
$$ x^2 y^2 +4a(x+y) = (xy+2t)^2 $$
we get  $$ t xy - ax -ay + t^2 = 0,  $$
$$ t^2 xy - tax -tay + t^3 = 0,  $$
$$  \color{red}{(tx-a)(ty-a) = a^2 - t^3}  $$
IF $a > 1$  and $t = a + \delta$  with $\delta \geq 0,$  we find
$$ ((a+\delta)x-a)((a+\delta)y-a) = a^2 - (a+\delta)^3  < 0 $$
since $a>1.$
However, the left hand side is non-negative, which is a contradiction.
$$ \color{red}{ t \leq a-1} $$
I will fill in the (lengthy) details in a bit.
I always have $x \geq y \geq 1$
IF $$ \color{blue}{  a^{2/3} < t \leq a-1} $$
we get
$$  (tx-a) (a-ty) = t^3 -a^2 > 0 $$
so $a-ty >0,$ $ty - a < 0,$
$$ ty < a  $$
$$ y < \frac{a}{t} < a^{1/3} $$
$$ a - ty \geq 1  $$
$$ tx-a \leq t^3 - a^2 $$
$$ tx \leq t^3 - (a^2 - a)$$
$$  x \leq t^2 - \frac{a^2 - a}{t} $$
DETAIL: As $t$ increases, $t^2$ increases, while $\frac{1}{t}$ decreases. Then $\frac{-1}{t}$ increases. We have $a \geq 2$  so that $a^2 - a > 0,$ so that
$\frac{-(a^2-a)}{t}$ increases. Together, $t^2 - \frac{a^2 - a}{t}$ increases and takes its maximum value at $t=a-1,$ that being $ a^2 - 3a + 1.$
Thus
$$ \color{magenta}{x \leq a^2 - 3a + 1}$$
$$xy + t < a^{7/3} -3a^{4/3} + a + a^{1/3} -1   $$
$$ z < a^{4/3} -3a^{1/3} + 1 + a^{-2/3} -\frac{1}{a}   $$
$$ \color{red}{ z < a^{4/3} }  $$ when  $a^{2/3} < t \leq a-1$
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I always have $x \geq y \geq 1$
IF $$ \color{blue}{1 \leq t <  a^{2/3} }$$
$$  (tx-a) (ty-a) = a^2 - t^3 > 0$$
$$  (tx-a) \leq a^2 -t^3$$
$$  tx  \leq a^2 + a - t^3 < a^2 + a$$
$$  x  \leq \frac{a^2 + a}{t}  $$
Meanwhile
$$ t^2 xy - ta(x+y)= -t^3 < 0 $$
$$txy < ta(x+y) \leq 2ax  $$
$$  ty < 2a
$$ y < \frac{2a}{t}   $$
Together
$$ xy < \frac{2 a^3 + 2a^2}{t^2}   $$
$$ xy + t <  \frac{2 a^3 + 2a^2}{t^2} + a^{2/3} $$
$$  z <  \frac{2 a^2 + 2a}{t^2} + \frac{1}{a^{1/3}} $$
IF $t \geq 2$
then $z < \frac{a^2 + a}{2}$
IF  $t=1$ we  have $$  (x-a)(y-a) = a^2 - 1 > 0   $$
If $x>a$ then $y>a.$      Then $y-a \geq 1$ and $x-a \leq a^2 - 1$
When $t=1$ we have $x \leq a^2 + a - 1.$
In general, when we have real $p \geq 1, q \geq 1,$ and $pq=c,$
the maximum of $p+q$ occurs when $p=1$ and $q=c$
so that $p+q \leq 1+c$
With  $  (x-a)(y-a) = a^2 - 1    $ we get $x-a+y-a \leq a^2.$ Thus
$$   x+y \leq a^2 + 2a$$
With $t=1,$ we know $z = x+y.$  With $t=1$
$$ \color{red}{z \leq a^2 + 2a } $$
DETAIL $$\color{fuchsia}{ z(az-xy) = x+y }$$
and $$ z =  \frac{xy+t}{a}  $$
so that when $t=1,$ we get $az = xy+1$  or $az-xy = 1,$ so that $z(az-xy) = x+y$ tells us $z=x+y,$ when $t=1$
A: Let $a$ be a positive integer that is not a cube, and let $x$, $y$ and $z$ be positive integers such that
$$(xz+1)(yz+1)=az^3+1.$$
Expanding the left hand side and rearranging a bit then shows that
$$az^2-xyz-(x+y)=0,\tag{1}$$
so $z$ is an integral root of a quadratic with discriminant $x^2y^2+4a(x+y)$. In particular this discriminant is a perfect square, so there exists a positive integer $v$ such that
$$x^2y^2+4a(x+y)=(xy+2v)^2,$$
and with a bit of rearranging we find the curious identity
$$(a-xv)(a-yv)=a^2-v^3.$$
We see that $v<a$ as otherwise the right hand side is negative, whereas the left hand side is not. Applying the quadratic formula to $(1)$ shows that
$$z=\frac{xy+\sqrt{x^2y^2+4a(x+y)}}{2a}=\frac{xy+(xy+2v)}{2a}=\frac{xy+v}{a},$$
where we have the $+$-sign because $z$ is positive. It follows that
$$z<\frac{xy}{a}+1,$$
so now to prove that $z<a^2+2a$ it suffices to show that $xy<a(a+1)^2$.
A: 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 , $z \le x+y < t^2 \le (a-1)^2 < a^2 +2a$
