Diophantine Equation $x^2y^2-4x-4y=z^2$ Is it true that the Diophantine equation 
$$
x^2y^2-4x-4y=z^2\; ; \; x,y>0
$$
has the only solutions $(x,y)=(2,2),(2,3),(1,5),(3,2),(5,1)$?
What is the general solution in all integers?
 A: Note that the perfect squares next to $x^2y^2$ are 
$$x^2y^2 \pm 2xy +1.$$
Therefore if $(x , y, z)$ with $x, y > 0$ is an integral point on $z^2 = x^2y^2 - 4(x + y)$, then 
$$-4(x + y) \leq 1 - 2xy.$$
It follows that any solution must be of the form $(1, y)$, $(2, y)$, $(x, 1)$, $(x, 2)$, $(x, 3)$, $(x, 4)$, $(x, 5)$ or $(x,6)$.
We can use a similar argument to see which of those forms yield solutions of $x^2y^2 - 4(x + y) = z^2$.
In the family $(1, y)$ we have
$$z^2 = (y - 2)^2 - 8$$
yielding only the point $(x, y) = (1, 5)$.
In the family $(2, y)$ we have
$$z^2 = (2y - 1)^2 - 9$$
yielding the points $(x, y) = (2, 2)$ and $(2,3)$.
In the families $(x, 1)$ and $(x ,2)$ we only have the points $(5, 1)$, $(2,2)$ and $(3, 2)$ by symmetry in the equation.
In the family $(x , 3)$ we have
$$z^2 = (3x - 1)^2 + 2x - 13$$
yielding the point $(x, y) = (2,3)$.
In the family $(x, 4)$ we have
$$z^2 = (4x - 1)^2 + 4x - 17$$
but there is no $x$ for which this can happen.
In the family $(x , 5)$ we have
$$z^2 = (5x - 1)^2 + 6x - 21$$
yielding the point $(x, y) = (1,5)$.
In the family $(x, 6)$ we have
$$z^2 = (6x - 1)^2 + 8x - 25$$
but there is no $x$ for which this can happen.
So the list of solutions in your question is indeed complete for $x,y > 0$.

Over all integers you have infinitely many solutions. Here's an outline of the (hopefully) complete scenario.
When $x, y < 0$, we can adapt the method above to obtain no solution $(x,y)$ besides the infinite families $(-1, -t, \pm(t - 2))$ and $(-t, -1, \pm(t - 2))$.
When $x = 0$ we have the integral points $(0, -t^2, 2t)$, $t \in \mathbf{Z}$. Similarly, when $y = 0$ we have the solutions $(-t^2, 0, 2t)$.
When $x$ and $y$ have opposite signs, we can regard this problem as trying to find the solutions to
$$z^2 = a^2b^2 -4a + 4b, \quad a, b > 0.$$
When $-4a + 4b = 0$, we obtain the family of solutions $(x, y, z) = (-t, t, t^2)$, $t \in \mathbf{Z}$.
When $-4a + 4b > 0$, we want $4b - 4a \geq 2ab + 1$ but that can never happen.
When $-4a + 4b < 0$, we want $4b - 4a \leq -2ab + 1$, in which case $b = 1$ and the equation becomes $z^2 = (a - 2)^2$. This corresponds to integral points $(x,y,z) = (2 + t, -1, t)$, $t \in \mathbf{Z}$, $t \geq 0$.

Summing up, these should be all solutions:


*

*when $x, y > 0$: here.

*$(t, -t, \pm t^2)$, for $t \in \mathbf{Z}$.

*$(-t^2, 0, 2t)$ for $t \in \mathbf{Z}$.

*$(0,-t^2, 2t)$ for $t \in \mathbf{Z}$.

*$(2 + t, -1, t)$ for $t \in \mathbf{Z}$.

*$(-1, 2 + t, t)$ for $t \in \mathbf{Z}$.

A: Write $(xy-z)(xy+z)=4(x+y)$.
For sufficiently large x,y, the LHS is larger.
Note that $xy$ and $z$ have the same parity.
Consider values for $xy-z=2,4,6$.
When $xy-z \geq 4$, this already implies $2xy-4 \leq x+y$.
It should be easy to get rid of $xy-z \geq 6$.
A: For equation: ${{x}^{2}}\,{{y}^{2}}-4y-4x={{z}^{2}}$
$$x=\frac{{{s}^{2}}+h}{hs-1},y=h,z=\frac{h\,{{s}^{2}}-2s-{{h}^{2}}}{hs-1}$$
