How to solve a bi-quadratic equation in two variables? The equation is of the form $$ x(x-1)y(y-1) \approx k $$
$x$ and $y$ both have to integral. 
Any heads up / hints as to how to approach this type of an equation.
Thank you. I'm new in here so please point out if I've made a mistake somewhere.
 A: We can solve this equation as follows, say, over a field $K$. Assume first that $k=0$. Then $x,y\in \{0,1\}$. So we have only finitely many solutions. Suppose now that $k\neq 0$ and $x\neq 0,1$. Then $y^2-y=\alpha$ with $\alpha=k/(x(x-1))$, with $x\neq 0,1$ arbitrary. This quadratic equation in $y$ can be solved, of course. In case that $K$ is infinite, we obtain infinitely many solutions.
The remaining cases work the same way.
Update: The question was changed to integral solutions. Then we would start with a prime factorization of $k$, to obtain all possible divisors of $k$. We could also solve the quadratic equation $y^2-y=\alpha$ for integral solutions, if we have chosen $x$ before ($x=2$ always works with $x(x-1)=2$ for $k$ even).
A: If $x$ is integral then $x(x-1)$ is even. Therefore $k$ should be a multiple of $4$. Let $k=4m$, then the equation becomes:
$$\frac{x(x-1)}{2}\cdot \frac{y(y-1)}{2} = m.$$
Hence, the equation has at least a solution iff $m$ is the product of two triangular numbers. 
It seems that there is no easy way for detecting these numbers: see the OEIS sequence A085780.
