TITLE: integer solutions for $x(x-1)+y(y-1)=xy$
QUESTION [7 upvotes]: I am in high school right now and I would like to learn how to approach this sort of problems. I think this is called a diophantine eqution. Thanks a bunch
This is what I deduces, just so you know the work done (though I doubt its useful)
$x|x(x-1)+y(y-1) \rightarrow x|y(y-1)$
$y|x(x-1)+y(y-1) \rightarrow y|x(x-1)$

REPLY [7 votes]: HINT:
On rearrangement, $$x^2-x(1+y)+y^2-y=0$$ which is a Quadratic equation in $x$ 
The discriminant is $(1+y)^2-4(y^2-y)=1-3y^2+6y=4-3(y-1)^2$
For real $x, 4-3(y-1)^2\ge0\implies (y-1)^2\le\frac43\implies -\frac2{\sqrt3}\le y-1\le \frac2{\sqrt3}$
As $\frac2{\sqrt3}<2, -2<y-1<2\iff-1<y<3$
As $y$ is integer, $0\le y\le2$
Test for each value