Solve $a^{2}+b^{2}=(a b)^{2}$
We will start by solving the more general problem $x+y=$ $xy$ for positive $x, y$. By symmetry, we can assume that $x \leq y$. If $x>2$ there is no solution. Indeed: $x+y \leq y+y=2y<xy$. It then suffices to treat the three cases $x=0,1,2$. For $x=0$ we deduce that $y=0$.

For $x=1$ this gives $y=y+1$; no solution.

For $x=2$ we have $y=2$.

We can return to the initial problem by looking for $x$ and $y$ to be squares. Then only the solution $x=0, y=0$ remains.