5. Find all positive integers $x, y$ such that $\frac{x^{3}+y^{3}-x^{2} y^{2}}{(x+y)^{2}}$ is a non-negative integer.
5. (1) When $x=y$, the original expression $=\frac{2 x-x^{2}}{4} \geqslant 0 \Rightarrow 0 \leqslant x \leqslant 2 \Rightarrow x=y=2$.
(2) By symmetry, assume $x \geqslant y$, when $x>y$, according to the problem, the original expression $=\frac{x^{2}-x y+y^{2}}{x+y}-\frac{x^{2} y^{2}}{(x+y)^{2}}=x+y-\frac{3 x y}{x+y}-\frac{x^{2} y^{2}}{(x+y)^{2}}$ is an integer, then $\frac{3 x y}{x+y}+\frac{x^{2} y^{2}}{(x+y)^{2}}$ is an integer. Let $\frac{x y}{x+y}=\frac{p}{q},(p, q)=1$, then $\frac{3 p}{q}+\frac{p^{2}}{q^{2}}$ is an integer, then $3 p+\frac{p^{2}}{q}$ is an integer, so $q=1$, thus $y-\frac{x y}{x+y}=\frac{y^{2}}{x+y}$ is an integer, then $y^{2} \geqslant x+y$, and $x^{3}+y^{3} \geqslant x^{2} y^{2}$, so $y^{3} \geqslant x^{2} y^{2}-x^{3}=x^{2}\left(y^{2}-x\right) \geqslant x^{2} y>y^{3}$, contradiction. Therefore, $x>y$ is impossible. In conclusion: $x=y=2$.