11. Find the integer solutions of the indeterminate equation $x^{5}+y^{5}=(x+y)^{3}$.
11. Clearly, the integer pairs $(x, y)$ that satisfy $x+y=0$ are all solutions to the equation. Now consider the solutions to the equation when $x+y \neq 0$. In this case, dividing both sides by $x+y$ gives
$$x^{4}-x^{3} y+x^{2} y^{2}-x y^{3}+y^{4}=(x+y)^{2},$$

Rearranging and simplifying, we get
$$\left(x^{2}+y^{2}\right)^{2}+x^{2} y^{2}=(x+y)^{2}(x y+1)$$

This indicates that $x y+1>0$, i.e., $x y \geqslant 0$. If $x, y \geqslant 0$, then by the power mean inequality, we have
$$(x+y)^{3}=x^{5}+y^{5} \geqslant 2\left(\frac{x+y}{2}\right)^{5},$$

Thus, $x+y \leqslant 4$. If $x, y \leqslant 0$, similarly, we have $x+y \geqslant-4$. Therefore, we always have $x y \geqslant 0$ and $|x+y| \leqslant 4$. Discussing separately for $|x+y|=1,2,3,4$, we find that the solutions are $(x, y)=$ $(0, \pm 1),( \pm 1,0),(2,2),(-2,-2)$.

In summary, the integer solutions to the equation are $\{(t,-t) \mid t \in \mathbf{Z}\} \cup\{(0, \pm 1),( \pm 1,0)$, $(2,2),(-2,-2)\}$.