1.3. Find all pairs of integers $(x, y)$ satisfying $x^{5}+y^{5}=(x+y)^{3}$.
1.3 The integer pair $(x, y)$ is a solution of the given equation if and only if $x+y=0$ or $(x, y)=(0 \pm 1),( \pm 1,0)$, or $\pm(2,2)$. Clearly, if $x+y=0$, then $(x, y)$ is a solution. Assume now that $(x, y)$ is a solution with $x+y \neq 0$.
We first show that $x y \geq 0$. Dividing both sides of $x^{5}+y^{5}=(x+y)^{3}$ by $x+y$ yields
$$
x^{4}-x^{3} y+x^{2} y^{2}-x y^{3}+y^{4}=(x+y)^{2} \text {. }
$$

This is equivalent to
$$
\left(x^{2}+y^{2}\right)^{2}+x^{2} y^{2}=(x+y)^{2}(x y+1),
$$
and it follows that $x y \geq 0$.
Next we show that $|x+y| \leq 4$. The convexity of the function $f(t)=t^{5}$ on $[0, \infty)$ implies that for nonnegative $x$ and $y$,
$$
\frac{x^{5}+y^{5}}{2} \geq\left(\frac{x+y}{2}\right)^{5}, \text { or equivalently, } x^{5}+y^{5} \geq \frac{1}{16}(x+y)^{5} \text {. }
$$

If $x+y>4$, then $x^{5}+y^{5}>(x+y)^{3}$. Similarly, if $x$ and $y$ are both nonpositive with $x+y<-4$, then $x^{5}+y^{5}<(x+y)^{3}$.
Finally, examining the cases where $x y \geq 0$ and $|x+y|=1,2,3$ or 4 , we find the solutions $(x, y)=(0 \pm 1),( \pm 1,0)$, or $\pm(2,2)$.