Find all the integers pairs $(x, y)$ which satisfy the equation

$$
x^{5}-y^{5}=16 x y
$$
If one of $x, y$ is 0, the other has to be 0 too, and $(x, y)=(0,0)$ is one solution. If $x y \neq 0$, let $d=\operatorname{gcd}(x, y)$ and we write $x=d a, y=d b, a, b \in \mathbb{Z}$ with $(a, b)=1$. Then, the given equation is transformed into

$$
d^{3} a^{5}-d^{3} b^{5}=16 a b
$$

So, by the above equation, we conclude that $a \mid d^{3} b^{5}$ and thus $a \mid d^{3}$. Similarly $b \mid d^{3}$. Since $(a, b)=1$, we get that $a b \mid d^{3}$, so we can write $d^{3}=a b r$ with $r \in \mathbb{Z}$. Then, equation (1) becomes

$$
\begin{aligned}
a b r a^{5}-a b r b^{5} & =16 a b \Rightarrow \\
r\left(a^{5}-b^{5}\right) & =16
\end{aligned}
$$

Therefore, the difference $a^{5}-b^{5}$ must divide 16. Therefore, the difference $a^{5}-b^{5}$ must divide 16. This means that

$$
a^{5}-b^{5}= \pm 1, \pm 2, \pm 4, \pm 8, \pm 16
$$

The smaller values of $\left|a^{5}-b^{5}\right|$ are 1 or 2. Indeed, if $\left|a^{5}-b^{5}\right|=1$ then $a= \pm 1$ and $b=0$ or $a=0$ and $b= \pm 1$, a contradiction. If $\left|a^{5}-b^{5}\right|=2$, then $a=1$ and $b=-1$ or $a=-1$ and $b=1$. Then $r=-8$, and $d^{3}=-8$ or $d=-2$. Therefore, $(x, y)=(-2,2)$. If $\left|a^{5}-b^{5}\right|>2$ then, without loss of generality, let $a>b$ and $a \geq 2$. Putting $a=x+1$ with $x \geq 1$, we have

$$
\begin{aligned}
\left|a^{5}-b^{5}\right| & =\left|(x+1)^{5}-b^{5}\right| \\
& \geq\left|(x+1)^{5}-x^{5}\right| \\
& =\left|5 x^{4}+10 x^{3}+10 x^{2}+5 x+1\right| \geq 31
\end{aligned}
$$

which is impossible. Thus, the only solutions are $(x, y)=(0,0)$ or $(-2,2)$.