Junior Balkan Mathematical Olympiad

Solution:
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 $d = \gcd(x, y)$ 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 $(-2, 2)$.