3 *. Find all pairs of natural numbers $x, y$ such that $x^{3}+y$ and $y^{3}+x$ are divisible by $x^{2}+y^{2}$.
3. Let's first prove that $x$ and $y$ are coprime. Assume the opposite. Then $x$ and $y$ are divisible by some prime number $p$. Let $p$ enter the prime factorizations of $x$ and $y$ with powers $a \geqslant 1$ and $b \geqslant 1$ respectively (see fact 10). Without loss of generality, we can assume that $a \geqslant b$. Then the maximum power of $p$ that divides $x^{3}+y$ is $b$ (since $x^{3}$ is divisible by $p^{3 a}$ and therefore by $p^{b+1}$, but $y$ is divisible by $p^{b}$ and not by $p^{b+1}$). On the other hand, $x^{2}+y^{2}$ is divisible by $p^{2 b}$, so $x^{3}+y$ cannot be divisible by $x^{2}+y^{2}$. This contradiction shows that $x$ and $y$ are coprime (see fact 5).

Next, by the condition, $x\left(x^{2}+y^{2}\right)-\left(x^{3}+y\right)=y(x y-1)$ must be divisible by $x^{2}+y^{2}$. Note that $y$ and $x^{2}+y^{2}$ cannot have a common divisor greater than 1 (since $x$ and $y$ are coprime), so $x y-1$ must be divisible by $x^{2}+y^{2}$ (see fact 9). But if we assume that $x y-1>0$, this is impossible, since $x^{2}+y^{2} \geqslant 2 x y>x y-1$. Therefore, $x y=1$, which means $x=y=1$.