8. (12) Find all natural numbers $n$ that have a divisor $d$ such that $n^{2}+d^{2}$ is divisible by $n d+1$.

## Senior League
8. Answer: $n=k^{3}, k \in \mathbb{N}$.

For such $n$, we can take $d=k$. We will prove that $n=d^{3}$. Note that if $n^{2}+d^{2}=(n d+1) m$, then $m-d^{2}$ is divisible by $n$, from which either $m \geqslant n+d^{2}$, but then $n^{2}+d^{2} \geqslant(n d+1)\left(n+d^{2}\right)$, which is false, or $m=d^{2}$, in which case $n=d^{3}$.

## Senior League