Menchikov A.B.

Find all pairs of natural numbers $a$ and $k$ such that for any natural number $n$, coprime with $a$, the number

$a^{k^{n}+1}-1$ is divisible by $n$.
If $a=1$, then $a^{k^{n}+1}-1=0$, and therefore, it is divisible by $n$.

Let $a \geq 2$. Take $n=a^{k}-1$, then $a^{k} \equiv 1(\bmod n)$, and consequently, $0 \equiv a^{k^{n}+1}-1 \equiv\left(a^{k}\right)^{k^{n-1}} \cdot a-1 \equiv a-1$ $\left(\bmod a^{k}-1\right)$.

This can only be true if $k=1$, but in this case, $a^{k^{n}+1}-1=a^{2}-1$ must be divisible by all $n$, which is impossible. Therefore, pairs where $a \geq 2$ are not suitable.

## Answer

$a=1, k-$ any natural number.