Problem: Find all pairs of positive integers $(x,p)$
such that $p$ is a prime and $x^{p-1}$ is a divisor of $(p-1)^{x}+1$.

Solution: If $p = 2$ then $x \in \{1,2\}$,
and if $p = 3$ then $x \in \{1,3\}$, since this is IMO 1990/3.
Also, $x = 1$ gives a solution for any prime $p$.
We show that there are no other solutions.

Assume $x > 1$ and let $q$ be smallest prime divisor of $x$.
We have $q > 2$ since $(p-1)^x+1$ is odd.
Then
\[ (p-1)^x \equiv -1 \pmod q \implies (p-1)^{2x} \equiv 1 \pmod q \]
so the order of $p-1 \bmod q$ is even and divides $\gcd(q-1,2x) \le 2$.
This means that
\[ p-1 \equiv -1 \pmod q \implies p = q. \]

In other words $p \mid x$ and we get $x^{p-1} \mid (p-1)^{x}+1$.
By exponent lifting lemma, we now have
\[ 0 < (p-1) \nu_{p}(x) \le 1 + \nu_p(x). \]
This forces $p=3$,
which we already addressed.