Let $p$ be prime and $n$ divide $p-1$. Let $gcd(a,p)=1$, and $a^{(p-1)/n} \equiv 1 \mod p$. Show that there are $n$ solutions to $x^n \equiv a \mod p$ for $x$ if $a$ is a constant.
For now I have come up with that there are $p-1$ solutions to $x^{p-1} \equiv 1 \mod p$
Let $g $ be a primitive root modulo $p$ and $u\in\Bbb Z $ such that $a=g^u $.
By assumption, there exists an integer $m $ such that $p-1=nm $.
From $a^m\equiv 1\pmod p $ follows $mu\equiv 0\pmod {mn} $, hence there exists an integer $v $ such that $u=vn $.
If $x=g^y $, then we have $x^n\equiv a\pmod p $ if and only if $yn\equiv u\pmod {mn} $, that's $y\equiv v\pmod m$. Consequently, $x^{v+km }$ for $0\leq k\leq n-1$ are the required $n $ solutions.