prove that $N(x^m=a)=N(x^d=a)$.

Question

Let $p>2$ be a prime, $m\in \mathbb N$, $d=$ GCD$(m,p-1)$.
Let $N(x^n=a)$ denote the number of solutions to the equation $x^n=a$ in $\mathbb F_p$.
prove that $N(x^m=a)=N(x^d=a)$.

I am familiar with the claim that if $n\mid p-1$ then $N(x^n=a)=\sum_{\chi\in S}\chi(a) $ where $S=\{\chi:\chi^n=\epsilon\}$, $\chi$ is a character on $\mathbb F_p$ and $\epsilon$ is the identity character.
I manged to show that $N(x^d=a)=N(x^{\alpha m}=a)$ where $d=\alpha m+\beta (p-1)$.
This question appeared in "A Classical Introduction to Modern Number Theory"

Answer 1

The special case $a=0$ is obvious, so let us assume $a \neq 0$. Recall that the multiplicative group of the field is cyclic and let $g$ be a generating element (a primitive root). Further let $k$ be an integer such that $g^k = a$.

We have $x^n = a$ if and only if $x = g^s$ with $sn \equiv k \pmod{p-1}$.

So you are trying to show that the number of solutions to

$$mX \equiv k \pmod{p-1}$$ is the same as that to $$dX \equiv k \pmod{p-1}$$ where $d = \gcd (m,p-1)$. Let $m = cd$. Then the first congruence becomes $$d(cX) \equiv k \pmod{p-1}$$

Since $c$ and $p-1$ are co-prime, "multiplication by $c$" is bijective and your claim follows.