Exercise 8.1

Proof. Let $d=m\wedge (p-1)$. We prove that $N(x^m=a)=N(x^d=a)$ for all $d\in {𝔽}_p$.

$\text{∙}$ If $a=0$, $0$ is the only root of $x^m-a$ or $x^d-a$, so $N(x^m=a)=N(x^d=a)=1$.

$\text{∙}$ If $a\in {𝔽}_p^{\text{∗}}$ and $x^m=a$ has a solution, then we know from the demonstration of Proposition 4.2.1 that $N(x^m=a)=d$, and $N(x^d=a)=d\wedge (p-1)=d$, thus $N(x^m=a)=N(x^d=a)$.

$\text{∙}$ If $a\in {𝔽}_p^{\text{∗}}$ and $x^m=a$ has no solution, then (Prop. 4.2.1) $a^{(p-1)∕d}\ne 1$, thus $x^d=a$ has no solution : $N(x^m=a)=0=N(x^d=a)$.

Using Prop. 8.1.5, as $d\mid p-1$, we obtain

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