Prove that the equation $y^2 = x^5 + 1$ has exactly p solutions modulo p.

Question

Condition on p: $ p \ncong 1(mod5) $

I understand that I'm trying to count the number of times that $x^5 + 1$ is a quadratic residue modulo p. I just can't quite figure out exactly why the number of solutions modulo p is equal to p itself.

Thanks in advance.

Answer 1

Since $(\mathbb{Z}/p\mathbb{Z})^{\times} \cong \mathbb{Z}/(p-1)\mathbb{Z}$ and $p-1$ is not divisible by $5$, exponentiating $x \mapsto x^5$ (and therefore $x \mapsto x^5+1$) is a bijection.