# Prove that the congruence $x^{5} \equiv a \pmod p$ has a solution for every integer $a$

Let $$p$$ be a prime such that $$5 \nmid p-1$$. Prove that the polynomial congruence $$x^{5} \equiv a \pmod p$$ has a solution for every integer $$a$$.

I struggle to solve the case where $$p \nmid a$$. I've thought about using the existence of primitive roots modulo $$p$$ to apply the theorem which holds that if $$g$$ is a primitive root modulo $$p$$, then the set $$\{g,g^{2},...,g^{p-1}\}$$ runs through all the invertible congruence classes modulo p, but I haven't have luck.

• Consider image of $\{1,\ldots,p-1\}$ when applying $f:\;x\to x^5$. If it's not $\{1,\ldots,p-1\}$, then, by the pigeonhole principle, there are some $a\ne b$ that $a^5\equiv b^5\pmod{p}$. Derive a contradiction. Commented Jun 23, 2020 at 9:28

Hint. If $$\gcd(p-1,5)=1$$ then there are $$x,y\in \mathbb{Z}$$ such that $$(p-1)x+5y=1$$.
$$gcd(5,p-1)=1$$ implies that there exists integers $$u,v$$ such that $$5u+v(p-1)=1$$. We deduce that $$a=a^{5u}a^{5(p-1)}$$ mod $$p$$ Little Fermat implies that $$a^{p-1}=1$$.
• $a=a^{5u}a^{v(p-1)}$ would be more precise but doesn't matter much as $a^{p-1}\equiv 1\pmod{p}$ anyway.) Commented Jun 23, 2020 at 9:37