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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.

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    $\begingroup$ 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. $\endgroup$
    – Alexey Burdin
    Jun 23, 2020 at 9:28

2 Answers 2

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Hint. If $\gcd(p-1,5)=1$ then there are $x,y\in \mathbb{Z}$ such that $(p-1)x+5y=1$.

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$gcd(5,p-1)=1$ 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 $a^{p-1}=1$.

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    $\begingroup$ $a=a^{5u}a^{v(p-1)}$ would be more precise but doesn't matter much as $a^{p-1}\equiv 1\pmod{p}$ anyway.) $\endgroup$
    – Alexey Burdin
    Jun 23, 2020 at 9:37

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