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Let $p$ be a prime, $a$ and integer such that $(a,p)=1$ and $n$ a positive integer $(n,p-1)=1$. Prove that $x^n\equiv a$ mod $p$ has a unique solution.

So I know $n$ is odd since $p-1$ is even.

So I thought since $a$ is in the reduced residue system, then if $g$ is a primitive root, that $g^n\not\equiv 1$, so $(g^n,p)=1$ and I believe it's unique since the only other solution possible should be $-g$ but since $n$ is odd, $-g^n\equiv -a$. But I'm not sure that I can assume a primitive root exists.

AColoredReptile
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    here's a constructive proof: since $n$ is relatively prime to $p-1$, there is $m$ such that $nm\equiv1\bmod p-1$, and then $(x^n)^m\equiv x\equiv a^m\bmod p$ by Fermat's little theorem – J. W. Tanner Dec 11 '19 at 05:15
  • by the way, $\mathbb Z/p\mathbb Z$ always has a primitive root if $p$ is prime – J. W. Tanner Dec 11 '19 at 05:16
  • See the answers by Nicky and I in the linked dupe for the general idea. – Bill Dubuque Dec 11 '19 at 05:37
  • @J.W.Tanner Is my method not going to work? I get that $g$ generates the reduced residue system but I guess I need to prove any element in the reduced residue system is a unique $nth$ power of a unique primitive root? – AColoredReptile Dec 11 '19 at 05:58

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Note that $(n,p-1) = 1$. Therefore, there exist $X,Y$ such that $nX +(p-1)Y = 1$.

If $x^n \equiv y^n \mod p$, let $y^{-1}$ be the inverse of $y \mod p$. We get from commutativity that $(xy^{-1})^{n} \equiv 1 \mod p$.

We also have $x^{p-1} \equiv y^{p-1} \mod p$. Now, again taking the inverse we get $(xy^{-1})^{p-1} \equiv 1 \mod p$.

Try to show from here that $x \equiv y \mod p$.

Let $xy^{-1} = b$. Then, $b^{n} \equiv 1 \mod p$ and $b^{p-1} \equiv 1 \mod p$. Take the first one power $X$, the second one power $Y$ (where $X,Y$ can be negative integers, no problem, it is all defined nicely) and multiply to see that $b \equiv 1 \mod p$. Thus $x \equiv y \mod p$.

What this shows is that the map $x \to x^n$ from integers mod $p$ to integers mod $p$ is injective. Since the size of the sets is the same and finite, it is also surjective. Thus, we get that every for every $a \mod p$ there is unique $x$ such that $x^N \equiv a \mod p$.

Sarvesh Ravichandran Iyer
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  • Does my attempt using primitive roots work? Since in this case I don't need that $a$ is any integer only that it is in the reduced residue class. – AColoredReptile Dec 11 '19 at 06:26
  • Even if you assume existence of a primitive root, the fact that $(g^n , p ) \equiv 1$ does not imply that $g^n \equiv a \mod p$. Also, the only other solution being $-g$ is incorrect : since we are working modulo $p$, there could be problems more than sign that result in non-injectivity. The condition $(n,p-1) = 1$ is important for this reason. – Sarvesh Ravichandran Iyer Dec 11 '19 at 06:31