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I have been working through Charles Pinter's book on abstract algebra and have come into some confusion with one of the problems.

It is stated as follows:
Suppose $m$ has a primitive root, and let $n$ be relatively prime to $\phi(m)$ $(n>0)$.
Prove that if a is relatively prime (coprime) to $m$, then $x^n\equiv a(mod \ m)$ has a solution.

My proof attempt is then as follows:
Since $gcd(n,\phi(m))=1$, there are some $s,t \in \mathbb{Z}$ such that $s n + t \phi(m) = 1$.
We then have that $$a = a^1 = a^{s n + t \phi(m)} = a^{s n}a^{t \phi(m)} = (a^s)^n(a^{\phi(m)})^t\equiv(a^s)^n(mod \ m)$$ Since, by Euler's thoerem and the fact that $gcd(a,m)=1$, we have that $a^{\phi(m)}\equiv1(mod \ m)$ and so $(a^{\phi(m)})^t\equiv1^t\equiv1(mod \ m)$. Thus, it seems that $a^s$ is a valid solution.

My confusion arises from the fact that I have nowhere used that m has a primitive root. Have I made a mistake, or is the result still true without that assumption?

Bill Dubuque
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Flumpo
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    You are right. It holds for any $m$. If there were any concerns about the possibly non-integer values of $a^{s n}$ or$ a^{t \phi(m)}$ (as $s$ or $t$ may be negative), consider that in your last display-line, it is all in the $group$ of equivalence classes $\pmod m$ of integers that are co-prime to $m$, so it's OK. – DanielWainfleet Aug 05 '22 at 17:30
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    Yes, the result holds more generally - see the Theorem in the linked dupe. – Bill Dubuque Aug 05 '22 at 21:28

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