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Came across this question in The Art and Craft of Problem Solving.

Question: Let $p$ be a prime. Show that if $x^k\equiv1 \pmod{p}$ for all nonzero $x$, then $p-1$ divides $k$.

I know that if $k>(p-1)$, then can use Fermat's Little Theorem to 'reduce' to $k\leq(p-1)$. Then since a primitive root exists, there is some $a$ with ord(a) $=p-1$. But the existence of primitive root needs a bit of group theory, and I am wondering if there is a more elementary solution?

My question: Is there alternative ways of solving this question, preferably a more elementary way.

I know there is a similar question here but I think the answer posted has mistakes. Let p be a prime and k a positive integer such that $a^k$mod p = a mod p for all integers a. Prove that p - 1 divides k - 1.

eatfood
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    The existence of a primitive root follows fairly easily (though not trivially) from this statement, so I doubt there any proof of this that is significantly easier than just proving that a primitive root exists. – Eric Wofsey Aug 29 '18 at 03:50
  • But the existence of primitive root needs a bit of group theory With only number theory you can do it. In group theory we usually apply concept of number theory instead! – tarit goswami Aug 29 '18 at 05:08
  • @EricWofsey yes that is my thought process too – eatfood Aug 29 '18 at 16:38
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    One possibility is to use that $\Bbb{Z}/p\Bbb{Z}$ is a field, so the equation $x^k=1$ cannot have more than $k$ solutions. Here the assumption is that there are $p-1$ solutions, so $k\ge p-1$. A matter of taste whether that is more elementary than calling upon the existence of a primitive root. – Jyrki Lahtonen Sep 02 '18 at 04:14
  • Your Q is not equivalent to the Q in your link. If $x^k\equiv 1 \mod p$ for all $x>0$ then $p^k\equiv 1 \mod p,$ which is impossible for prime $p$ and $k>0.$ – DanielWainfleet Sep 02 '18 at 05:09
  • Lat month I saw on this site a brief, very elementary proof by Gauss that a primitive root exists. I'm trying to remember it. – DanielWainfleet Sep 02 '18 at 05:11
  • @JyrkiLahtonen Thats a very neat way to do it! I feel its more 'natural' cause it directly deals with the number of solutions – eatfood Sep 02 '18 at 09:59
  • Can this be proved without using primitive roots or using the roots of a finite field? My textbook does not include either of those but has this question. – farleyknight Oct 30 '18 at 02:07

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