Number of solutions to the congruence $x^{k} \equiv 1 \bmod \ p$

Question

I want to find the number of solutions to the congruence $$x^{k} \equiv 1\bmod \ p.$$ I know the answer is $$\gcd(k,p-1)$$ but I need proof for it.

thanks

http://math.stackexchange.com/questions/320910/number-of-solutions-to-the-congruence-xq-equiv-1-mod-p — lab bhattacharjee, Dec 08 '13 at 18:49
http://mathworld.wolfram.com/DiscreteLogarithm.html and then http://www.proofwiki.org/wiki/Solution_of_Linear_Congruence — lab bhattacharjee, Dec 08 '13 at 18:51

score 3 · Answer 1 · answered Dec 08 '13 at 19:10

Let $d=\gcd(k,p-1)$. Write $d=ku+(p-1)v$. Fermat's theorem then implies that $x^{k} \equiv 1\bmod \ p$ iff $x^{d} \equiv 1\bmod \ p$.

The non-zero classes mod $p$ form a cyclic group. The set of solutions of $x^{d} \equiv 1\bmod \ p$ corresponds to the subgroup of order $d$.

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