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Can you please show the proof of

"The solutions to $x^m \equiv 1 \bmod p$ will all be solutions to $x^{mn} \equiv 1 \bmod p$ for any $n$."

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    $\begingroup$ If $x^m\equiv 1$ mod $p$, then $x^{mn} \equiv 1^n \equiv 1$ mod $p$. $\endgroup$ – TonyK Jul 6 '15 at 9:36
  • $\begingroup$ $$x^{mn}=(x^m)^n$$ See math.stackexchange.com/questions/188657/… $\endgroup$ – lab bhattacharjee Jul 6 '15 at 9:39
  • $\begingroup$ I have asked the question to clarify if $p\nmid(2^a-1)$ that $p\nmid 2^{ab}-1$. Is this correct? $\endgroup$ – Kurtul Jul 6 '15 at 11:33
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Hint: Do you know the proof of $$x \equiv a \mod m \implies x^n \equiv a^n \mod m$$

For any integer $n>0$ and given $x,a,m$? If not, try to prove this by induction first.

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  • $\begingroup$ $x^n-a^n=(x-a)(x^{n-1}+x^{n-2}a+\cdots+a^{n-1})$ proves it. $\endgroup$ – user26486 Jul 6 '15 at 11:39

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