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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2If $x^m\equiv 1$ mod $p$, then $x^{mn} \equiv 1^n \equiv 1$ mod $p$. – TonyK Jul 06 '15 at 09:36
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$$x^{mn}=(x^m)^n$$ See http://math.stackexchange.com/questions/188657/why-an-bn-is-divisible-by-a-b – lab bhattacharjee Jul 06 '15 at 09:39
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I have asked the question to clarify if $p\nmid(2^a-1)$ that $p\nmid 2^{ab}-1$. Is this correct? – Kurtul Jul 06 '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.
wythagoras
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