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I'm trying to prove intuitively that there exists a number $k$ such that

$$a^k \equiv 1 (m),$$

iff $a$ and $m$ are coprime.

One of the ways of the proof is easy: If $a^k \equiv 1 (m)$, then that means $a^k = pm+1$ for some integer $p$. Since all consecutive numbers are coprime, then $a^k$ and $pm$ are coprime, which directly implies that $a$ and $m$ are also coprime.

However, I don't know how to prove the other direction of the statement, that is, if $a$ and $m$ are coprime, then there exists a number $k$ such that $a^k \equiv 1 (m)$, without invoking permutation properties of the reduced residue system modulo $m$ as shown in here.

Bill Dubuque
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elcocodrilotito
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  • Euler's theorem establishes even a concrete solution. The pigeonhole principle allows to easily prove the existence : Since there must be powers with equal residue , we can divide by a suitable power to get the desired result because $a$ must be invertible mod $m$ – Peter Feb 10 '23 at 15:20
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    The permutation-based view is the most intuitive way to view this. You may find it more intuitive viewed graphically in terms of orbits, e,g, see the argument here, generalized to recurrences in any finite ring. – Bill Dubuque Feb 10 '23 at 16:05
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    Those who fail to master these simple ubiquitous facts about such periodicity often end up (painfully) reinventing the wheel (cycle). – Bill Dubuque Feb 10 '23 at 16:09

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