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October 22nd, 2018, 11:55 AM   #11
Joined: Jan 2016
From: Athens, OH

Posts: 91
Thanks: 47

This may be more information than you want, but here goes.
Given positive integers a and m, find all positive integers n with $a^n$ congruent to 1 mod m.
First a and m must be relatively prime; otherwise, no such n exists.
So for a prime to m, $a^n$ is 1 mod m for $n=\phi(m)$ -- here $\phi$ is Euler's totient function. Let k be the smallest positive integer with $a^k=1$ mod m; k is called the order of a modulo m. Then k divides $\phi(m)$ and any solution n is a multiple of k.

For your specific problem, $\phi(53)=52$. As was noted in the last post, the order of 5 turns out to be 52. This is a lot of arithmetic to do by hand. I have a little computer program that computes the order of a mod m; if you're interested, post again.

Lastly, for m a prime, the problem of finding a with the order of a being m-1 is a hard problem. Such an a is called a primitive root. As far as I know, there is no easy way to find primitive roots.
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