Let $n=am+1$ where $a $ and $m>1$ are positive integers and let $p$ be the least prime divisor of $m$. Prove that if $a<p$ and $ m \ | \ \phi(n)$ then $n$ is prime.
This question is a generalisation of the question at Let $n=apq+1$. Prove that if $pq \ | \ \phi(n)$ then $n$ is prime.. Here the special case when $m$ is a product of two distinct odd primes has been proven. The case when $m$ is a prime power has also been proven here https://arxiv.org/abs/2005.02327
How do we prove that the proposition holds for an arbitrary positive integer integer $m>1 $? ( I have not found any counter - examples).
Note that if $n=am+1$ is prime, we have $\phi(n)= n-1=am$. We see that $m \ | \ \phi(n) $. Its the converse of this statement that we want to prove i.e. If $m \ | \ \phi(n) $ then $n$ is prime.
from Hot Weekly Questions - Mathematics Stack Exchange
David Jones
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