Euler's Theorem when $m$ is square-free

Question

Suppose that $m$ is square-free, and that $k$ and $\bar{k}$ are positive integers such that $k\bar{k} \equiv 1\pmod{\phi(m)}$. Show that $a^{k\bar{k}} \equiv a \pmod m$ for all integers $a$. In the case that $a \in$ the multiplicative group $\bmod m$, it is a simple application of Euler's Theorem. But when it's not in the mult. group, why does the condition that $m$ is square-free allow the the statement $a^{k\bar{k}} \equiv a \pmod m$ to hold?

Answer 1

Let $m=pq$ the product of two distinct primes. Now the ring $\Bbb{Z}_{m}=\Bbb{Z}_p\times \Bbb{Z}_q$, so that every element $a \in \Bbb{Z}_{m}$ can be uniquely written as $(a_p,a_q)$ where $a_p \in \Bbb{Z}_p$ and $a_q\in \Bbb{Z}_q$ (indeed, $a_p= a\pmod p$ and $a_q=a\pmod q$) with pointwise addition and multiplication. Even if $a$ is singular in $\Bbb{Z}_m$ (i.e when $a_p$ or $a_q$ is singular) we have for example $(a_p,0)^{k\bar{k}}=(a_p^{k\bar{k}},0)=(a_p,0)$ by Eulers theorem. The condition for $m$ to be squarefree assures us that $a_p$ and $a_q$ are not both singular. This proof can easily be generalized if $m$ is the product of several distinct primes.