The following problem is one of the exercises in Topics in the Theory of Numbers (Erdős et al.)
Show that if the positive integer $m$ has at least two distinct odd prime divisors, and $c$ is relatively prime to $m$, then $c^{\varphi(m)/2} \equiv 1 \pmod{m}$.
I am attempting to solve this problem using any of the tools developed before this point in the book.
My current attempt involves using Euler's product formula to write $\varphi(m) = \prod_{p \mid m} p^{k - 1} (p - 1).$ Applying Fermat's little theorem gives me $c^{p - 1} \equiv 1 \pmod{p}$ for every $p \mid m$. I then reason that since at least two of the prime factors are odd, we have that $(p - 1) \mid \phi(m) / 2$, so that $c^{\phi(m) / 2} \equiv 1 \pmod{p}$ for every $p \mid m$. This is where I get stuck. I am unsure how to lift the result to modulo $m$. As I understand, Hensel's lemma may be useful in this regard, but it has not yet been discussed before this point in the book. Is there a way to salvage my attempted proof without Hensel's lemma? Otherwise, does someone have a different idea that uses similar elementary theorems?