prove that if $u \equiv w \bmod \Phi(m)$ and $\operatorname{gcd}(a, m)=1$ then $a^{u} \equiv a^{w} \bmod m$

Question

Let $u \equiv w \bmod \Phi(m)$. if $\operatorname{gcd}(a, m)=1$ then how to elementary prove that $a^{u} \equiv a^{w} \bmod m$ ?

Answer 1

Note that if $\gcd(a, m) = 1$, by Euler's theorem $a^{\varphi(m)} \equiv 1 \pmod{m}$.

Then as $\varphi(m) \mid (u-w)$, we have $a^{u-w} \equiv 1 \pmod{m}$ and $a^u \equiv a^w \pmod{m}$.