Modulo Euler's Totient Function?

Question

Prove or disprove: $2^a \equiv 2^b \pmod{m} \text{ iff } a \equiv b \pmod{\phi{(m)}}$.

Through some testing, I found that this is true. How can I prove this?

Answer 1

It is valid for $m$ odd ($gcd(m,2)=1$) due to Euler's theorem (https://en.wikipedia.org/wiki/Euler%27s_theorem). If $ a \equiv b \mod m$ $$ \Rightarrow a-b=k\phi(m)$$ $$ \Rightarrow 2^{a-b}=2^{k\phi(m)}=\left(2^{\phi(m)}\right)^k\equiv 1^k \equiv 1\mod m, \gcd(2,m)=1$$. It is not valid for $m$ even, for example $m=6, \phi(6)=2, a=0, b=2$.