So this is my problem: Prove that if $m$ is a positive odd integer, then $2 + 2^2 + . . . + 2^{\Phi(m)} \equiv 0\pmod m$
My work thus far:
I know by Euler's theorem that if $(a, m)=1$ then $a^{\Phi(m)}\equiv 1\pmod m$ and from checking with random odd integers that $\sum_{n=1} ^{\Phi(m)} 2^n$ will be divisible by $m$ with no remainder but I am having a hard time bridging these two thoughts.
Any ideas how to proceed? Thanks or the help!
Just sum up the geometric sequence.
You get $2 + 2^2 + \cdots + 2^{\Phi(m)} = 2^{\Phi(m) + 1} - 2$. At this point you can use Euler's theorem.