$2^i \equiv 2^j \pmod n$ implies $2^{j−i }\equiv1$ if $n$ is odd; also if $n$ is even? 
Show that, if $0 \leq i < j$ and $2^i \equiv 2^j \pmod n$ and $n$ is odd then $2^{j−i} \equiv 1 \pmod n$. Is this necessarily true if $n$ is even?

I have tried to prove this by using Fermat's little theorem but it does not work.

If $n$ is odd, then $n \land 2^{i} = 1$, so you can "divide both sides" by $2^i$. 
This is not true if $n$ is even: $2^3 \equiv 2^2 \pmod 4$, but $2 \not \equiv 1 \pmod 4 $
