Prove $(2^{n}+1)=Q(2^{m}+1)+2^{r}+1$ if r is even and $n=qm+r$, $rProve $(2^{n}+1)=Q(2^{m}+1)+2^{r}+1$ if r is even and $n=qm+r$, $r<m$

I think you meant $q$ is even and $Q$ is integer, in that case as $2^m+1 \mid 2^{qm}-1$ so, $2^n+1 \equiv 2^r+1 (\mod 2^m+1)$. (Note even if $q$ isn't even and your problem is correct then $2^m+1\mid 2^{qm}-1 \implies q$ is even) 
