TITLE: Prove that $2^n +1$ in never a perfect cube
QUESTION [7 upvotes]: Prove that $2^n +1$  in never a perfect cube
I've been thinking about this problem, but I don't know how to do it. I know that if $m^3=2^n+1$, then  $m$ should be an odd number, but I 'm not able to get to a contradiction.

REPLY [5 votes]: Andre's answer is neat, I've got a similar solution. Assume $m^3 = 2^n + 1$ then $m$ is odd therefore $m=2p+1$ thus $$8p^3 + 12p^2 + 6p + 1 = 2^n + 1 \implies 2p(4p^2 + 6p + 3) = 2^n$$ Note that $2p$ and $4p^2 + 6p + 3$ must be powers of $2$, however the latter is odd. Then, again $4p^2 + 6p + 3=1 \implies p=-1 \text{ or } p=-1/2$ both of which are impossible because $m > 0$