Find all prime numbers p such that 2p+1 is a perfect cube. $$2p+1=n^3$$
$$ 2p = n^3 - 1$$
$$ 2p = (n-1)(n^2+n+1)$$
The only number that fits the criteria is 13, how can I prove this?

HINT: The factorization of $2p$ shows that $n-1$ must be either $2$ or $p$. (The other two cases, $n-1=1$ and $n^2+n+1=1$, are trivially eliminated.) You already know what happens when $n-1=2$. Otherwise, $n-1$ must be an odd prime $p$. In that case $n$ is even. What does that tell you about $n^3-1$?
