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?
 A: 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 $n^3-1$?
A: HINT:
$n$  must be odd $=2m+1$(say)
$\implies p=4m^3+6m^2+3m=m(4m^2+6m+3)$
Both $m,4m^2+6m+3$ can not be $>1$
A: Your final equation has $2p$ written as a product, and you already know there are very few ways to do this for $2p$ -- namely, one of the factors have to be either $1$ or $2$. So just check each of the cases
$$ n-1 = 1 \quad\text{or}\quad n-1=2 
\quad\text{or}\quad n^2+n+1 = 1 \quad\text{or}\quad n^2+n+1 = 2 $$
A: Since $p$ is a prime, the only factors possible for $2p$ is $2$ and $p.$
This must correspond to $(n-1$) and $(n^2 + n +1)$. We know using common sense that $n^2 + n +1$ is greater than $n-1,$ also that since $p$ is a prime, $p>2. $
Hence,$n-1=2$ which means $n=3.$
$p=n^2 + n +1$. Here, replace the value of $n$ with $3$(As obtained from the previous step) and you'll get $p=13$
