<s>[INST] Find all prime numbers $p$ such that $16p+1$ is a perfect cube What I have attempted: 
Suppose  $16p+1=k^3$ where $k \in Z$ then $16p=k^3-1=(k-1)(k^2+k+1)$ so we can say that $k=17$ and thus $p=17^3+17+1=4931$ which is prime. 
How would I find the remaining numbers?
 [/INST] You had $$16p=k^3-1=(k-1)(k^2+k+1)$$
Because $k$ is odd, $k-1$ is even and $k^2+k+1$ is odd. If $k^2+k+1$ is odd, then $k-1$ must be a multiple of $16$. But for $k-1$ to be a multiple of $16$ other than $16$, $p$ would have to not be a prime. Therefore, $k-1 = 16$ and $k = 17$.
That means that $k^2+k+1$ must be our prime. So plug in $k=17$ to get $$p = 17^2 + 17 + 1 = 307$$
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