<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] Since $k^3\equiv1\pmod{16}\implies k\equiv1\pmod{16}$, if $16p+1$ is a perfect cube, we must have
$$
\begin{align}
16p+1
&=(16k+1)^3\\
&=4096k^3+3\cdot256k^2+3\cdot16k+1
\end{align}
$$
Thus, we get $p=256k^3+48k^2+3k=(256k^2+48k+3)k$, which can only be prime if $k=1$, that is $p=307$ and thus
$$
17^3=16\cdot307+1
$$
is the only case.
</s>