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?

  • 5
    $\begingroup$ observe that $k^2+k+1$ is always odd. $\endgroup$ – Frieder Jäckel Dec 3 '17 at 1:18
  • 1
    $\begingroup$ I believe you mean $p=17^{\color{#C00}{2}}+17+1=307$ $\endgroup$ – robjohn Dec 3 '17 at 2:35

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.


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 $$p = 17^2 + 17 + 1 = 307$$


Since $k^2+k+1$ is always odd it is necessary that $16|k-1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.