0
$\begingroup$

Consider a positive odd integer $N>1$ of form

$N=T^2+27U^2, T,U\in\mathbb{Z}, T, U\neq 0$

which cannot be divided by 3.

Question: Suppose N can be divided by a prime number $p$, $p\equiv1(mod 3)$. Is it possible that $p={T_0}^2+27{U_0}^2,T_0,U_0\in\mathbb{Z}$?

  • 0
    Yes, it is possible. For example $N = p = 43 = 27 + 16$. But I don't that is, what you are looking for.2012-03-26

1 Answers 1

1

It isn't always possible...try $N = 28$, it is divisible by $7$ and this number cannot be written in that form.

A general result is that $p = x^2 + 27y^2$ if and only if $p\equiv 1$ mod $3$ and $2$ is a cube mod $p$.