Prime factors of special numbers

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 $p={T_0}^2+27{U_0}^2,T_0,U_0\in\mathbb{Z}$?

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Yes, it is possible. For example $N = p = 43 = 27 + 16$. But I don't that is, what you are looking for. – martini Mar 26 '12 at 9:56

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 $p$.