# Primes of the form : $\sqrt{x^2+n\cdot y^2}$ , where $n$ is an Idoneal number

How to prove following statement :

Conjecture :

Let $n$ be an Idoneal number with property : $n \equiv 0 \pmod 4$ , and let $q$ be a prime number with

property : $q \equiv a \pmod n$ , $a \in \mathbb N$ , expressible as $q=\sqrt{x^2+n\cdot y^2}$ ; $x,y > 0$

then every odd prime $p$ such that : $p^2>n$ and $p\equiv a \pmod n$ can be expressed as :

$p=\sqrt{x^2+n\cdot y^2}$ ; $x,y > 0$ .

For example :

$p=\sqrt{x^2+4\cdot y^2}$ if and only if : $p\equiv 1 \pmod 4$

$p=\sqrt{x^2+8\cdot y^2}$ if and only if : $p\equiv 1 \pmod 8$ or $p\equiv 3 \pmod 8$

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It seems like your conjecture is incomplete? What is $N$? –  Thomas Andrews Jan 9 '12 at 15:40
@ThomasAndrews,set of natural numbers . Is something else unclear ? –  pedja Jan 9 '12 at 15:43
The confusion is in using $a$ and $q$, when you could have just stuck with $q$. Namely, "if $q$ is a prime of the form, then if $p\equiv q\pmod n$... ". $a$ is irrelevant. –  Thomas Andrews Jan 9 '12 at 15:56
@ThomasAndrews,In fact conjecture is quite simple : If one prime number with property $a \pmod n$ can be expressed in the form :$\sqrt{x^2+n\cdot y^2}$ then all prime numbers with property $a \pmod n$ can be expressed in this form . –  pedja Jan 9 '12 at 15:58
@ThomasAndrews,I am sure that my formulation of conjecture isn't perfect but I think that is clear enough... –  pedja Jan 9 '12 at 16:03