# for every prime $p>7$, there are integers $x,y$ such that $p=x^2+7y^2$, if and only if, $p \equiv 1,2,4 \pmod7$.

I'd really love your help with showing that for every prime $p>7$, there are integers $x,y$ such that $p=x^2+7y^2$, if and only if, $p \equiv 1,2,4 \pmod7$.

$x^2+7y^2$ is the norm of the Euclidean domain $a+b\sqrt{-7}$ and $p$ is composite in the domain iff it is a norm, i.e $x^2+7y^2=1$ (I'm not sure I need this now).

as well the squares$\pmod7$ (The numbers that left the same residue after squaring) are $0,1$ so I guess $x^2+y^2 \equiv 0,1,2 \pmod7$, Is this correct? How should I go on?

Thanks a lot!

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The squares $\bmod 7$ are $0, 1, 2, 4$. – Qiaochu Yuan Jun 27 '12 at 19:55
There's a book by David Cox called 'primes of the form $x^2 + n y^2$; as I recall, you can classify primes of this form for $n=1,2,3,4,7$ by using the theory of quadratic forms over $\mathbf{Z}$ and quadratic reciprocity (i.e. no need for class field theory), and I think all of those cases are cut up into exercises in his book. Maybe there's an easy tricky way for $n=7$ though, who knows. – vgty6h7uij Jun 27 '12 at 19:59
@Jozef: If your comment was directed at Yuan, I think he meant that $x^2 \equiv 0,1,2, \text{or } 4 \!\!\mod 7$ – Eric Stucky Jun 27 '12 at 20:09
@vgty6h7uij $n=4$ is obvious, since primes $x^2+(2y)^2$ are of the form $1\bmod 4$, which motivates $p \equiv 1,2,4 \pmod7$. Nice question, interesting comment. – draks ... Jun 27 '12 at 21:30
@vgty6h7uij: Dear vgt, Certainly no class-field theory is needed. The quad. integer ring of disc. $-7$ is a PID, and one sees that $p > 7$ splits iff $p$ is a square mod 7. (Quad. reciprocit.) This gives the result. So the "easy tricky way for $n = 7$" is just that we have a PID in this case. Regards, – Matt E Jun 28 '12 at 3:57

You can refer to this part of Granville's course notes. In particular from Proposition 4.1: An integer $n$ is properly represented by a binary quadratic form of discriminant $-7$ iff $-7$ is a square modulo $4n$. Since $-7\equiv 1\pmod 4$ and $\left(\frac{-7}{p}\right)=\left(\frac{p}{7}\right)$ for all primes $p$, $-7$ is a square mod $4p$ if $p\equiv 1,2,4 \pmod 7$.

There's only one reduced binary form with discriminant $-7$ (an exercise in the notes), so all such primes can be written as $x^2+xy+2y^2$. If $x$ is even or $x$ and $y$ are both odd, then this represents an even number, so we must have $y=2z$ for some integer $z$.

So there are $x,z\in\mathbb{Z}$ with $p=x^2+2xz+8z^2=(x+z)^2+7z^2$.

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This is precisely the method to which I was referring in my comment to OP's question, if anyone's interested; the reason the same strategy works for $n=1,2,4,5,7$ is that for these $n$, you also have a unique reduced form of discriminant equal to that of $X^2+nY^2$; for other $n$, the uniqueness is violated (if I remember well). – vgty6h7uij Jun 28 '12 at 14:50

Here is a standard algebraic number theory approach; I don't know if it is at the right level for the OP.

• $\mathbb Z[(1 +\sqrt{-7})/2]$ is a PID (well-known, and easily checked e.g. by the Minkowski bound).

• A prime $p \neq 7$ splits in this ring iff $p$ is a square mod $7$. (Quadratic reciprocity.)

• This gives the result, except that one gets $p = x^2 + 7 y^2$ where $x$ and $y$ might be half-integers rather than integers. But multiplying both sides by $4$ and looking mod $8$ rules this latter possibility out (as long as $p \neq 2$).

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