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I'm trying to show this statement:

$$p \equiv 1,3 \pmod 8, \; \; \exists \; u,v \in \mathbb Z : u^2 + 2 v^2 = p.$$

I believed I proved it the other direction using the ring $\mathbb Z{\sqrt{-2}}$. This is my reasoning so far: Suppose that $-2 \equiv x^2 \pmod p$ then $p$ divides $x^2 + 2 = (x - i\sqrt{2})(x + i\sqrt{2})$. This shows that $p$ is not a prime, that is $p = mn $. Taking the norms, we get $N(p) = N(m)N(n) = x^2 + 2y^2$ We can see that $p \equiv 1,3 \pmod 8$. How do I prove using the forward direction?

user111019
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  • http://math.stackexchange.com/questions/685958/let-p-be-prime-and-frac-3p-1-prove-that-p-is-of-the-form-p-a23b2/685983#685983 -- will this help ?? – r9m Feb 24 '14 at 20:58
  • user129017 I am taking a class that only assumes that you've taken 1 calculus and 1 linear algebra course. The number theory is being introduced under the context of writing expository proofs. I am familiar with Legendre's symbol but not all of its properties and I have no idea what a lattice is. – user111019 Feb 24 '14 at 21:02
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    what about Jack D'Aurizio's proof ? It only uses fermat's descent (Elementary-Numbertheory) – r9m Feb 24 '14 at 21:05

1 Answers1

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As this has class number one there are many approaches. Solve $$ \beta^2 \equiv -8 \pmod p. $$ If your $\beta$ is odd, replace by $p-\beta$ which is even, anyway call it $b.$ Now you have $$ b^2 \equiv -8 \pmod {4p} $$ with $b$ even, or $$ b^2 - 4pt = -8. $$ So, the binary quadratic form $$ \langle p,b,t \rangle $$ or $$ f(x,y) = p x^2 + b x y + t y^2 $$ has discriminant $-8.$ Gauss reduce this, the result is $$ \langle 1,0,2 \rangle $$ because that is the only reduced form for that discriminant.

ALMOST DONE

So, reduction: there is a 2 by 2 matrix with determinant 1, write it as $$ P = \; \left( \begin{array}{rr} \alpha & \beta \\ \gamma & \delta \end{array} \right), $$ also name $$ M = \; \left( \begin{array}{rr} p & b/2 \\ b/2 & t \end{array} \right), $$ and $$ R = \; \left( \begin{array}{rr} 1 & 0 \\ 0 & 2 \end{array} \right). $$

The relationship found by doing reduction step by step is $$ P^T M P = R. $$

Let $Q = P^{-1}.$ Then $$ Q^T R Q = M. $$

The left-hand column of $Q$ gives the representation of your original prime $p.$ Multiply it out.

Will Jagy
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