establish that every prime number p of the form $ 8k + 1$ or $8k +3$ can be written as $ p = a^2 + 2b^2$ for some choice of integers a and b.

Question

The question is:

Establish that every prime number p of the form $ 8k + 1$ or $8k +3$ can be written as $ p = a^2 + 2b^2$ for some choice of integers a and b.

And the Hint says:

Mimic the proof of theorem 13.2, which is given below:

My trial is:

For the case: $p = 8k + 1$

I used the theorem that says the Legendre symbol $(2/p) = 1$ if $p \equiv 1 \pmod 8$

And I end up after using Thue lemma that $$2x_{0}^2 \equiv y_{0}^2,$$

But this will lead to $p = 2a^2 - b^2$ which is not the required, could anyone help me in fixing this please?

I feel that I need the Legendre symbol $(2/p) = 1$ if $p \equiv 1 \pmod 8$ to be $(- 2/p) = 1$ if $p \equiv 1 \pmod 8$ instead, but How can I do this ?

What about the second case when $p \equiv 3 \pmod 8$?

Answer 1

Since the Legendre symbol is multiplicative we have: $$\left(\frac{-2}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{-1}{p}\right)$$ The first one is equal to $1$ iff $p \equiv 1,7 \pmod{8}$ and the second iff $p \equiv 1 \pmod{4}$. So they are both one when $p \equiv 1 \pmod{8}$ and both $-1$ when $p \equiv 3 \pmod{8}$. In both cases their product is one which means that $-2$ is a quadratic residue mod $p$ i.e. there exists an integer $a$ such that: $$a^2 \equiv -2 \pmod{p}$$ Now you mimic the given proof and everything should work out well.

Answer 2

Comment:An experimental approach; following conditions cover a set of these types of primes:

For $8k+1$:

Let $k=k_1+2$ ⇒ $p=8(k_1+2)+1=8k_1+8+8+1=2[4(k_1+1)]+3^2$

If $k_1+1=c^2$ then we have:

$p=3^2+2\times (2c)^2$

Examples; $k_1=3$ ⇒ $p=41=3^2+2 (2\times 2)^2=3^2+2\times 4^2$

$k_1=15$ ⇒ $c^2=15+1=4^2$ ⇒ $p=137=3^2+2\times 8^2$

For $p=8k+3$ it can be seen that:

$p=8k+3 ≡ c^2k^2 \mod 3^2$;

Type 1: $p=(3d)^2 +2(ck)^2$

Type 2: $p=(ck)^2 +2(3d)^2$

apart from ck=1 or $(3d)^0=1$ we have following values for ck and 3d:

$ck=1, 5^2, 7^2, 11^2, (2n+1)^2$; $2n+1 ≠ 3 t$

$3d = 3, 9, 15, . . .(2n+1)3$

Examples:

$p=8\times 52+3=419=9^2+2\times 13^2$

$p=8\times 58+3=467=15^2 +2\times 11^2$

In both cases $p≡ c^2k^2 \mod 3^2$