Show that if $p \equiv 1 \pmod 6$ then there exist integers $a$ and $b$ such that $p = a^2 + 3b^2$. 
Show that if $p \equiv 1 \pmod 6$ then there exist integers $a$ and $b$ such that $p = a^2 + 3b^2$.

Let $p$ be a prime such that $p \equiv 1,3 \pmod 8$.
There exists $c \in \mathbb{Z}$ such that $c^2 \equiv −2 \pmod p$.
Define $L=\{(x,y) \in \mathbb{Z^2} \mid x\equiv cy \pmod p\}$ and we know that  $(a,b) \in L \implies a^2+2b^2 \equiv 0 \pmod p$.
Let $S = \{(x,y) \in \mathbb{R^2} \mid x^2+2y^2<2p\}$. S has area $\pi p \sqrt{2}$ and there exist integers $a$ and $b$ such that $p = a^2 + 2b^2$ using Minkowski's theorem.

I have to use a similar argument to the above and use the hint below.
Hint: first prove that $2p = a^2 + 3b^2$ has no solutions for such primes $p$.

 A: HINT
The hint can be proven so: $2p \equiv 2 \equiv a^2 \pmod 3$, a contradiction. 
Pigeonhole Principle
Also, note that since $-3$ is a quadratic residue, this implies there exists such $a$ that $a^2 \equiv -3 \pmod p$. 
By Thue's Lemma, we get that there exists such $-\sqrt{p}< x,y < \sqrt{p}$ that $x \equiv ay \pmod {p}$. 
This would imply there exists such $x,y$ that $0<x^2+3y^2<4p$ and $x^2+3y^2 \equiv 0 \pmod p$. 
By  Minkowski's theorem
$p \equiv 1$ $\bmod 6$, then $\left( \frac{-3}{p} \right) = 1$. Let $p$ be in one of these residue classes and let $a$ be a square root of $-3$ modulo $p$. 
Let $\Lambda \subset \mathbb{Z}^2$ be the lattice of pairs $(x,y)$ such that $x \equiv ay \bmod p$.
For any $(x,y) \in \Lambda$, we have $x^2+3 y^2 \equiv (ay)^2 + 3 y^2 \equiv (-3+3) y^2 \equiv 0 \bmod p$. Since $\Lambda$ has index $p$ in $\mathbb{Z}^2$, the fundamental domain of $\Lambda$ has area $p$. 
The ellipse $\{ (x,y) : x^2+3y^2 < N \}$ has area $\frac{\pi N}{\sqrt{3}}$. Now take $N =\frac{4\sqrt{3}}{\pi}p <3p$
