Let $p$ be prime and $\left(\frac{-3}p\right)=1$. Prove that $p$ is of the form $p=a^2+3b^2$ 
Let $p$ be prime and $\left(\frac{-3}p\right)=1$, where $\left(\frac{-3}p\right)$ is Legendre symbol. Prove that $p$ is of the form $p=a^2+3b^2$.

My progress:
$\left(\frac{-3}p\right)=1 \Rightarrow$ $\left(\frac{-3}p\right)=\left(\frac{-1}p\right)\left(\frac{3}p\right)=(-1)^{\frac{p-1}2}(-1)^{\left\lfloor\frac{p+1}6\right\rfloor}=1 \Rightarrow$ $\frac{p-1}2+\left\lfloor\frac{p+1}6\right\rfloor=2k$  I'm stuck here. This is probably not the way to prove that.
Also tried this way:
$\left(\frac{-3}p\right)=1$, thus $-3\equiv x^2\pmod{p} \Rightarrow$ $p|x^2+3 \Rightarrow$ $x^2+3=p\cdot k$
stuck here too.
Any help would be appreciated.
 A: $(\frac{p}{3})=(\frac{-3}{p})(\frac{p}{3})=(\frac{-1}{p})(\frac{3}{p})(\frac{p}{3})=(-1)^{\frac{p-1}{2}}(\frac{3}{p})(\frac{p}{3})=(-1)^{\frac{p-1}{2}}(-1)^{\frac{p-1}{2}\frac{3-1}{2}} = 1$
Hence,$(\frac{-3}{p})=1$ iff, $p\equiv1\mod3$.
Since, there is $u \in \mathbb{Z}$ such that, $-3\equiv u^2\pmod{p}$
Consider the lattice defined by $L=\{(a,b)\in\mathbb{Z}^2\, : \,a\equiv ub\pmod p\}$ generated by $(u,1)$ and $(0,p)$. $L$ has index $p$ in $\mathbb{Z}^2$, and area of its fundamental domain is $p$. Now, consider an ellipse $E_n$ defined by $x^2+3y^2=n$, then the area of $E_n=\frac{\pi n}{\sqrt3}>1.8n$
Choose, $n=2.3 p$, then Area of $E_{n}>4p$ and $E_n\cap L$ has a non zero point $(a,b)$.
Now, $a^2+3b^2\equiv(ub)^2+3b^2\equiv b^2(u^2+3)\equiv0 \pmod p$.
Since, $(a,b)\in E_n \implies a^2+3b^2<2.3p$ we have $a^2+3b^2=p,2p$.
But, $a^2+3b^2=2p \implies a^2\equiv 2p \pmod 3 \equiv 2 \pmod 3$ contradiction !!
Therefore, $a^2+3b^2=p$.
A: First part:
$$\left(\frac{-3}{p}\right)=1 \text{ if and only if }\; p\equiv{1}\!\!\!\!\pmod{3}.\tag{1}$$
This can be achieved through the Gauss quadratic reciprocity theorem in the most general form, or through the following lines. If $p=3k+1$, by the Cauchy theorem for groups there is an order-3 element in $\mathbb{F}_p^*$, say $\omega$; from $\omega^3=1$ follows $\omega^2+\omega+1\equiv 0\pmod{p}$, hence:
$$(2\omega+1)^2 = 4\omega^2+4\omega+1 = 4(\omega^2+\omega+1)-3 = -3,$$
and $-3$ is a quadratic residue $\pmod{p}$. On the other hand, if $-3$ is the square of something $\pmod{p}$, say $-3\equiv a^2\pmod{p}$, then:
$$\left(\frac{a-1}{2}\right)^3\equiv\frac{1}{8}(a^3-3a^2+3a-1)\equiv\frac{1}{8}\cdot 8\equiv{1},$$
and $\frac{a-1}{2}$ is an order-3 element in $\mathbb{F}_{p}^*$. From the Lagrange theorem for groups it follows that $3|(p-1)$.

Second part:
$$\text{If }p\equiv 1\pmod{3},\qquad p=a^2+3b^2.\tag{2}$$
Since by the first part we know that $-3$ is a quadratic residue $\pmod{p}$, there exists an integer number $c\in[0,p/2]$ such that:
$$ c^2+3\cdot 1^2 = k\cdot p.\tag{3}$$
The trick is now to set a "finite descent" in order to have $k=1$. Let $d$ the least positive integer such that $c\equiv d\pmod{k}$. Regarding $(3)$ mod $k$, we have:
$$ d^2+3\cdot 1^2 = k\cdot k_1.\tag{4}$$
Since the generalized Lagrange identity states:
$$(A^2+3B^2)(C^2+3D^2)=(AC+3BD)^2 + 3(BC-AD)^2,\tag{5}$$
by multiplying $(3)$ and $(4)$ we get:
$$ (cd+3)^2 + 3(c-d)^2 = k^2 pk_1.$$
Since $cd+3\equiv c^2+3\equiv 0\pmod{k}$ and $c\equiv d\pmod{k}$, we can rewrite the last line in the following form:
$$ \left(\frac{cd+3}{k}\right)^2+3\left(\frac{c-d}{k}\right)^2 = k_1\cdot p.\tag{6}$$
Now a careful analysis of the steps involved in the algorithm reveals that $k_1<k$, so the descent is able to reach $k_i=1$, or:
$$ p = a^2 + 3b^2$$
as wanted.
A: Let $\omega\in\Bbb C$ satisfy $\omega^2+\omega+1$. Then $\Bbb Z[\omega]$ is a UFD, since $N(a+b\omega)=a^2-ab+b^2$ defines an Euclidean norm.
Since $\left(\frac{-3}{p}\right)=1$, there is an $s$ such that $p\mid s^2+3$. Working over $\Bbb Z[\omega]$ we have $$p\mid (s+1+2\omega)(s+1+2\omega^2)$$ If $p$ is prime we would have both $p\mid s+1$ and $p\mid 2$. This is impossible, so $p$ cannot be prime. Therefore we must have $p=\alpha\beta$, where $\alpha,\beta\in\Bbb Z[\omega]$. Taking norms, we get $$p^2=N(p)=N(\alpha)N(\beta)$$ Since neither $\alpha,\beta$ are units, we have $N(\alpha),N(\beta)\neq 1$. This forces $N(\alpha)=N(\beta)=p$. Let $\alpha=m+n\omega$, then $$p=N(\alpha)=m^2-mn+n^2$$ Without loss of generality we may assume that $n$ is even, as $p$ is an odd prime. We can then write:
$$4p=4m^2-4mn+n^2+3n^2\\4p=(2m-n)^2+3n^2\\p=\left(m-\frac{n}{2}\right)^2+3\left(\frac{n}{2}\right)^2$$ Choose $a=m-\frac{n}{2}$ and $b=\frac{n}{2}$ and we get $$p=a^2+3b^2$$
