Odd prime divisor of $ 3x^2+y^2$ (where $x$, $y$ are relatively prime) is again of the same form Euler proved that any prime divisor $\neq2$ of an integer of the form $ 3x^2+y^2$ is again of the same form.
I know there are proofs available using quadratic reciprocity. I was curious to know whether this can be done without using quadratic reciprocity or more elementarily.
In fact I would be happy if someone could provide the actual proof done by Euler.
Any help would be appreciated. Thanks in advance.
 A: This can indeed be done elementarily, using some clever ideas.
We'll start with a geometric theorem called Minkowski's theorem, which may seem completely unrelated, but will make more sense later. This theorem states that any convex set in the plane with area exceeding $4$ and which is symmetric about the origin contains a lattice point apart from $(0,0)$. The proof is as follows:
Consider the shape "modulo" a $2 \times 2$ square, i.e. we simply map every point in the set into a $2 \times 2$ square by taking the $x$ and $y$ coordinates modulo $2$. Since the area is more than $4$, some point gets mapped to twice. So there are two points $(x,y)$ and $(x+2a, y+2b)$, both in the original shape, for $a, b$ integers, not both zero.
By symmetry about the origin, we know that $(-x,-y)$ is in the set. By convexity, all points on the line joining $(-x,-y)$ and $(x+2a, y+2b)$ are in the set, and in particular their midpoint is in the set. But their midpoint is $(a,b)$, which is a lattice point, and we have proven the theorem.
In particular, the theorem is true under affine transformations, so we have the more general result that if vectors $v, w$ generate a lattice (which is the set of vectors $av+bw$ for integers $a, b$), such that the area of the parallelogram formed by $v, w$ is $A$, then any convex symmetric set with area more than $4A$ contains a point in the lattice.
Now back to the original problem. Suppose that $p$ divides $3x^2+y^2$, for coprime $x, y$ (which we can assume to be nonzero by coprimality). If $p$ divides $x$, then we get $p$ divides $y$ as well, which is a contradiction. So $x$ is coprime to $p$, and so an inverse of $x$ modulo $p$ exists. 
By multiplying by the inverse of $x$ in modulo $p$, we have $p$ divides $n^2+3$ for some integer $n$. Now consider the lattice generated by $(n, 1)$ and $(p, 0)$. Any such point has form $(cn+dp, a)$, and we observe that
$$(cn+dp)^2 + 3\cdot c^2 \equiv c^2n^2 + 3c^2 \equiv c^2(n^2+3) \equiv 0 \pmod p.$$
So all such points $(a,b)$ in this lattice are solutions to $p|3a^2+b^2$. But the parallelogram formed by these vectors has area $p$. and the ellipse $x^2+3y^2 < 4 p$ has area $\frac{4 \pi p}{3} > 4p$ (since $\pi > 3$). So there is a lattice point inside this ellipse.
Let this lattice point be $(a, b)$. Then $a^2+3b^2 < 4p$, but since $(a,b)$ is in the lattice, $p|a^2+3b^2$. So $a^2+3b^2$ is $p, 2p$ or $3p$.
In the first case, we are done. In the second case, if $2p = a^2+3b^2$, then since squares are $0$ or $1$ modulo $4$, the right hand side ends up either being odd, or divisible by $4$, which is a contradiction since $p$ is odd (unless $p = 2$, but the $p=2$ case is trivial).
In the third case, $3p = a^2+3b^2$. But $3p$ and $3b^2$ are both divisible by $p$, so $a^2$ must be too, which means $3|a$. Writing $a = 3c$, we get $3p = 9c^2 + 3b^2$, and dividing out by $3$ concludes the proof.
