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.

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.
