I'm trying to show a positive integer $n$ is properly represented by the binary quadratic form $x^2+xy+y^2$ if and only if $n$ is not divisible by 9 or any prime of the form $3k+2$? It boils down to showing $-3$ is a square mod $4n$ if and only if $n=3^a\cdot\prod p^b$ where $a=0$ or $1$ and $p$ are primes of the form $3k+1$ but I'm stuck.
I'm currently going through Niven's An Introduction to the Theory of Numbers, Chapters 3.6-3.7.
Here is a direct approach in the style of Gauss' proof of Fermat's theorem on sums of 2 squares. Let $\omega$ be a primitive 3-rd root of 1. Introduce the imaginary quadratic field $K=\mathbf Q(\omega)=\mathbf Q (\sqrt {-3})$ and its ring of integers $A=\mathbf Z[\omega]$ (the so called Eisenstein ring, known to be a PID; beware, $A\neq \mathbf Z[\sqrt {-3}]$). The norm map $N:K^*\to {\mathbf Q}^*$ defined by the formula $N(x-y\omega)=(x-y\omega)(x-y{\omega}^2)=x^2+xy+y^2$ is a multiplicative function. So, restricting $N$ to $A-(0)$ and using the prime factorization of the integer $n$, you can bring back your problem to the determination of the powers $p^a$ of rational primes which are norms from $A-(0)$.
The classical theorem on ramification/decomposition of rational primes $p$ in a quadratic field easily yields the following : 1) If $p \equiv 2$ mod $3$, then $p$ is inert, i.e. remains prime in $A$ ; 2) If $p\equiv 1$ mod $3$, then $p$ splits in $A$, i.e. $p=\pi.\bar\pi$, where $\pi$ and $\bar\pi$ are two non-associate conjugate primes of $A$ ; 3) If $p=3$, then $1-\omega$ is prime in $A$ and $3=-{\omega}^2(1-\omega)^2$. Let us show for example the case 2): suppose that $p\equiv $ mod $3$ ; then $(\frac {-3}p)=1$ and the classical theorem for $\mathbf Q (\sqrt {-3})$ states that $p$ splits.
Conclusion: the only powers $p^a$ which are norms from $A-(0)$ are powers of $p\equiv 1$ mod $3$ and powers of $9$.
$4n=(x+2y)^2+3x^2$
For odd $p,$
$$p|n\implies ((x+2y)/x)^2\equiv-3\pmod p$$
Use For what odd prime p is -3 a quadratic residues? Non-residue?
