Determining all odd primes that can be expressed in the form $x^2+xy+5y^2$ 
Determine all  odd primes that can be expressed in the form $x^2+xy+5y^2$.

Its discriminant $d=-19$. And it is in its reduced form. But how to approach to find all such odd primes. any suggestions?
Please provide a hint based on quadratic forms only, as I am doing an elementary number theory course.
 A: $\mathbb{Q}(\sqrt{-19})$ is one of the few imaginary quadratic fields with class number one: in our case, $x^2+xy+5y^2$ is the only reduced binary quadratic form of discriminant $-19$. It follows that the numbers represented by such a quadratic form give a semigroup, and the primes represented by such a quadratic form are the primes for which $-19$ is a quadratic residue, i.e., by quadratic reciprocity, the primes belonging to some arithmetic progressions $\!\!\pmod{76}$.
For a wonderful reference, see D.A. Cox - Primes of the form $x^2+ny^2$, or these notes by M. Bates.
A: This is the norm form of the number field $ K = \mathbf Q((1 + \sqrt{-19})/2) $, which has class number $ 1 $. Therefore, the primes represented by this form are precisely the primes that are either split or ramified in $ \mathcal O_K $. $ 2 $ is inert, and an odd prime is split if and only if $ -19 $ is a quadratic residue modulo that prime, which, by quadratic reciprocity, comes down to determining the residue class of that prime modulo $ 4 \times 19 = 76 $.
A: If $(-19|p) = 1$ for an odd prime $p,$ this means that there is a solution to $\beta^2 \equiv -19 \pmod p.$ If $\beta$ is even, replace it by $\beta \mapsto p - \beta,$ which is now odd. We now have
 $$\beta^2 \equiv -19 \pmod {4p}.$$
This is a good thing. 
$$ \beta^2 = -19 + 4pt $$
for some (nonzero) integer $t.$ Or,
$$ \beta^2 - 4pt = -19. $$
This means that the positive quadratic form
$$ \langle p, \beta, t \rangle  $$
has discriminant $-19.$
A finite sequence of reduction steps takes this to a reduced form. As the only reduced form is $\langle 1,1, 5 \rangle,$ we have produced a 2 by 2 matrix of integers $R$ with determinant $1.$ With $H$ the Hessian matrix of $$ p x^2 + \beta x y + t y^2 $$ and
$G$ the Hessian matrix of $x^2 + xy + 5 y^2,$ we have
$$ R^t H R = G.  $$
Name
$$  Q = R^{-1}, $$ 
this is also determinant $1,$ with
$$  Q^t G Q = H.  $$
The left column of $Q$ gives a representation of $p$ by $x^2 + xy + 5 y^2.$
A: Write $f(x, y)$ for your $x^2+xy+5y^2$.
As Jack D'Aurizio pointed out, $f$ is the only reduced binary quadratic form of discriminant $-19$. So if any quadratic form of discriminant $-19$ represents $n$, $f$ does.
The pertinent result is: An integer $n$ is properly represented by a quadratic form of discriminant $D$ iff $D$ is a square modulo $4n$. (This is in e.g. [Granville], Proposition 4.1.)
If $f$ represents $n$ improperly, then $f(t, u)=n$ where $t$ and $u$ have a common factor $>1$, in which case $n$ also has that factor. If $n$ is prime, that could only happen if $n\mid t$ and $n\mid u$, in which case $f(t, u)\geqslant n^2>n$.
So, for prime $p$, some quadratic form of discriminant $-19$ represents $p$ iff $f$ represents $p$, iff $f$ represents $p$ properly, iff $-19$ is a square modulo $4p$ (by the above result). By quadratic reciprocity, this is equivalent to $4p$ being a square modulo $-19$, which happens iff $p$ is a square modulo 19 (i.e. 0, 1, 4, 5, 6, 7, 9, 11, 16, 17). For an odd prime $p$, this is equivalent to $p$ being a square modulo 38 (i.e. 1, 5, 7, 9, 11, 17, 19, 23, 25, 35). The residue classes 0 and 19 are used only for $p=19=f(-1, 2)$.
[Granville] Andrew Granville. Binary quadratic forms. https://dms.umontreal.ca/~andrew/Courses/Chapter4.pdf
