Primes represented by $x^2+14y^2$ and $2x^2+7y^2$. This is a well known example in binary quadratic forms which I have started studying. Consider binary quadratic forms of discriminant $D=-56$. By reduction it turns out that for odd primes $p$ which are congruent to one of $1,9,15,23,25,39\,(\!\!\bmod\,56)$ there are two proper equivalence classes of forms represented by the reduced forms $x^2+14y^2$ and $2x^2+7y^2$. I haven't read genus theory much but I know that these two forms are in same genus.
My question is that is it possible that a same prime $p$ congruent to one of the above residues can be represented by both the forms $x^2+14y^2$ and $2x^2+7y^2$. In other words is there a way to separate the primes of the given residues which are represented by these two forms etc. Can I say that a prime $p$ congruent to the above residues can be represented by precisely one of the above two forms but not by both??? Is it possible to come to a conclusion with only reduction theory (this is upto what I have studied till now)???
For the case $D=-20$, however the reduced forms $x^2+5y^2$ and $2x^2\pm 2xy+3y^2$ represent different primes which can easily be seen by examining the resudues  mod 20 which are coprime to 20, which happen to be disjoint for the two forms. But unfortunately this fails for $D=-56$ and the forms in question, as the residues mod 56 for $x^2+14y^2$ and $2x^2+7y^2$ are identical i.e
$1,9,15,23,25,39\,(\!\!\bmod\,56)$.
Reference: Advanced Algebra by Knapp (page 16).
Thanks in advance.
 A: The answer is yes, class of the quadratic form is uniquely determined by the prime $p$.
One way to put it - it is about the factorization of the principal ideal $(p)$ in the ring $R=\mathbb{Z}[\sqrt{-14}]$ to a product of 2 prime ideals, $(p)=P_1P_2$. To $P_1$ one can then assign a quadratic form, namely $q(x,y)=|xu+yv|^2/p$, where $u,v$ is a $\mathbb Z$-basis of $P_1$. Both $P_1$ and $P_2$ represent $p$, simply because $p\in P_1$ (and also $p\in P_2$). So in the end this is about the bijection between the ideal classes and the classes of quadratic forms (in your case $P_1$ has always the same class as $P_2$ - you get from $P_1$ to $P_2$ via $x\to x$, $y\to-y$, which doesn't change the 2 quadratic forms you mention, and so it definitely doesn't change their classes).
edit - without ideals:
If a quadratic form with a negative even discriminant $D$ (in your case $D=-56$) represents a prime $p$, you can bring it to the form $px^2+2bxy+cy^2$, with $D/4=b^2-pc$ (if $q(\mathbf u)=p$ for some $\mathbf u\in\mathbb Z^2$, the vector $\mathbf u$ has relatively prime components, so we can change the basis of $\mathbb Z^2$ to make it the 1st basis vector).
Replacing $x$ by $x+ky$ (for suitable $k\in\mathbb Z$) we can achieve that $0\leq b<p$. But since $b^2\equiv D/4$ mod $p$, and since this congruence has only 2 solutions, there are only 2 such $b$'s. So $p$ is represented by at most 2 classes of quadratic forms. One can pass between these forms by $x\to x$, $y\to -y$ (to get $b\to-b$), i.e., in the end, we can pass between their classes by any transformation with determinant $-1$.
And your forms ($x^2+14 y^2$ and $2x^2+7y^2$) are invariant under such a transformation.
