Quadratic residue character determined by a binary quadratic form Let $F = ax^2 + bxy + cy^2$ be a binary quadratic form over $\mathbb{Z}$.
We say $D = b^2 - 4ac$ is the discriminant of $F$.
Let $m$ be an integer.
If $m = ax^2 + bxy + cy^2$ has a solution in $\mathbb{Z}^2$, we say $m$ is represented by $F$.
My question
Is there any other proof of the following theorem other than the Gauss's original proof?
Since this theorem is important, I think having different proofs would be nice.
It would be also nice if some one would post a modern form of the Gauss's proof, because not everybody can have an easy access to the book. 
Theorem(Gauss: Disquisitiones Arithmeticae, art.229)
Let $F = ax^2 + bxy + cy^2$ be a binary quadratic form of discriminant $D$.
Suppose $D$ is not a square integer.
Let $p$ be an odd prime divisor of $D$.
Let $m$ and $k$ be integers which are not divisible by $p$.
Suppose $m$ and $k$ are represented by $F$.
Then $\left(\frac{m}{p}\right) = \left(\frac{k}{p}\right)$.
Remark
The above result and this question suggest that the repesentations of integers by an integral binary quadratic form might have a connection with the quadratic reciprocity law.
 A: (Edited to incorporate material from comment)
We assume $ax^2+bxy+cy^2=m$, and $p$ is an odd prime divisor of the discriminant $D$. 
If $p$ divides $a$, then it also divides $b$, so $cy^2\equiv m\pmod p$, so  ${m\overwithdelims()p}={c\overwithdelims()p}$. So, let's assume $p$ does not divide $a$. Then we get $$\displaylines{4a^2x^2+4abxy+4acy^2=4am\cr(2ax+by)^2+(4ac-b^2)y^2=4am\cr(2ax+by)^2\equiv4am\pmod p\cr}$$ and we see that ${m\overwithdelims()p}={a\overwithdelims()p}$.
A: The following proof is basically Gauss's. 
Let $(p, r)$ be an integer solution of $m = ax^2 + bxy + cy^2$.
Let $(q, s)$ be an integer solution of $k = ax^2 + bxy + cy^2$.
Let $f(px + qy, rx + sy) = Ax^2 + Bxy + Cy^2 $.
Then
$A = ap^2 + bpr + cr^2$
$B = 2apq + b(ps + qr) + 2crs$
$C = aq^2 + bqs + cs^2$
Hence $A = m$ and $C = k$.
Since $B^2 - 4AC = D(ps- qr)^2$, $B^2 - 4mk = D(ps- qr)^2$.
Hence $4mk \equiv B^2$ (mod $D$).
In particular $4mk \equiv B^2$ (mod $p$).
Hence $\left(\frac{4mk}{p}\right) = 1$.
Hence $\left(\frac{mk}{p}\right) = 1$.
Hence $\left(\frac{m}{p}\right) = \left(\frac{k}{p}\right)$ as desired.
