6
$\begingroup$

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.

  • 0
    I've done some editing. Perh$a$ps you can let $m$e k$n$ow if the$r$e's a$n$ything specific still in need of correction.2012-09-08

2 Answers 2

5

(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 ${m\overwithdelims()p}={a\overwithdelims()p}$.

  • 0
    If $p$ divides $a$, then it also divides $b$, so $cy^2\equiv m\pmod p$, so $(c|p)=(m|p)$.2012-09-05
0

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.