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.

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    $\begingroup$ I cannot imagine why someone would downvote this question, esp. only 1 minute after the question was posted. $\endgroup$ – Bill Dubuque Sep 5 '12 at 1:15
  • $\begingroup$ @BillDubuque I think you can help in his related question, particulary on a rationale to get that slick multiplication equality in the proof I posted. $\endgroup$ – Pedro Tamaroff Sep 5 '12 at 1:15
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    $\begingroup$ I thought when you came back to edit this question you might have something to say about the answer that has been up for several days. $\endgroup$ – Gerry Myerson Sep 8 '12 at 6:27
  • $\begingroup$ @GerryMyerson I'm afraid that careless readers might misunderstand your answer. If you correct your answer, I will accept it with appreciation. Regards, $\endgroup$ – Makoto Kato Sep 8 '12 at 8:27
  • $\begingroup$ I've done some editing. Perhaps you can let me know if there's anything specific still in need of correction. $\endgroup$ – Gerry Myerson Sep 8 '12 at 12:28

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

  • $\begingroup$ Thanks. Could you explain why $a$ is not divisible by $p$? $\endgroup$ – Makoto Kato Sep 5 '12 at 2:31
  • $\begingroup$ If $p$ divides $a$, then it also divides $b$, so $cy^2\equiv m\pmod p$, so $(c|p)=(m|p)$. $\endgroup$ – Gerry Myerson Sep 5 '12 at 2:42

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.


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