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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
5
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.