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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 $F$. If $m = ax^2 + bxy + cy^2$ has a solution $(s, t)$ such that gcd$(s, t) = 1$, we say $m$ is properly 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 is meaningful.

It would be also nice if some one would post a modern form of the Gauss's proof, since not everybody can have an easy access to the book.

Theorem(Gauss: Disquisitiones Arithmeticae, art.154) Let $ax^2 + bxy + cy^2$ be a binary quadratic form over $\mathbb{Z}$. Let $D$ be its discriminant. Let $m$ be an integer. Suppose $m$ is properly represented by $ax^2 + bxy + cy^2$. Then $D$ is a quadratic residue modulo $4m$.

EDIT The Gauss's DA is notorious for its difficult read. This was even so for his contemporaries. Dirichlet devoted a lot of time to simplify DA. There is a legend that Dirichlet always carried DA in his travels. Gauss's proof often uses a "magic" equation which seems to come out of nowhere. One of the reasons is that, as he wrote, he could not afford elaborate proofs due to lack of enough available pages for an economical reason. So I think it would be nice if there is a more natural proof.

Makoto Kato
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    Doesn't he prove this in his book? – Pedro Sep 04 '12 at 23:20
  • @PeterTamaroff You know there are usually several different correct proofs of a mathematical theorem. – Makoto Kato Sep 04 '12 at 23:22
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    You should be saying what kind of proof you want. I'm reading the proof just now. – Pedro Sep 04 '12 at 23:23
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    Gauss's proof of art.154 in my copy seems to be very clear and concise - what problem do you have with his proof, exactly? – Old John Sep 04 '12 at 23:24
  • @JohnSenior I think having a different proof doesn't hurt. – Makoto Kato Sep 04 '12 at 23:25
  • Yes, But you haven't answered my question: what problem do you have with his clear and concise proof? – Old John Sep 04 '12 at 23:26
  • @PeterTamaroff A different proof. If there are several different proofs, that would be nice. – Makoto Kato Sep 04 '12 at 23:27
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    @JohnSenior Do I have to have a problem with his proof in searching for other proofs? – Makoto Kato Sep 04 '12 at 23:29
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    But do you want what: a more contemporary proof? Shorter? Clearer? More convoluted? Using other theory? Avoiding certain axioms? Let's put down some variables! – Pedro Sep 04 '12 at 23:30
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    It might be nice if you actually answered a question. – Old John Sep 04 '12 at 23:31
  • @PeterTamaroff Any correct different proof will do. – Makoto Kato Sep 04 '12 at 23:33
  • @JohnSenior Since the theorem is important, I think it's natural to looking for the different proofs. – Makoto Kato Sep 04 '12 at 23:35
  • Still no answer - I sign off – Old John Sep 04 '12 at 23:37
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    I have no idea why this question was downvoted. I don't think asking different proofs should be frowned upon. – Makoto Kato Sep 04 '12 at 23:45
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    What is frowned upon is that you don't give any hint of what you really want. I don't think there are many people that are going to read this and go "OK, let's find some proof, using some theory, that Makoto will like." Most seem to be cool with Gauss' proof. Maybe you can answer what we ask you so you can get what you want. "But do you want what: a more contemporary proof? Shorter? Clearer? More convoluted? Using other theory? Avoiding certain axioms? Let's put down some variables!" – Pedro Sep 04 '12 at 23:54
  • @PeterTamaroff I said any correct proof will do. – Makoto Kato Sep 04 '12 at 23:57
  • @MakotoKato But not Gauss'? – Pedro Sep 04 '12 at 23:59
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    @PeterTamaroff Please read my question. – Makoto Kato Sep 04 '12 at 23:59
  • OK, I can do something for you then. – Pedro Sep 05 '12 at 00:00
  • Someone should ask a meta question about how to ask for alternative proofs of results with well-known or readily available proofs. – William Sep 05 '12 at 00:05
  • @William Why don't you do that if you think so? – Makoto Kato Sep 05 '12 at 00:21
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    I confess I am greatly saddened by the 6 downvotes to this question. I cannot imagine why anyone would downvote it. Please be more constructive. If you think the question needs improvement then please say why and/or help to improve it. – Bill Dubuque Sep 05 '12 at 01:21
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    I understand that MSE, MO, and many other venues are viewed as opportunities for "competition", adversarial attitudes, and so on, but I wish it were otherwise. That is, I would wish that questions would not be asked as "challenges", but as requests for information. I am oh-so-well aware of the competition element in mathematics... but this is of no consequence to furthering it. – paul garrett Sep 05 '12 at 01:25
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    @Peter Many of the proofs in Gauss's Disq. Arith. are very old-fashioned or far from optimal; nowadays they can be presented much more slickly and with greater insight. So it is quite natural to pose a question such as that above. Further, comparing old proofs to new proofs often has great pedagogical value. – Bill Dubuque Sep 05 '12 at 01:27
  • @JohnSenior Please see my prior comment. – Bill Dubuque Sep 05 '12 at 01:28
  • @paulgarrett I see nothing at all "challenging" in MK's question. But, alas, I cannot say the same for some of the comments. – Bill Dubuque Sep 05 '12 at 01:30
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    @BillDubuque I concurr with you. I upvoted this, and I'm really trying to help. I hope I'm coming off as that, and not otherwise. Would you give a hand with the mysterious product Gauss introduces? I recall you do some of that moves in your proofs too! – Pedro Sep 05 '12 at 01:31
  • @JohnSenior The Gauss's proof is correct. I have no problem with it except that I have no idea how he came up with the crucial equation of his proof. – Makoto Kato Sep 05 '12 at 02:11
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    @PeterTamaroff it is Gauss composition, which takes two forms of the same discriminant and gives a third. It respects "equivalence." On the level of equivalence classes of forms, it makes a group. The difficult part of that proof is associativity. Dirichlet made a big simplification with his "united" pairs of forms. Many modern books do Gauss composition, some better than others. I like Buell. For one thing he does full detail on positive forms and indefinite forms in the same volume. – Will Jagy Sep 05 '12 at 02:15
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    I'm also upvoting this question, more as a protest for the easy downvotes than anything else. It is always nice to have several proofs of some given result , e.g. see the galore of proofs of $,\zeta(2)=\pi^2/6,$ . – DonAntonio Sep 05 '12 at 02:17
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    @WillJagy Interesting! – Pedro Sep 05 '12 at 02:25
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    I voted to close this question. Any request for different proofs, etc., should include some motivation. Saying "not everybody can have an access to the book" is not motivation at all, and has never been a part of what math.SE is about. –  Sep 05 '12 at 03:13
  • @SteveD "Any request for different proofs, etc., should include some motivation." Could you explain why this is so? Anyway, I wrote that the theorem is important. That's one of my motivations. – Makoto Kato Sep 05 '12 at 03:28
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    @MakotoKato: it is like if I asked a question, "reading my linear algebra book, it said every vector space has a basis. Can I see different proofs of this result?". Why do you want different proofs? If questions that simply say "need different proofs of X" were allowed, the site would be inundated with such requests. –  Sep 05 '12 at 03:31
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    @SteveD My question is not of that sort. I took for granted that my question should be of interest for anyone who has the basic knowledge of elementary number theory. – Makoto Kato Sep 05 '12 at 03:36
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    @MakotoKato: fair enough. But you asked for different proofs, with the original reason being that not everyone had access to the book. Indeed, the highest voted answer as of now is simply the exact proof from the book you mentioned. I don't see why this question needed to be asked in the first place, but I will agree it has evolved into something worthwhile via the comments and subsequent questions. –  Sep 05 '12 at 18:19
  • @SteveD Do you still think this question should be closed? If yes, why? – Makoto Kato Sep 07 '12 at 02:06
  • @JohnSenior I answered your question. So I think there is no reason for you to sign off unless you have no problem with my question. – Makoto Kato Sep 07 '12 at 02:06
  • @SteveD Did you reset your vote for close? – Makoto Kato Sep 08 '12 at 09:49
  • I certainly did not. Again, it was the great community here and their thoughtful and insightful comments that made this page worth visiting. The question had nothing to do with it. –  Sep 08 '12 at 14:35
  • @SteveD You said the lack of the motivation was the reason for the vote to close. I wrote the motivation. – Makoto Kato Sep 08 '12 at 14:57
  • @SteveD I'm waiting for your reply. This is important for this thread. – Makoto Kato Sep 08 '12 at 23:27
  • It is not important for this thread at all, where are you getting that? Please stop pinging me, my interest in this thread has waned. –  Sep 09 '12 at 02:42
  • @SteveD You voted to close this thread. It is obviously important to this thread(it takes only 5 votes to close the thread). I'm surprised you don't recognize this. Please use your privilege with discretion. If the reason for the vote is no longer applicable to the question, please reset the vote. – Makoto Kato Sep 09 '12 at 03:57
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    It is a privilege I have earned, and it not your right to question what I do with it, just as I don't question why you ask just so many questions! I voted to close this question because it was worthless, and now by the magic of the community, has turned into something worthwhile. That has nothing to do with your original question. In fact, as activity has faded, I would indeed like to see this question closed. Even more so because you can't stop responding to me after I've clearly asked you to do so. –  Sep 09 '12 at 16:55
  • @SteveD I asked proofs more natural than the Gauss's original proofs because his proof seems to come out of nowhere. Please explain why my question is worthless. – Makoto Kato Sep 09 '12 at 20:51

3 Answers3

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"In this section, we'll deal primarily with bivariate functions of $x$ and $y$ of the form:

$$\tag 1 f(x,y)=ax^2 + 2bxy + cy^2$$(...) When we're not interested in the indeterminates $x,y$ we'll refer to $(1)$ as $(a,b,c)$

We'll say a number $M$ is represented in the form $(a,b,c)$ if $x,y$ can be given values such that $M=f(x,y)$.

Theorem. If the number $M$ can be represented in the form $(a,b,c)$ such that $(x,y)=1$, then $b^2-ac$ will be a quadratic residue modulus $M$.

Proof Let $m,n$ be the values of the indeterminates, that is $$ am^2 + 2bmn + cn^2 = M$$ and take $\mu$ and $\nu$ such that $m\mu+n\nu=1$. Then by multipying out we can easily show that

$$\left( {a{m^2} + 2bmn + c{n^2}} \right)\left( {a{\nu ^2} - 2b\nu \mu + c{\mu ^2}} \right) = $$

$${\left( {\mu \left( {mb + nc} \right) - \nu \left( {ma + nb} \right)} \right)^2} - \left( {{b^2} - ac} \right){\left( {m\mu + n\nu } \right)^2}$$

or $$M\left( {a{\nu ^2} - 2b\nu \mu + c{\mu ^2}} \right) = {\left( {\mu \left( {mb + nc} \right) - \nu \left( {ma + nb} \right)} \right)^2} - \left( {{b^2} - ac} \right)$$

Thus it will be the case$${b^2} - ac \equiv {\left( {\mu \left( {mb + nc} \right) - \nu \left( {ma + nb} \right)} \right)^2}\bmod M$$ as we claimed.

Pedro
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    I don't know how much simpler it can get, really. – Pedro Sep 05 '12 at 00:10
  • Do you think you could come up with the proof by yourself? As for me, I don't. – Makoto Kato Sep 05 '12 at 00:19
  • Is this now a matter of personal achievement? – Pedro Sep 05 '12 at 00:42
  • @MakotoKato To come up with a proof of a theorem by yourself that someone else already proved is probably not that useful. However a proof that is aesthetically pleasing is helpful for educational purpose, or an argument that is understandable can be applied to solve new questions. – William Sep 05 '12 at 00:43
  • @William I have no idea how Gauss came up with the above equation(the long one). – Makoto Kato Sep 05 '12 at 00:46
  • @MakotoKato I think Bill Dubuque might be of great help here. – Pedro Sep 05 '12 at 01:13
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    @PeterTamaroff the identity is Gauss composition, probaly the simplest occurrence. Taking $\Delta = b^2 - a c,$ the "principal form" of that discriminant i the one that represents $1,$ here $\langle 1,0, -\Delta \rangle.$ You wrote $ \langle a,2b,c \rangle \circ \langle a, -2b, c \rangle = \langle 1,0, -\Delta\rangle $ with the relevant values of the "variables" included. Versions exist with the middle coefficient odd. See Duncan A. Buell, Binary Quadratic Forms. Page 57 for "united" forms, pages 61-65 in general. The forms become a group, historically the first nontrivial one. – Will Jagy Sep 05 '12 at 01:47
  • @WillJagy I guess you can really help here with the modern part of the story. Care to leave an answer? – Pedro Sep 05 '12 at 02:27
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    @Peter See Will's comment on composition of forms. A more readable modern presentation of Gauss is George B. Mathews Theory of Numbers (free via Google Books). One can also clarify related proofs by translating them into the (linear!) language of ideals / modules, see here. – Bill Dubuque Sep 05 '12 at 02:38
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By the proposition of this question, there exist integers $l, k$ such that $ax^2 + bxy + cy^2$ and $mx^2 + lxy + ky^2$ are equivalent. Hence $D = l^2 - 4mk$. Hence $D \equiv l^2$ (mod $4m$).

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Makoto Kato
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Let $f(x, y) = ax^2 + bxy + cy^2$. Suppose $m = ax^2 + bxy + cy^2$ has a solution $(p, r)$ such that gcd$(p, r) = 1$. There exist integers $s, q$ such that $ps - qr = 1$. Suppose $f(pu + qv, ru + sv) = m'u^2 + luv + kv^2$, where $u, v$ are indeterminate variables. Then

$m' = ap^2 + bpr + cr^2$

$l = 2apq + b(ps + qr) + 2crs$

$k = aq^2 + bqs + cs^2$

Hence $m' = m$.

Since $D = l^2 - 4mk$, $D \equiv l^2$ (mod $4m$).

Makoto Kato
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