Talk:Brahmagupta–Fibonacci identity

It is true this identity can be interpreted using norms for Q(i) but wouldn't it be more simple just to relate to the module property of complex numbers? It would also extend the property to real numbers. Since this identity is, as far as I know, generally used in the context of integer or rational numbers I would keep the norm comment also.Ricardo sandoval 06:41, 8 April 2007 (UTC)

I change the article accordingly but maybe the section "Interpretation via norms" can be compressed since its essentially the same as the relation to the complex numbers part. Ricardo sandoval 18:49, 29 April 2007 (UTC)

Not Correct

As I understand it, Brahmagupta's Identity is:

$\ (x^{2}-Ny^{2})(x'^{2}-Ny'^{2})=(xx'+Nyy')^{2}-N(xy'+x'y)^{2}$

which is a generalization of an earlier identity of Diophantus: The identity was used by Brahmagupta to generate solutions of Pell's equation.

Diophantus' Identity

$\ (x^{2}+y^{2})(x'^{2}+y'^{2})=(xx'-yy')^{2}+(xy'+x'y)^{2}$

Some one should correct this. Fowler&fowler«Talk» 14:21, 7 June 2007 (UTC)

PS the identity can also be formulated by using the Brahmagupta matrix (although Brahmagupta, himself, didn't really define a matrix). Fowler&fowler«Talk» 14:27, 7 June 2007 (UTC)

Hmm, you seem to be right. Here's page 72 of Stillwell. Brahmagupta's identity seems more general than the one here, which corresponds to N=-1. But if it was already known to Diophantus, why did Fibonacci's name get associated with it, so many centuries later? This is weird. Shreevatsa (talk) 23:35, 25 June 2010 (UTC)

Not only did this give a way to generate infinitely many solutions to x2 − Ny2 = 1 starting with one solution,

actually, you need two solutions. — Preceding unsigned comment added by 109.58.34.100 (talk) 12:33, 15 July 2012 (UTC)

Actually, it doesn't look like it. Why would you need two? 110.23.118.21 (talk) 10:35, 29 July 2016 (UTC)

references and punctuation

My understanding is, that if you source the sentence then reference comes after the punctuation. If however you just reference a term/name/phrase within a sentence then reference belongs next to that term/name/phrase and before any following punctuation. I guess in this case you can see it as either referencing either just the name or the whole sentence. However seeing it as just referencing the name would match the first name sourcing above it (brahmagubta-fibonacci) in style.--Kmhkmh (talk) 05:11, 18 July 2017 (UTC)

Well, your understanding goes against wikipedia conventions. With the exceptions of dashes and closing parentheses references should always come after punctuation. See MOS:PUNCTFOOT. Sapphorain (talk) 06:14, 18 July 2017 (UTC)

WP never ceases to amaze.--Kmhkmh (talk) 07:01, 18 July 2017 (UTC)

minus sign in concrete example

As of this writing the definition and first example on the page is the following:

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${\begin{aligned}\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)&{}=\left(ac-bd\right)^{2}+\left(ad+bc\right)^{2}&&(1)\\&{}=\left(ac+bd\right)^{2}+\left(ad-bc\right)^{2}.&&(2)\end{aligned}}$

For example,

$(1^{2}+4^{2})(2^{2}+7^{2})=26^{2}+15^{2}=30^{2}+1^{2}.$

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I think it would be clearer if the minus sign was not left off of the 26, resulting in:

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${\begin{aligned}\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)&{}=\left(ac-bd\right)^{2}+\left(ad+bc\right)^{2}&&(1)\\&{}=\left(ac+bd\right)^{2}+\left(ad-bc\right)^{2}.&&(2)\end{aligned}}$

For example,

$(1^{2}+4^{2})(2^{2}+7^{2})=(-26)^{2}+15^{2}=30^{2}+1^{2}.$

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Yes, both of these options result in the same value when evaluated, but it just seems like it makes it less clear that the $26^{2}$ comes from $(ac-bd)^{2}=((1\times 2)-(4\times 7))^{2}=(2-28)^{2}=(-26)^{2}=(26)^{2}$ if the minus sign is left out.

Ryan 1729 (talk) 00:32, 25 October 2022 (UTC)