Talk:Pierre de Fermat

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This article is within the scope of WikiProject Mathematics.
Mathematics grading:	Start Class	High Importance	Field: Mathematicians
A vital article
Tompw (talk) 19:46, 1 March 2007 (UTC)

Contents 1 Origin 2 Birthdate 3 last theorem... 4 Cultural depictions of Pierre de Fermat 5 2 methods of FLT proof and Pythagorean triples

[edit] Origin

Basque origin is confirmed in the most recent EB but not the 1911 EB. VivaEmilyDavies 18:58, 13 Apr 2005 (UTC)

The claim needs to be substantiated. I am removing it for now. Fermat does not sound like a Basque name at all. Besides, Pierre Fermat was born in Lomagne, an area in the very north of Gascony far away from the Basque Country. If it is, say, his great-grand-father that came from the Basque Country, it's a bit far fetched to present Fermat as a mathematician of Basque origin. In Southwest France, he's always presented as a celebrity of Languedoc (even though he was actually born in Lomagne, at the border of Languedoc, but he served all his life as a lawyer in Languedoc). Hardouin 20:29, 21 July 2005 (UTC)

One of Fermat's forefathers immigrated around 1500 to the North of Gascony most probably from Catalonia. The Cistercian monastary of Grand Selve (which had founded Fermat's birthplace Beaumont-de-Lomagne in 13th century) had excellent relations to Catalonia (daughter monastaries) and invited families from Catalonia to come to Beaumont, one of those numerous bastides (fortified villages with market right) which were founded in the South of France after the Albigensian wars to repopulate the depopulated country. Klaus Barner 10:31 10 November 2006

[edit] Birthdate

"Fermat's date of birth is usually given as 1601; recently it has been suggested that the correct date is 1607." The preceding two statements appears on page 3 of the February 2005 issue of Mathematics Magazine.

The birthdate on this page (August 20th) is not consistent with the Wikipedia births page for August 17th.

Recently, an anonymous contributor (203.103.111.231) has changed the birthdate from August 17 to August 20, with only a nonspecific comment "incorrect date of birth". He has not given any source for the claim and all sources I am able to check right now (i.e. internet only) give August 17. So I have reverted the change. If someone has a specific source, feel free to change it back, and cite the source. --Mormegil 13:57, 19 October 2005 (UTC)

There is strong evidence that Fermat was born in 1607:

Barner, Klaus: How old did Fermat become? (English)

[J] NTM (N.S.) 9, No.4, 209-228 (2001). [ISSN 0036-6978]

reviewed in: http://www.emis.de/cgi-bin/zmen/ZMATH/en/quick.html?type=html&an=1001.01006&format=complete

There is a nice paragraph on this in the german wikipedia. If there is interest, I can translate.

Simon Singh's "Fermat's Last Theorem" book says he was born on August 20th, 1601. Yet, I'm not confident about its validity in comparison to the other sources, so I didn't update the term. AilaG 22:33, 8 February 2006 (UTC)

[edit] last theorem...

It is my understanding that it is now widely acccepted that Fermat did not have a proof of his so called last theorem. Any one have anything to say about that?

Yet, it has not been proven as a fact. If you have any proof that the claim is serious (i.e. that there is a serious debate about it) I think you can add it, along with the names and titles of those who raised the claim. It is common for people to just guess that he didn't have the proof, and I personally agree with that guess, but Wikipedia's about facts, not smart guesses. AilaG 22:26, 8 February 2006 (UTC)

Fermat's Enigma is also known as Fermat's last theorem, correct? Evil Deep Blue 00:04, 31 May 2006 (UTC)

[edit] Cultural depictions of Pierre de Fermat

I've started an approach that may apply to Wikipedia's Core Biography articles: creating a branching list page based on in popular culture information. I started that last year while I raised Joan of Arc to featured article when I created Cultural depictions of Joan of Arc, which has become a featured list. Recently I also created Cultural depictions of Alexander the Great out of material that had been deleted from the biography article. Since cultural references sometimes get deleted without discussion, I'd like to suggest this approach as a model for the editors here. Regards, Durova 16:00, 18 October 2006 (UTC)

[edit] 2 methods of FLT proof and Pythagorean triples

2 methods of FLT proof and Pythagorean triples X^n+Y^n=Z^n Fermat had made a proof that the equation cannot have nonzero natural number solution in the even number n that is greater or equal 4. Therefore we need to make a proof that the equation cannot have nonzero natural number solution in the odd and prime number n. Y+A=X+B=Z, A=Z-Y, B=Z-X X-A=Y-B=Z-A-B=X+Y-Z G=(X-A)/(AB)^(1/n)=(Y-B)/(AB)^(1/n)=(Z-A-B)/(AB)^(1/n)=(X+Y-Z)/(AB)^(1/n) X=G(AB)^(1/n)+A, Y=G(AB)^(1/n)+B, Z=G(AB)^(1/n)+A+B {G(AB)^(1/n)+A}^n+{G(AB)^(1/n)+B}^n={G(AB)^(1/n)+A+B}^n When n=1, G=0. When n=2, G=2^(1/2). When n>2, G=function(A,B) is the positive real number. X=(2AB)^(1/2)+A, Y=(2AB)^(1/2)+B, Z=(2AB)^(1/2)+A+B (X,Y,Z) are the irrational numbers or all Pythagorean triples in all natural number (A,B). We can translate the upper form into this. AB=2k^2, B=2k^2/A X=2k+A, Y=2k(k+A)/A, Z=2k+A+2k^2/A XY=2k(2k+A)(k+A)/A When A is the odd number, k=hA, XY=2A^2h(2h+1)(h+1) and hk=A, XY=2k^2(2+h)(1+h)/h When A is the even number, 2k=hA, XY=A^2h(h+1)(h+2)/2 and 2hk=A, XY=2k^2(1+h)(1+2h)/h Therefore XY cannot be the power numbers in all Pythagorean triples.

• * * * * 1st method of FLT proof * * * * *

G(AB)^(1/n) is the irrational number in all natural number (A,B), so (X,Y,Z) are the irrational numbers. {G(AB)^(1/n)+A}^n+{G(AB)^(1/n)+B}^n={G(AB)^(1/n)+A+B}^n When A=B, 2{GA^(2/n)+A}^n={GA^(2/n)+2A}^n {2^(1/n)-1}GA^(2/n)={2-2^(1/n)}A G=[2^{(n-1)/n}+…+2^(2/n)+2^(1/n)]A^{(n-2)/n} We make new form with [2^{(n-1)/n}+…+2^(2/n)+2^(1/n)]. [2^{(n-1)/n}+…+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)]/2 This form is the irrational number in all natural number (A,B). G(AB)^(1/n) divide and multiply by [2^{(n-1)/n}+…+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)]/2, and now we can get two forms. And when A=B, q=1. G(AB)^(1/n)=q[2^{(n-1)/n}+…+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)]/2 q=2G(AB)^(1/n)/[2^{(n-1)/n}+…+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)] If G(AB)^(1/n) is the natural number (N) in some (a,b), G(ab)^(1/n)=N can not have [2^{(n-1)/n}+…+2^(2/n)+2^(1/n)], and G(AB)^(1/n) can not have [2^{(n-1)/n}+…+2^(2/n)+2^(1/n)]. So when A=B, q cannot be 1. That is an apparent contradiction. Therefore G(AB)^(1/n) is the irrational number in all natural number (A,B). 1st method. end.

• * * * * 2nd method of FLT proof * * * * *

X^n+Y^n=Z^n {X^(n/2)}^2+{Y^(n/2)}^2={Z^(n/2)}^2 When n=2, we can display {X^(n/2),Y^(n/2),Z^(n/2)} with (a,b). a=Z^(n/2)-Y^(n/2), b=Z^(n/2)-X^(n/2) X^(n/2)=(2ab)^(1/2)+a, Y^(n/2)=(2ab)^(1/2)+b, Z^(n/2)=(2ab)^(1/2)+a+b When n is the prime number and (X,Y,Z) is co prime, the ab is the irrational number. ab=Z^n-(YZ)^(n/2)-(XZ)^(n/2)+(XY)^(n/2) We multiply X^(n/2) and Y^(n/2). (XY)^n=2a^3b+2ab^3+13(ab)^2+6ab(a+b)(2ab)^(1/2) If (X,Y,Z) is the natural number, (XY)^n is the natural number, but 2a^3b+2ab^3+13(ab)^2+6ab(a+b)(2ab)^(1/2) is the irrational number. That is an apparent contradiction. Therefore (X,Y,Z) must be the irrational number. 2nd method. end.

Someone should rewrite the above in a 'proper math notation'. While the above may certainly be correct, it is nigh unreadable in its current state. --Nthitz 20:36, 6 January 2007 (UTC)