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[edit] OriginBasque origin is confirmed in the most recent EB but not the 1911 EB. VivaEmilyDavies 18:58, 13 Apr 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.
The lead sentence (currently) states Fermat's birthdate as "17 August 1601 or 1607/8", but no explanation is given anywhere in the article on why two dates are shown. Could someone please provide some explanatory text (perhaps in the "Life and Work" section)? Also, the link in the citation following the "1607/8" date (ref 1) is dead and should probably be replaced. — Loadmaster (talk) 17:33, 22 September 2009 (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?
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 FermatI'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.
[edit] PronunciationIt would be nice if there were a note in the beginning about how to pronounce "Fermat." Since he was from France, I'm fairly confident it's not "fur-mat", but I'm not actually sure how it's pronounced. Ketsuekigata 21:57, 26 April 2007 (UTC) [edit] Possible plagiarism (brought up by 68.38.6.68)http://en.wikipedia.org/w/index.php?title=Pierre_de_Fermat&oldid=169141767 68.38.6.68 believes that the content from this page has been copied from [1], can anyone verify this? —Preceding unsigned comment added by Teh roflmaoer (talk • contribs) 14:50, 4 November 2007 (UTC) [edit] CorrespondenceI am tempted to remove the text Fermat was ... a recluse. His only contact with the wider mathematical community aside from a brief exchange of letters with Blaise Pascal, was Marin Mersenne. from the Life and work section (I agree that he was secretive). Although this section is sourced, Singh is not a particularly reliable source here. For example, Fermat carried on correspondence with Roberval, Frenicle, Carcavi, Digby, Gassendi, and Huygens. Any objections to its removal? — Myasuda (talk) 03:54, 26 February 2008 (UTC) [edit] Occitan and BasqueHello, At the top of this talk page there's a little discussion about Fermat's Basque origin. This indeed appears in the current article about him in Britannica online. It is nevertheless rather unusual, because his last name sounds much more Occitan or Catalan than Basque or French. Indeed, Occitan and Catalan Wikipedias call him "an Occitan mathematician", although without a clear reference, and this article here in English includes him in the category of Occitan people, again without a clear reference. He is, of course, best known as a French mathematician, but if the information about his Basque or Occitan origin is true, the article may say that he is "a French mathematician of Basque and Occitan origin". --Amir E. Aharoni (talk) 07:45, 18 August 2009 (UTC) Categories: C-Class biography articles | C-Class biography (core) articles | Core biography articles | Top-priority biography articles | C-Class biography (science and academia) articles | Top-priority biography (science and academia) articles | Science and academia work group articles | Biography articles with comments | WikiProject Biography articles | C-Class France articles | Top-importance France articles | France articles with comments | Mathematics articles related to mathematicians | Frequently viewed mathematics articles | Start-Class mathematics articles | High-Priority mathematics articles | Vital mathematics articles | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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