tag:blogger.com,1999:blog-12535639.post111575502733439130..comments2018-06-24T13:49:09.957-07:00Comments on Fermat's Last Theorem: Fermat's One ProofLarry Freemanhttp://www.blogger.com/profile/06906614246430481533noreply@blogger.comBlogger19125tag:blogger.com,1999:blog-12535639.post-55293484985483816192013-12-04T07:43:42.001-08:002013-12-04T07:43:42.001-08:00a = (2pq)d
b = (p2 - q2)d
Is that "d" i...a = (2pq)d<br />b = (p2 - q2)d<br /><br />Is that "d" is necessary? Or is it just might be replaced with 1?Martynas Riaukahttps://www.blogger.com/profile/12989294632526534858noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-14628546583588135412013-06-06T14:20:24.023-07:002013-06-06T14:20:24.023-07:00Hi larry,
I am doing some reading on how XN+YN=Z...Hi larry,<br /><br /> I am doing some reading on how XN+YN=ZN+1 where n is the raised power. Is this significant in anyway?Andy Linhttps://www.blogger.com/profile/11771475170532018819noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-14405565941352855292013-06-04T10:05:59.685-07:002013-06-04T10:05:59.685-07:00Hi Andy,
The main idea is infinite descent. If t...Hi Andy,<br /><br />The main idea is infinite descent. If this were true, then there would also exist a set of positive variables where it would also be true.<br /><br />But for integers, this is impossible. There is not always an infinite number of smaller positive integers.<br /><br />The explanation behind infinite descent can be found <a href="http://fermatslasttheorem.blogspot.com/2005/05/infinite-descent.html" rel="nofollow">here</a>Larry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-8112503905866895192013-06-04T08:59:14.769-07:002013-06-04T08:59:14.769-07:00i understand where z^2 = P^4 + Q^4 comes from but ...i understand where z^2 = P^4 + Q^4 comes from but why is it not trueAndy Linhttps://www.blogger.com/profile/11771475170532018819noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-6002168500347231122013-06-04T07:40:10.534-07:002013-06-04T07:40:10.534-07:00@Lvka,
Your example is not correct.
If one side ...@Lvka,<br /><br />Your example is not correct.<br /><br />If one side is a^2 and the other side is 2*b^2, then the hypotenuse would not be an integer.<br /><br />sqrt(a^4 + 4*b^4) is not an integer.<br /><br />See <a href="http://fermatslasttheorem.blogspot.com/2005/05/pythagorean-triples-solution.html" rel="nofollow">Pythagorean Triples</a> for more information.Larry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-82641564388324960342013-06-04T07:37:09.924-07:002013-06-04T07:37:09.924-07:00The argument is up above.
Here is the summary:
(...The argument is up above.<br /><br />Here is the summary:<br /><br />(1) Assume that a right triangle with integer sides has an integer area.<br />(2) Then the following equation would be true:<br /><br />z^2 = P^4 + Q^4<br /><br />(3) But this can't be true so the conclusion follows.Larry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-20226039310908037332013-06-04T06:43:07.761-07:002013-06-04T06:43:07.761-07:00Yes but I need to understand how to prove this. Yes but I need to understand how to prove this. Andy Linhttps://www.blogger.com/profile/11771475170532018819noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-58886370045789090612013-06-03T20:36:59.293-07:002013-06-03T20:36:59.293-07:00I believe that Fermat is arguing that for a right ...I believe that Fermat is arguing that for a right triangle with integer sides, it's area cannot be the square of an integer.<br /><br />-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-9884747390376242332013-06-03T20:13:21.969-07:002013-06-03T20:13:21.969-07:00So that a 14 year old can understand.....So that a 14 year old can understand.....Andy Linhttps://www.blogger.com/profile/11771475170532018819noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-72868606946097988042013-06-03T20:03:30.883-07:002013-06-03T20:03:30.883-07:00Larry, can you explain why a right triangle cannot...Larry, can you explain why a right triangle cannot be equal to a square in simpler terms?<br /> BTW Thank you for the last answerAndy Linhttps://www.blogger.com/profile/11771475170532018819noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-28262895347006087882012-10-18T11:15:56.914-07:002012-10-18T11:15:56.914-07:00There seems to be something wrong here...
If a ri...There seems to be something wrong here...<br /><br />If a right triangle has the sides a^2 and 2*b^2, its area is (ab)^2, which is obviously the square of an integer...Lvkahttps://www.blogger.com/profile/09663692507774640889noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-90331633166183464562010-06-01T16:14:41.978-07:002010-06-01T16:14:41.978-07:00Hi Scouse Rob,
Thanks for your comment! I've...Hi Scouse Rob,<br /><br />Thanks for your comment! I've added the link to 13(b) of Pythagorean Triples: Solution.<br /><br />-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-25729901635433418962010-06-01T03:55:55.192-07:002010-06-01T03:55:55.192-07:00In Step (2), we also need the fact that p,q are of...In Step (2), we also need the fact that p,q are of different parity. (in addition to being relatively prime.)<br /><br />This fact is proved in 13(b) of the Pythagorean Triples: Solution!<br /><br />It would be helpful if there was some reference back to this parity proof, it took me a little while to back track and find it.<br /><br />Thanks<br /><br />RobScouse Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-81554235046654841002010-05-09T22:55:00.657-07:002010-05-09T22:55:00.657-07:00This comment has been removed by the author.Sathimanthahttps://www.blogger.com/profile/11955966599375080711noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-57761987718080246572008-12-11T10:56:00.000-08:002008-12-11T10:56:00.000-08:00Hi Jeremy,Great question. The proofs works becaus...Hi Jeremy,<BR/><BR/>Great question. The proofs works because we know that P^2 and Q^2 have different parity (that is, one is odd and one is even).<BR/><BR/>I will update the proof to make this point more clear.<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-87379540323188817922008-12-11T08:24:00.000-08:002008-12-11T08:24:00.000-08:00The proof you give is a complete proofof the fact ...The proof you give is a complete proof<BR/>of the fact that there is no right triangle whose area is a perfect square.<BR/>However, this is not a complete<BR/>proof of the fact that there are no<BR/>positive integer solutions to p^4 - q^4 = z^2.<BR/><BR/>The issue is the claim that P^2 - Q^2 and P^2 + Q^2 are coprime. If exactly one of P or Q is odd (which will be the case if these P and Q come from a triangle), they will be coprime. However, if both are odd, then P^2 - Q^2 and P^2 + Q^2 are both even, and so they aren't coprime!Jeremyhttps://www.blogger.com/profile/05269132567028412492noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-27325560035417936762008-08-04T10:39:00.000-07:002008-08-04T10:39:00.000-07:00I suppose I have answered my own question. Looks l...I suppose I have answered my own question. Looks like my observation is completely inconsequential.<BR/><BR/>Its obvious (upon further inspection of course) that M and N can not reach 1 and 0, since the loop I described above cannot be 'stepped' into by decrementing squares. <BR/><BR/>Not only is there not an infinite set of integers below an arbitrary integer, not all integer squares can be reached. Infinite descent holds even stronger.Andrewhttps://www.blogger.com/profile/17873772073113703774noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-11947910607787720362008-08-04T10:20:00.000-07:002008-08-04T10:20:00.000-07:00I agree with your proof. However, your conclusion ...I agree with your proof. However, your conclusion that there is not a set of infinitly desending integers such that (M^2 + N^2) and (M^2 - N^2) bothers me. <BR/><BR/>With reasonable assumptions, M and N would reach 1 and 0 respectively, and would cease to infinitly decend. 1^2 - 0 is a square, as is 1^2 +0. Their product (1)*(1) is also a square. This is where the continuing descent stops. <BR/><BR/>I contend that the proof still stands, but for a different reason.Andrewhttps://www.blogger.com/profile/17873772073113703774noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1143035091138574252006-03-22T05:44:00.000-08:002006-03-22T05:44:00.000-08:00I just want to say I really appreciate this - than...I just want to say I really appreciate this - thank you very much! <BR/><BR/>I'll be reading the whole thing over the course of time!Terry Smithhttps://www.blogger.com/profile/09611379550701673137noreply@blogger.com