tag:blogger.com,1999:blog-12535639.post6655198572339029837..comments2024-02-26T06:55:41.876-08:00Comments on Fermat's Last Theorem: Sturm's Theorem: Sturm ChainsLarry Freemanhttp://www.blogger.com/profile/06906614246430481533noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-12535639.post-30272015350704611402009-06-15T08:47:45.947-07:002009-06-15T08:47:45.947-07:00Hi Larry
I have a 1 page proof of Fermat's Las...Hi Larry<br />I have a 1 page proof of Fermat's Last Theorem which I believe is the original proof that Fermat made.Check it out. Here is the link<br /><br />http://www.youtube.com/watch?v=gwbHORdjcbQ<br /><br />MichaelReliable Home Inspectionhttps://www.blogger.com/profile/09397205998512916607noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-33319460412072433002009-02-04T16:43:00.000-08:002009-02-04T16:43:00.000-08:00Hi Ross,Actually z doesn't have to be odd. See my...Hi Ross,<BR/><BR/>Actually z doesn't have to be odd. See my example where z=10.<BR/><BR/>You are right. My proof only relates to integers. <BR/><BR/>The Pythagorean Triples are usually limited to integers because it is harder to come up with integers than real numbers that satisfy x^2 + y^2 = z^2.<BR/><BR/>For example, regardless of the value of x or y, I could set z = sqrt(x^2 + y^2) to get the third triple.<BR/><BR/>I think that your method is interesting because it is not obvious that it works.<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-89109592074312848242009-02-04T13:21:00.000-08:002009-02-04T13:21:00.000-08:00Got it. I'm a bit confused (and looking for reason...Got it. I'm a bit confused (and looking for reasons to not work on a grant): isn't this proof for something that is different (or at least just a subset) of 'my' equation? A rule you postulate is that Z must be odd, but I don't see why that is fundamental unless you limit the set to whole numbers.Ross_Cagan@me.comhttps://www.blogger.com/profile/15911288803575229490noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-61635180728554221562009-02-03T20:31:00.000-08:002009-02-03T20:31:00.000-08:00Hi Ross,Here's a solution that I blogged about in ...Hi Ross,<BR/><BR/>Here's a solution that I blogged about in an earlier blog entry.<BR/><BR/>Let d,p,q be any integers that you want.<BR/><BR/>z = d[p^2 + q^2]<BR/>y = d[p^2 - q^2]<BR/>x = d[2pq]<BR/><BR/>For example, if p = 2 and q = 1 and d = 2, we get:<BR/>z = (2)(5) = 10<BR/>y = (2)(3) = 6<BR/>x = (2)(4) = 8<BR/><BR/>36 + 64 = 100.<BR/><BR/>The proof can be found <A HREF="http://fermatslasttheorem.blogspot.com/2005/05/pythagorean-triples-solution.html" REL="nofollow">here</A>.<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-88400588695183155372009-02-03T16:29:00.000-08:002009-02-03T16:29:00.000-08:00We (my daughter and I) were wondering: what other ...We (my daughter and I) were wondering: what other equations are used to calculate P triplets?Ross_Cagan@me.comhttps://www.blogger.com/profile/15911288803575229490noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-71995196217289456482009-02-03T05:07:00.000-08:002009-02-03T05:07:00.000-08:00Thanks, Larry. An elegant proof.Thanks, Larry. An elegant proof.Ross_Cagan@me.comhttps://www.blogger.com/profile/15911288803575229490noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-35606280475891084922009-02-02T18:29:00.000-08:002009-02-02T18:29:00.000-08:00Hi Ross,I checked your formulas and they work.Here...Hi Ross,<BR/><BR/>I checked your formulas and they work.<BR/><BR/>Here are my notes:<BR/><BR/>Z^2 = (y+1)^2 = y^2+2y+1 = [Ax+A]^2 + 2[Ax+A] + 1 <BR/><BR/>X^2+Y^2 = X^2 + [Ax+A]^2 <BR/><BR/>2A = x - 1<BR/><BR/>2[Ax+A]+1 = 2Ax + 2A + 1 = x[x-1] + [x-1] + 1 = x^2-x + x - 1 +1 = x^2<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-39724624187139940072009-02-02T17:39:00.000-08:002009-02-02T17:39:00.000-08:00Hi Larry,I stumbled on your blog and have a questi...Hi Larry,<BR/><BR/>I stumbled on your blog and have a question. I was noting to my daughter my magic equation for calculating Pythagorean Triplets:<BR/><BR/>y = (Ax)+A<BR/>z = y+1<BR/>where...<BR/>A= (0.5x-0.5)<BR/><BR/>It works with any real number I've tried (e.g., 8, 31.5, 32.5) including fractions, negative numbers, etc, though I haven't tried many. True? I was relating the story of how describing it to my 6th grade teacher eventually got me kicked out of class (long story).<BR/><BR/>Your input would be appreciated.<BR/><BR/>Thanks...Ross<BR/>ross.cagan@mssm.eduRoss_Cagan@me.comhttps://www.blogger.com/profile/15911288803575229490noreply@blogger.com