rayb07 Posted November 3, 2008 Share Posted November 3, 2008 Posted below is a link to an attempted proof of FLT that I haven't been able to find a flaw in: http://groups.google.com/group/fermats-last-theorem-flt-2008 Link to comment Share on other sites More sharing options...
AlphaNumeric Posted November 10, 2008 Share Posted November 10, 2008 The proof claims that x^p + y^p can be divided by (x+y) to give an integer. That's clearly false. x=3, y=4, z=5 are pairwise coprime and p=2 gives 3^2 + 4^2 = 5^2 but x+y = 7 and clearly 7 doesn't go into 5^2. Therefore the logic employed in the first paragraph is wrong and kills the 'proof'. Link to comment Share on other sites More sharing options...
rayb07 Posted November 10, 2008 Author Share Posted November 10, 2008 The proof claims that x^p + y^p can be divided by (x+y) to give an integer. That's clearly false. x=3, y=4, z=5 are pairwise coprime and p=2 gives 3^2 + 4^2 = 5^2 but x+y = 7 and clearly 7 doesn't go into 5^2. Therefore the logic employed in the first paragraph is wrong and kills the 'proof'. I found the flaw. Line [14] does not follow from [13]. The equation x^p + ((x + y) - x)^p = z^p behaves differently for even values of p, giving 2x^p = z^p (mod (x + y)), which has counterexamples. Link to comment Share on other sites More sharing options...
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