rayb07 Posted November 3, 2008 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
AlphaNumeric Posted November 10, 2008 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'.
rayb07 Posted November 10, 2008 Author 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.
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