Why is it that 19 must divide either x,y,or z in the equation x^6+y^6=Z^6? This problem is really starting to bug me :/

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- Oct 13th 2009, 07:48 PMNives[SOLVED] On Divisibility by 19
Why is it that 19 must divide either x,y,or z in the equation x^6+y^6=Z^6? This problem is really starting to bug me :/

- Oct 13th 2009, 08:33 PMNives
Nevermind guys, I just solved it :). srry

- Oct 14th 2009, 04:13 AMaman_cc
- Oct 14th 2009, 07:59 AMBingk
Um ... I thought that $\displaystyle a^n+b^n=c^n$ has no positive integer solutions a,b,c when n is greater than 2 (Fermat's last theorem??). But in the case of even powers, it doesn't matter if the integers are positive or negative ... so in your case, the only solution would be x = y = z = 0 ... and any number divides zero ...

- Oct 14th 2009, 08:16 AMartvandalay11
- Oct 14th 2009, 08:23 AMBingk
Ah, yeah ... ugh, I forgot that case also :) ... so basically, either x or y equals zero (or both equal zero) will have solutions ... and since 19 divides zero ....