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 :/

Printable View

- Oct 13th 2009, 08:48 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, 09:33 PMNives
Nevermind guys, I just solved it :). srry

- Oct 14th 2009, 05:13 AMaman_cc
- Oct 14th 2009, 08:59 AMBingk
Um ... I thought that 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, 09:16 AMartvandalay11
- Oct 14th 2009, 09:23 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 ....