# [SOLVED] On Divisibility by 19

• Oct 13th 2009, 07:48 PM
Nives
[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 PM
Nives
Nevermind guys, I just solved it :). srry
• Oct 14th 2009, 04:13 AM
aman_cc
Quote:

Originally Posted by Nives
Nevermind guys, I just solved it :). srry

Would you mind sharing how you did it, please. I am stuck on it for a while now. Thanks
• Oct 14th 2009, 07:59 AM
Bingk
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 AM
artvandalay11
Quote:

Originally Posted by Bingk
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 ...

You can have a solution of $\displaystyle x=0$ and $\displaystyle y=z$ for all $\displaystyle y,z\in\mathbb{Z}$....
• Oct 14th 2009, 08:23 AM
Bingk
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 ....