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 :/
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 ...