# Thread: [SOLVED] On Divisibility by 19

1. ## [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 :/

2. Nevermind guys, I just solved it . srry

3. 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

4. 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 ...

5. 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 $\displaystyle y,z\in\mathbb{Z}$....

6. 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 ....