# Thread: congruence, relation x^2+y^2=Z^2, then 60|xyz

1. ## congruence, relation x^2+y^2=Z^2, then 60|xyz

If x,y,z is integer such that $\displaystyle {x}^{2}+{y}^{2}={z}^{2}$ show that 60|xyz

2. Note that you can set

$\displaystyle x=m^2-n^2$

$\displaystyle y=2mn$

for some integers, m,n.

From that it follows that z equals:

$\displaystyle z^2=(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2$
$\displaystyle z=m^2+n^2$

Now, quite frankly I forgot (or at least can't remember at the moment) how to prove that the solutions are of that form, but you can easily verify by putting anything for n and m... you will get pythagorean triples.

Anyway, what you want to prove can now be written as:

$\displaystyle xyz=(m-n)(m+n)*2mn(m^2+n^2)=60k$, for some k.
$\displaystyle mn(m-n)(m+n)(m^2+n^2)=30k$.

And all that's left is to prove that the left side is divisible by 30. Just think about the possibilities... you'll find it obvious that it's divisible by 2, and for 3 and 5 you'll just have to write it down... or maybe not