Let $a,b,c,x,y,z \in \mathbb{Z}>1$
How do I prove if $x,y,z$ are square-free integers and: $$a\sqrt{x}\pm b\sqrt{y}=c\sqrt{z}$$
Then $\gcd(x,y,z)>1$?
I know for some of you it may be simple. As for me, it's not that trivial. Please any hint will be appreciated. Thanks.
Square the whole expression. You get that $$ a^2x\pm b^2y+2ab\sqrt{xy}=c^2z $$ or in other words, that $xy$ is a perfect square (otherwise the left-hand side wouldn't be an integer).
That means, since $x$ and $y$ are both square-free, that $x=y$. The above equation then says that $$ (a\pm b)^2x=c^2z $$ And again, by square-free-ness, we must have $x=z$. Therefore the full relation is that $x=y=z$. What you wanted to prove follows as long as $1$ isn't considered square-free.