The diophantine equation $x^2+y^2=3z^2$ I tried to solve this question but without success:

Find all the integer solutions of the equation: $x^2+y^2=3z^2$

I know that if the sum of two squares is divided by $3$ then the two numbers are divided by $3$, hence if $(x,y,z)$ is a solution then $x=3a,y=3b$. I have $3a^2+3b^2=z^2$ and that implies
$$
\left(\frac{a}{z}\right)^2+\left(\frac{b}{z}\right)^2=\frac{1}{3}
$$
so I need to find the rational solutions of the equation $u^2+v^2=\frac{1}{3}$ and I think that there are no solutions for that because $\frac{1}{3}$ doesn't have a rational root, but I dont know how to explain it.
Thanks

How about using  infinite descent ?
As for any integer $a\equiv0,\pm1\pmod4\implies a^2\equiv0,1$
We have $x^2+y^2\equiv0\pmod3$
So, $3|(x,y)$
Let $x=3X,y=3Y$  
and so on
