Q:

Proving that the only integer solution of $2x^2+3y^2=z^2$ is $(0,0,0)$

I'd like to prove that the only integer solutions of $$2x^2+3y^2=z^2$$ is $(0,0,0)$. 
By working in $\mathbb{Z}_2$ and $\mathbb{Z_3}$, I have gone as far as proving that in $\mathbb{Z}$, any integer solutions must have $x,y,z$ each being multiples of 3.  
But I am not quite sure how to then deduce that in $\mathbb{Z}$, they must therefore be 'zero'-multiples of 3.  I do not need a full solution, but a hint or a pointer would be helpful. 

A:

Hint:
Let $(x_0;y_0;z_0)$ - the smallest solution.  Then 
$$\left(\frac{x_0}3;\frac{y_0}3;\frac{z_0}3 \right) -$$ solution.  Contradiction

A:

First prove that x, y, z must be equal to 0 modulo 3 indeed. 
Then what happens ?  $x=3x'$, $y=3y'$, $z=3z'$ with x', y', z' being integers.  What can you show and how to conclude ? 
SPOILER : this is the answer
If x, y, z are not (0, 0, 0) you can divide by 3 an infinite number of times and have non 0 integers, which is impossible

