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

Question

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.

Answer 1

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

Answer 2

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

Answer 3

HINT.-Using the identity $(a^2-b^2)^2+(2ab)^2=(a^2+b^2)^2$ we can put$$\sqrt2 x=2ts\\\sqrt 3y=t^2-s^2$$ where $t$ and $s$ are quadratic irrational.

Hence $$ts\in \mathbb Q(\sqrt 2)\\t^2-s^2\in \mathbb Q(\sqrt 3)$$ It is not difficult to verify the impossibility of these inclusions.