How can I show $z^2 = 6x^2 + 2y^2$ has no non-trivial integer solutions? EDIT: I missed a part of the question as there was a typo in my notes (this is part of my working on a proof using the Bruck–Ryser–Chowla theorem which was presented incorrectly in my version of the lecture notes). I have updated the question and shown some new working.

I was thinking of supposing that $x^2$, $y^2$ & $z^2$ have no common factors (because if they did, just divide out by that factor), and then considering possible solutions to $z^2 - 2y^2 = 6x^2$ modulo 6. See, we know $z^2 \equiv 2y^2 \mod 6$ which gives us six options.


*

*$z^2 \equiv 2y^2 \equiv 0 \mod 6$

*$z^2 \equiv 2y^2 \equiv 1 \mod 6$

*$z^2 \equiv 2y^2 \equiv 2 \mod 6$

*$z^2 \equiv 2y^2 \equiv 3 \mod 6$

*$z^2 \equiv 2y^2 \equiv 4 \mod 6$

*$z^2 \equiv 2y^2 \equiv 5 \mod 6$


In each case, I was thinking I could arrive to a contradiction, but I'm having a little trouble with the last five cases.


*

*Let $z^2 = 6 \alpha$ and $2y^2 = 6 \beta$. Then $x^2$, $y^2$ & $z^2$ all have a factor of 3 which is impossible.

*Let $z^2 = 6 \alpha + 1$ and $2y^2 = 6 \beta + 1$ ...

*Let $z^2 = 6 \alpha + 2$ and $2y^2 = 6 \beta + 2$ ...

*Let $z^2 = 6 \alpha + 3$ and $2y^2 = 6 \beta + 3$ ...

*Let $z^2 = 6 \alpha + 4$ and $2y^2 = 6 \beta + 4$ ...

*Let $z^2 = 6 \alpha + 5$ and $2y^2 = 6 \beta + 5$ ...


Any comments on the general proof structure or help with the last two cases would be greatly appreciated! Cheers.

Assume $(x,y,z)$ is a solution to $z^2 = 6x^2 + 2y^2 $ with smallest positive $|z|$, then $\gcd(x,y) = 1$. Putting $z=2z_1$ gives
$$2z_1^ 2 = 3x^2 + y^2 $$ Since $x,y$ are both odd, RHS is divisible by $4$, so $z_1$ is even, that is $8\mid (2z_1^2)$. Hence $3x^2 + y^2 \equiv 0 \pmod{8}$. But the square of an odd number modolo $8$ must be $1$, a contradiction.
Hence $(0,0,0)$ is the only solution.
