Prove that $x$ and $y$ are divisible by $3$ when $x^2+6u^2=2y^2+3v^2$ 
a) Solve  $x^2\equiv 2y^2\pmod{3}$
b) Use part a to prove that $x$ and $y$ are divisible by $3$ when $x^2+6u^2=2y^2+3v^2$

My attempt:
$x^2\equiv 2y^2\pmod{3}\iff $$(y^{-1}x)^2 \equiv 2\pmod{3}$ but $\left(\frac{2}{3}\right)=-1$ since $3 \equiv 3\pmod{8}$. This means that 2 can't be Q.R. but at the same time I showed that $(xy^{-1})^2$ is clearly a square. Does it mean that the inverse doesn't exist?

Once you have $x \equiv y \equiv 0 \pmod 3, $ you get a little extra, since $x^2 \equiv y^2 \equiv 0 \pmod 9. $ This says something about the other two letters, $ 6 u^2 \equiv 3 v^2  \pmod 9. $ Then  $ 2 u^2 \equiv  v^2  \pmod 3. $ Finally, this part of the conclusion is $u \equiv v \equiv 0 \pmod 3. $
Stated carefully, the result is simply that if 
 $$ x^2 + 6 u^2 \equiv 2 y^2 + 3 v^2  \pmod 9, $$
then
$$ x \equiv u \equiv y \equiv v \equiv 0 \pmod 3. $$ Which means that if
$ x^2 + 6 u^2 \equiv 2 y^2 + 3 v^2  \pmod 9, $ then $$ \gcd(x,u,y,v) \neq 1. $$
Now assume we have any integer solution to $$ X^2 + 6 U^2 = 2 Y^2 + 3 V^2   $$
where the only requirement is that at least one of $X,U,Y,V$ is not equal to $0.$ We may then find
$$ T = \gcd(X,U,Y,V);  $$
in turn, we may define
$$ x = X/T, \; u = U/T, \; y = Y/T, \; v = V/T.  $$
It follows that $  \gcd(x,u,y,v) = 1.  $
It is still true that t least one of $x,u,y,v$ is not equal to $0.$ It is also still true that $x^2 + 6 u^2 = 2 y^2 + 3 v^2.$ However, this means $ x^2 + 6 u^2 \equiv 2 y^2 + 3 v^2  \pmod 9, $ from which we proved that $ \gcd(x,u,y,v) \neq 1. $ This contradicts the assumption that there is any solution in integers $X,U,Y,V$ with at least one of them nonzero.
The final result is that $x^2 + 6 u^2 = 2 y^2 + 3 v^2$ in integers implies
$x=u=y=v=0.$ Only trivial integer solutions. 
