<s>[INST] Proving or disproving $12\mid x$ given $x^2+2\mid y^2-2$ 
Let $x$, $y$ be positive integers such that $x^2+2\mid y^2-2$. Prove or disprove that $12\mid x$.

This conclusion comes when I was dealing with another problem, and I feel it is right because when $x=12$ and $y=32$,
$$\frac{32^2-2}{12^2+2}=7.$$
Let me explain the origin of this example  when $x=12$  can consider  $\sqrt{146k+2}\in \Bbb Z$, I try $k=1,2,\cdots$,when $k=7$ is such it
but when $x=24$, I can't find $y\le 48$ no example such it,because I want to find $k$ such $\sqrt{478k+2}\in \Bbb Z$
 [/INST] If $3\nmid x$ then $$ x^2\equiv 1\pmod 3\implies x^2+2\equiv 0\pmod 3$$ so $$ 3\mid y^2-2$$ which means $2$ is a square modulo 3. But this does not hold so $3\mid x$.

If $x$ is odd, then $x^2+2 = 4k(k+1)+3$, so there exists prime $p= 8l\pm 3$ such that $p\mid x^2+2$, but then $p\mid y^2-2$ and this is not true since by Legendre symbols we have $$\Big({2\over p}\Big) = (-1)^{p^2-1\over 8} =-1 $$
So $2\mid x$.

Added after Mike Bennett comment. So if $x=4k+2$ then $$ x^2+2\equiv 6\pmod 8$$ so again we have $p= 4l+3$ such that $p\mid x^2+2$, so $-2$ is a square modulo $p$. But since $p\mid y^2-2$, number $2$ is also square modulo $p$ so we have $$\Big({2\over p}\Big)= 1$$
Since Legendre symbol is multiplicative we have: $$ 1= \Big({-2\over p}\Big) =\Big({2\over p}\Big)\cdot \Big({-1\over p}\Big)   = 1\cdot   (-1)^{p-1\over 2}= -1$$
A contradiction. So $4\mid x$ and we are done.
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