Prove by contradiction that $\forall x,y \in \Bbb Z: x^2-4y \ne 2$

Question

Prove that for all $x,y \in \mathbb{Z}$, $x^2 - 4y \ne 2$. Using a contradictory method would be appropriate.

So, for this question, I assume, for the sake of a contradiction, that

There exists $x,y \in \mathbb{Z}$ such that $x^2 - 4y = 2$.

After this, I have to derive a contradiction somehow. I'm not quite certain where the contradiction is or how I am supposed to come up with one.

Any hints to help me on the right track would be appreciated. Thank you.

Answer 1

$x^2=2(2y+1)$, so $x$ has to be even. Then, let $x=2k$ where $k\in\mathbb Z$. So we have $$(2k)^2-4y=2\Rightarrow 2(k^2-y)=1.$$ The LHS is even, and the RHS is odd, which is a contradiction.

Answer 2

If the equality holds then we have in $\Bbb Z_4$

$$x^2=2$$ which's impossible: try all the possibilities $x=0,1,2,3\mod(4)$!