[Math] There exist no integers for which $x^2-4y=2$

Question

I am working on a new exercise in my textbook:

$$\text{Prove that: (P): }\;\nexists \;x,y \in \mathbb{Z}, x^2-4\cdot y = 2 $$

I am stuck and I would really like to see a correct proof so I can move on while understanding the "trick".

Thank you.

Answer 1

Suppose $x^2=4y+2$.

The RHS is divisible by $2$ but not by $4$. But if the LHS is divisible by $2$, it must be divisible by $4$.