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 $4$.

Answer 2

Since $x^2=4y+2=2(2y+1)$

Suppose that $x=2k$, then $x^2=4k^2$. It is obvious that $4k^2 \not= 2(2y+1)$.

Answer 3

A square can only be 0 or 1 mod 4. Looking at above mod 4, you get $x^2 = 2 \mod 4$. Hence no solution.

Answer 4

$x^2-4y=2$
$(x+2\sqrt{y})(x-2\sqrt{y})=2$
The only factors of 2 are $\{1,2\}$ and $\{-1,-2\}$
Thus, $(x+2\sqrt{y})=1$ or $(x+2\sqrt{y})=2$

I will start with $(x+2\sqrt{y})=1$. Then $2\sqrt{y}=1-x\implies\sqrt{y}=\frac{1-x}{2}$.
By subtitution, $(x-2\sqrt{y})=2$ implies that $\left(x-2\left(\frac{1-x}{2}\right)\right)=2\implies x-1+x=2\implies x=3/2$. Note that $x$ is not an integer.

Just use the same method for the the other factor and interchange the $2$ and the $1$.