- $x=0, y>0$
We have $2y^2=1 \Rightarrow y=\frac{1}{\sqrt{2}}\notin \mathbb{Q}$. So, this case is rejected.
- $x=0, y<0$
We have $2y^2=1 \Rightarrow y=-\frac{1}{\sqrt{2}}\notin \mathbb{Q}$. So, this case is rejected.
- $x=y=0$
This case is not possible, since that would mean $0=1$.
Are these cases correct?
What can we say about the cases:
- $x,y<0$
- $x<0, y>0$
- $x>0, y>0$
? @shaihorowitz