Is it a $\mathbb{Q}$-vector space? Let $V_2=\{(x,y)\in \mathbb{Q}^2 \mid x^2=-y^2\}$ with the addition and the scalar multiplication of $\mathbb{Q}^2$. 
To check if $V_2$ is a $\mathbb{Q}$-vector space, we must check the axioms, right? 
We have the following: 
$(x_1, y_1), (x_2, y_2)\in V_2 : x_1^2=-y_1^2, x_2^2=-y_2^2$ 
$x_1^2x_2^2=y_1^2y_2^2 \Rightarrow x_1x_2=\pm y_1y_2$  
$(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2) : (x_1+x_2)^2=x_1^2+2x_1x_2+x_2^2=-y_1\pm 2y_1y_2-y_2^2$ 
it doesn't imply that $(x_1+x_2)^2=-(y_1+y_2)^2$, right? 
Does it follow from that that $V_2$ is not closed under the addition, and so it is not a $\mathbb{Q}$-vector space?
 A: Ben correctly pointed out that if $x^2=-y^2$ and $x,y\in\mathbb{Q}$, then we must have $x=y=0$ (why?). So $V_2$ is the trivial vector space.
To make the question less trivial, let's look at $\mathbb{C}$ instead of $\mathbb{Q}$. Now $V_2$ has lots of elements, e.g. $(1, i)$. Is it a vector space?
Here your argument is relevant, but not complete. You're right that this version of $V_2$ is not a vector space, since it's not closed under addition; however, you haven't proved that. You need to exhibit a counterexample: find an example of vectors $(x_1, y_1)$ and $(x_2, y_2)$ in $V_2$ such that $(x_1+x_1, y_1+y_1)$ is not in $V_2$.
A: You are correct that $x_1^2=-y_1^2$ and $x_2^2=-y_2^2$ does not imply $(x_1+x_2)^2=-(y_1+y_2)^2$ when you manipulate them as formal polynomials.  However, remember that $x_1,y_1,x_2$, and $y_2$ must actually be rational numbers here, and there might not actually exist a counterexample in the rational numbers even if the formal manipulation does not allow you to simplify $(x_1+x_2)^2$ to $-(y_1^2+y_2)^2$.  What rational numbers $x$ and $y$ are there such that $x^2=-y^2$?  (Hint: There aren't very many!)
