Solution:

(ans. $a= \pm 1$ for two solutions, $a= \pm \sqrt{2}$ for three solutions.
The solutions are given by $x=\frac{a \pm \sqrt{2-a^{2}}}{2},\ y= \pm x$. There are two solutions if the quadratic equation involving $x$ has a single solution $\Rightarrow$ $2-a^{2}=0 \Rightarrow a= \pm \sqrt{2}$. If one value of $x$ is $0$, then there will at most be 3 solutions. Solving $x=0$ in $a$ yields $a= \pm 1$ and this gives exactly 3 solutions).