, 2011

If the pair $(x, y)$ is the solution of the given system, then the pair $(-x, y)$ is also the solution. We conclude that the unique solution of this system has to be of the form $(0, y)$.

Taking $x = 0$ in the given system we get
$$\begin{aligned} 1 &= y + a, \\ y^2 &= 1, \end{aligned}$$
so $y = 1$ or $y = -1$, and hence $a = 0$ or $a = 2$.

$$2^{|x|} + |x| = x^2 + y$$
$$x^2 + y^2 = 1.$$
It is easy to see that $(0, 1)$ is the solution. Let us show that there are no other solutions.
From the second equation it follows that $|x| \le 1$ and $|y| \le 1$. Because of $0 \le |x| \le 1$ we get that $|x| \ge x^2$.
We also get $2^{|x|} \ge 2^0 = 1 \ge |y| \ge y$, so $2^{|x|} + |x| \ge x^2 + y$.
For the equality to hold, we must have $|x| = 0$ and $|y| = y$, that is $x = 0$ and $y \ge 0$. Because of the second equation it follows that $(x, y) = (0, 1)$, so this system has no other solutions.
In the case $a = 2$ we can notice that the system
$$2^{|x|} + |x| = x^2 + y + 2$$
$$x^2 + y^2 = 1$$
has (at least) three solutions $(0, -1)$, $(1, 0)$, $(-1, 0)$.
We conclude that the system has a unique solution if and only if $a = 0$.