Hrvatska 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 $$\begin{array}{rl}1& =y+a,\\ y^2& =1,\end{array}$$ 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 $a=0$.