8. It is clear that $b \geq 0$. Rewrite the equation as:

$\sqrt{(x+2)^{2}+4 a^{2}-4}+\sqrt{(x-2)^{2}+4 a^{2}-4}=4 b$. If $x_{0}$ is a solution to this equation, then $-x_{0}$ is also a solution to this equation. From the uniqueness, it follows that $x_{0}=0$. Then substituting $x=0$ into the original equation, we get the necessary condition on the parameters: $4|a|=4 b$, i.e., $b=|a|$. Taking this into account, we obtain the equation

$\sqrt{4 b^{2}+4 x+x^{2}}+\sqrt{4 b^{2}-4 x+x^{2}}=4 b$. After squaring, we get:

$\sqrt{\left(4 b^{2}+x^{2}\right)^{2}-16 x^{2}}=4 b^{2}-x^{2}$. Squaring again, we get:

$16 b^{4}+8 b^{2} x^{2}+x^{4}-16 x^{2}=16 b^{4}-8 b^{2} x^{2}+x^{4} \Rightarrow 16 x^{2}\left(b^{2}-1\right)=0$. From this, it follows that,

if $b \neq 1$ ( $b \geq 0$ !), then $x=0$ will be the only solution to the equation. It remains to consider the case $b=1$. In this case, we get the equation:

$\sqrt{x^{2}+4 x+4}+\sqrt{x^{2}-4 x+4}=4 \Rightarrow|x+2|+|x-2|=4$. Using the interval method, we find that the solution to this equation will be the interval $[-2 ; 2]$, i.e., $x$ in this case is not unique.
Answer: $b \in [0 ; 1) \cup (1, +\infty)$.

## Olympiad "Sails of Hope" 2016. MIIT II round.

4 variant.