
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Given the equation:

\[
\sqrt{(x+2)^{2} + 4a^{2} - 4} + \sqrt{(x-2)^{2} + 4a^{2} - 4} = 4b
\]

If \(x_0\) is a solution to this equation, then \(-x_0\) is also a solution. From the uniqueness of the solution, \(x_0 = 0\). Substituting \(x = 0\) into the original equation, we derive the condition for the parameters: \(4|a| = 4b\), i.e., \(b = |a|\). Taking this into account, we obtain the equation

\[
\sqrt{4b^{2} + 4x + x^{2}} + \sqrt{4b^{2} - 4x + x^{2}} = 4b
\]

Squaring both sides, we get:

\[
\sqrt{(4b^{2} + x^{2})^{2} - 16x^{2}} = 4b^{2} - x^{2}
\]

Squaring again, we have:

\[
16b^{4} + 8b^{2}x^{2} + x^{4} - 16x^{2} = 16b^{4} - 8b^{2}x^{2} + x^{4} \Rightarrow 16x^{2}(b^{2} - 1) = 0
\]

Thus, if \(b \neq 1\) (\(b \geq 0\)!), then \(x = 0\) is the unique solution to the equation. We must also consider the case \(b = 1\). In this case, the equation becomes:

\[
\sqrt{x^{2} + 4x + 4} + \sqrt{x^{2} - 4x + 4} = 4 \Rightarrow |x+2| + |x-2| = 4
\]

Using the interval method, the solution to this equation is the segment \([-2, 2]\), meaning \(x\) is not unique in this case.

Present the answer in LaTex format: \boxed{Your answer}
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