
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
Rewrite the system in the form

$$
\left\{\begin{array}{l}
|x-1|-y=1-a^{4}-a^{4}(x-1)^{4} \\
(x-1)^{2}+y^{2}=1
\end{array}\right.
$$

and make the substitution $\mathrm{t}=\mathrm{x}-1$.

We get

$$
\left\{\begin{array}{l}
|t|-y=1-a^{4}-a^{4} t^{4} \\
t^{2}+y^{2}=1
\end{array}\right.
$$

If (t, y) is a solution of the system, then (-t, y) is also a solution. If the solution is unique, then $t=0$, and

$$
\left\{\begin{array}{l}
y=a^{4}-1 \\
y^{2}=1
\end{array}\right.
$$

For $\mathrm{y}=-1$, $\mathrm{a}=0$, and for $\mathrm{y}=1$, $\mathrm{a}^{4}=2$.

Let $\mathrm{a}=0$, then the system becomes

$$
\left\{\begin{array}{l}
|t|-y=1 \\
t^{2}+y^{2}=1
\end{array}\right.
$$

There are more than one solution, for example, $(0,-1)$, $(1,0)$ (where the first value in the pair is t).

Let $\mathrm{a}^{4}=2$, then the system becomes

$$
\left\{\begin{array}{l}
|t|-y=-1-2 t^{4} \\
t^{2}+y^{2}=1,
\end{array},<=>\left\{\begin{array}{l}
y=1+|t|+2 t^{4} \\
y^{2}=1-t^{2} .
\end{array}\right.\right.
$$

From the first equation, $\mathrm{y} \geq 1$, with $\mathrm{y}=1$ only when $\mathrm{t}=0$.

In the second equation, for $\mathrm{t} \neq 1$, $\mathrm{y}<1$. Thus, for $\mathrm{t} \neq 0$, there is no solution, and $(0,1)$ is the unique solution when $\mathrm{a}^{4}=2$.

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