In a number system with base $\mathrm{Az} A+1$

$$
(A A A A)^{2}=A A A C B B B C,
$$

different letters represent different digits. Determine the values of $A, B, C$.

---

Please note that the term "Az" in the base expression seems to be a placeholder or a typo. If it is meant to be a specific value, please provide the correct value or context. For the purpose of this translation, I have kept it as is.
On the right side, converted to the decimal system and then transformed:

$$
\begin{gathered}
\left(A A A A_{A+1}\right)^{2}=\left(A\left((A+1)^{3}+(A+1)^{2}+(A+1)+1\right)\right)^{2}=\left(A \frac{(A+1)^{4}-1}{A}\right)^{2}= \\
=\left((A+1)^{4}-1\right)^{2}=(A+1)^{8}-2(A+1)^{4}+1= \\
=(A+1)^{7} A+\left((A+1)^{4}\left((A+1)^{3}-2\right)+1\right.
\end{gathered}
$$

Here, since $A \geq 1$, $(A+1)^{3}-2>0$. Therefore, when writing the above number in the base $A+1$ numeral system, 0 appears at the $(A+1)^{3}$, $(A+1)^{2}$, and $(A+1)$ place values, and 1 appears at the $(A+1)^{0}$ place value; thus $B=0, C=1$. Based on the above, the original equation can be written as:

$$
\begin{gathered}
(A+1)^{7} A+(A+1)^{4}\left((A+1)^{3}-2\right)+1= \\
=A\left((A+1)^{7}+(A+1)^{6}+(A+1)^{5}\right)+(A+1)^{4}+1
\end{gathered}
$$

Subtracting 1 and then simplifying by $(A+1)^{4}$:

$$
(A+1)^{3}-2=A\left((A+1)^{2}+(A+1)\right)+1
$$

from which $A=2$. Our line of reasoning simultaneously shows that the triplet $A=2, B=0, C=1$ is a solution, and no other solution exists.