Question 127, Find the last two digits of $\left[(2+\sqrt{3})^{2^{2020}}\right]$.

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Question 127, Solution: Let $a_{n}=(2+\sqrt{3})^{2^{n}}+(2-\sqrt{3})^{2^{n}}$, then $\left\{a_{n}\right\}: a_{0}=4, a_{n+1}=a_{n}^{2}-2$. And $\left[(2+\sqrt{3})^{2^{2020}}\right]=a_{2020}-1$.
According to the recursive relation, we have:
$$
\begin{array}{l}
a_{0} \equiv 4(\bmod 100), a_{1} \equiv 14(\bmod 100), a_{2} \equiv 94(\bmod 100), a_{3} \equiv 34(\bmod 100) \\
a_{4} \equiv 54(\bmod 100), a_{5} \equiv 14 \equiv a_{1}(\bmod 100)
\end{array}
$$

Combining the recursive relation, we know that for $n \in Z^{+}$, $a_{n+4} \equiv a_{n}(\bmod 100)$.
Thus, we have
$$
\left[(2+\sqrt{3})^{2^{2020}}\right]=a_{2020}-1 \equiv a_{4}-1 \equiv 53(\bmod 100)
$$

In conclusion, the last two digits of $\left[(2+\sqrt{3})^{2^{2020}}\right]$ are 53.