Example 9 Find the last two digits of $\left[(\sqrt{29}+\sqrt{21})^{2012}\right]$.
(Adapted from the 25th IMO Preliminary Question)
Let $a_{n}=(\sqrt{29}+\sqrt{21})^{2 n}+(\sqrt{29}-\sqrt{21})^{2 n}$
$$
=(50+2 \sqrt{609})^{n}+(50-2 \sqrt{609})^{n} \text {. }
$$

Then the characteristic equation of the sequence $\left\{a_{n}\right\}$ is
$$
x^{2}-100 x+64=0 \text {. }
$$

Therefore, the sequence $\left\{a_{n}\right\}$ has the recurrence relation
$$
a_{n+2}=100 a_{n+1}-64 a_{n}(n=0,1, \cdots) \text {, }
$$

and $a_{0}=2, a_{1}=100$.
By mathematical induction, we know that for any $(n=0,1, \cdots), a_{n}$ is an integer.
$$
\begin{array}{l}
\text { Also, } 0<(\sqrt{29}-\sqrt{21})^{2 n}<1, \text { thus, } \\
{\left[(\sqrt{29}+\sqrt{21})^{2 n}\right]=a_{n}-1 .} \\
\text { Hence } a_{n} \equiv-64 a_{n-2} \equiv 6^{2} a_{n-2} \equiv 6^{4} a_{n-4} \\
\equiv \cdots \equiv 6^{n} a_{0}=2 \times 6^{n}(\bmod 100) .
\end{array}
$$

Since $6^{4}=1296 \equiv-4(\bmod 100)$, we have
$$
\begin{array}{l}
a_{1006} \equiv 2(-4)^{252} \equiv 2^{505} \\
\equiv\left(2^{24}\right)^{21} \times 2(\bmod 100) . \\
\text { Also, } 2^{12}=4096 \equiv-4(\bmod 100), \\
2^{24} \equiv 16 \equiv 2^{4}(\bmod 100), \\
\text { thus } a_{1006} \equiv\left(2^{4}\right)^{21} \times 2=2^{85}=\left(2^{24}\right)^{3} \times 2^{13} \\
\equiv 2^{25} \equiv 2^{5} \equiv 32(\bmod 100),
\end{array}
$$

which means the last two digits of $a_{1000}$ are $32$, and the last two digits of $\left[(\sqrt{29}+\sqrt{21})^{2 n}\right]=a_{1006}-1$ are 31.