Example 11 (25th IMO Shortlist) Find the last two digits of $\left[(\sqrt{29}+\sqrt{21})^{1984}\right]$.

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Consider $a_{n}=(\sqrt{29}+\sqrt{21})^{2 n}+(\sqrt{29}-\sqrt{21})^{2 n}=(50+2 \sqrt{609})^{n}+(50-$ $2 \sqrt{609})^{n}$, then the characteristic roots of $a_{n}$ are $r_{1,2}=50 \pm 2 \sqrt{609}, r_{1}+r_{2}=100, r_{1} r_{2}=64$, hence the characteristic equation is $r^{2}-100 r+64=0$, so $a_{n}$ satisfies the following recurrence relation:
$$
a_{n+2}=100 a_{n+1}-64 a_{n}(n=0,1,2, \cdots) .
$$

And $a_{0}=1+1=2, a_{1}=100$, by mathematical induction and (1) we know that for all $n=0,1,2, \cdots, a_{n}$ are all integers.
Also, $0<(\sqrt{29}-\sqrt{21})^{2 n}<1$, so $\left[(\sqrt{29}+\sqrt{21})^{2 n}\right]=a_{n}-1$, and by (1) we have
$$
a_{n} \equiv-64 a_{n-2} \equiv 6^{2} a_{n-2} \equiv 6^{4} a_{n-4} \equiv \cdots \equiv 6^{n} a_{0}=2 \cdot 6^{n}(\bmod 100) \text {. }
$$

Since $6^{4}=1296 \equiv-4(\bmod 100)$, we have
$$
a_{992} \equiv 2(-4)^{248} \equiv 2^{497}=\left(2^{24}\right)^{20} \cdot 2^{17},
$$

and since $2^{12}=4096 \equiv-4(\bmod 100), 2^{24} \equiv 16=2^{4}$, we have
$$
\begin{aligned}
a_{999} & \equiv\left(2^{4}\right)^{20} \cdot 2^{17} \equiv 2^{97} \equiv\left(2^{24}\right)^{4} \cdot 2 \equiv\left(2^{4}\right)^{4} \cdot 2 \\
& \equiv 2^{12} \cdot 2^{5} \equiv-4 \times 32 \equiv-128 \equiv 72(\bmod 100) .
\end{aligned}
$$

Thus, the last two digits of $a_{992}$ are 72, and therefore the last two digits of $\left[(\sqrt{29}+\sqrt{21})^{1984}\right]=a_{992}-1$ are 71.