[Solution] Let $x=\sqrt{29}+\sqrt{21}, y=\sqrt{29}-\sqrt{21}$, then
$$
\begin{array}{l}
x^{2}=50+2 \sqrt{609}, \\
y^{2}=50-2 \sqrt{609} .
\end{array}
$$

Let $a=x^{2}, \quad b=y^{2}$, then we have
$$
\begin{aligned}
a+b & =100, \\
a b & =64 .
\end{aligned}
$$

Let $S_{0}=a^{0}+b^{0}, S_{n}=a^{n}+b^{n}$, then we have
$$
\begin{aligned}
S_{n} & =(a+b) S_{n-1}-a b S_{n-2} \\
& =100 S_{n-1}-64 S_{n-2} .
\end{aligned}
$$

Since $S_{n}$ is a positive integer,
we have
$$
\begin{array}{l}
S_{n} \equiv 36 S_{n-2} \equiv 6^{2} S_{n-2} \equiv 6^{4} S_{n-4} \equiv \cdots \\
\equiv 6^{992} S_{n-992}(\bmod 100) \text {. } \\
\end{array}
$$

That is, $S_{992} \equiv 6^{992} S_{0}=2 \cdot 6^{992}(\bmod 100)$.
Since $\quad 6^{4} \equiv 1296 \equiv-4 \quad(\bmod 100)$,
then
$$
S_{992} \equiv 2(-4)^{248} \equiv 2^{497} \equiv\left(2^{22}\right)^{22} \cdot 2^{13} \text {. }
$$

Since
$$
2^{22}=2^{2} \quad(\bmod 100),
$$

then
$$
\begin{aligned}
S_{992} & \equiv\left(2^{2}\right)^{22} \cdot 2^{13}=2^{44} \cdot 2^{13} \equiv\left(2^{2}\right)^{2} \cdot 2^{13} \\
& \equiv 2^{17} \equiv 4096 \times 32 \equiv 72 \quad(\bmod 100) .
\end{aligned}
$$

That is, the last two digits of $S_{992}$ are 72.
Since
$$
0<\sqrt{29}-\sqrt{21}<1,
$$

we have
$$
\begin{array}{l}
0<(\sqrt{29}-\sqrt{21})^{1984}<1 . \\
{\left[(\sqrt{29}+\sqrt{21})^{1984}\right]=S_{992}-1 .}
\end{array}
$$

Its last two digits are 71.
$10 \cdot 81$ Find the last two digits of $\left[(\sqrt{29}+\sqrt{21})^{1984}\right]$.
(25th International Mathematical Olympiad Candidate Problem, 1984)