$4 \cdot 34$ An integer has exactly 4 prime factors, the sum of whose squares is 476. Find this integer.
(China National Training Team Practice Question, 1990)
[Solution] Given $476 \equiv 2(\bmod 3)$, and when $3 \nmid a$,
$$
a^{2} \equiv 1 \quad(\bmod 3) \text {. }
$$

Then there are two prime factors of 3, and the remaining two prime factors are set as $p, q$. By the problem statement,
$$
\begin{array}{l}
p^{2}+q^{2}+3^{2}+3^{2}=476 . \\
p^{2}+q^{2}=458<22^{2} .
\end{array}
$$

Assuming $p \geqslant q$, then
$$
\begin{array}{c}
2 p^{2} \geqslant 458 \\
15<\sqrt{229} \leqslant p \leqslant 21 .
\end{array}
$$

Thus, $p$ is 17 or 19.
When $p=19$, $q^{2}=97$.
There is no solution in this case.
When $p=17$, $q^{2}=169$, i.e.,
$$
q=13 \text {. }
$$

Therefore, the required integer is
$$
3 \cdot 3 \cdot 13 \cdot 17=1989
$$