3.8 Find 8 primes (not necessarily distinct) such that the sum of their squares is 992 less than 4 times their product.
(Kiev Mathematical Olympiad, 1974)
[Solution] Let the 8 prime numbers be $x_{1}, x_{2}, \cdots, x_{8}$.
Let $S=x_{1}^{2}+x_{2}^{2}+\cdots+x_{8}^{2}$,
$$
P=x_{1} x_{2} \cdots x_{8} .
$$

By the problem, we have
$$
4 P-S=992 .
$$

First, it is impossible for all $x_{i}$ to be odd.
Since the square of an odd number leaves a remainder of 1 when divided by 8, then

Also,
$$
\begin{array}{c}
S \equiv 0 \quad(\bmod 8) . \\
4 P \equiv 4 \quad(\bmod 8), \\
992 \equiv 0 \quad(\bmod 8) .
\end{array}
$$

Thus, equation (1) cannot hold.
Second, it is impossible for some of the $x_{i}$ to be odd primes and others to be 2.
Suppose there are $k$ odd primes and $8-k$ twos ($1 \leqslant k \leqslant 7$), then
$$
\begin{array}{c}
4 P \equiv 0 \quad(\bmod 8), \\
S \equiv k+4(8-k) \equiv 32-3 k(\bmod 8), \\
992 \equiv 0 \quad(\bmod 8) .
\end{array}
$$

Thus, equation (1) cannot hold either.
From the above, all $x_{i}(i=1,2, \cdots, 8)$ must be 2.
Upon verification, $x_{1}=x_{2}=\cdots=x_{8}=2$ satisfies the condition.