3. (1974 Kyiv Mathematical Olympiad) Find 8 prime numbers (not necessarily distinct), such that the sum of their squares is 992 less than 4 times their product.


3. Let these 8 primes 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
$$
4 P-S=992 \text {. }
$$

First, it is not possible for all $x_{i}$ to be odd.
Since the square of an odd number is congruent to 1 modulo 8, then
$S \equiv 0(\bmod 8)$.
Also, $4 P \equiv 4(\bmod 8)$,
$992 \equiv 0(\bmod 8)$.
Thus, equation (1) cannot hold.
Second, it is not possible 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
$4 P \equiv 0(\bmod 8)$,
$$
\begin{array}{l}
S \equiv k+4(8-k) \equiv 32-3 k(\bmod 8), \\
992 \equiv 0(\bmod 8),
\end{array}
$$

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.