The square of four positive integers sums to 2008, among which the unit digits of two numbers are 2, and the unit digits of the other two numbers are 0. Find these four numbers.
(Tian Yonghai, Suihua Education Institute, Heilongjiang Province, 152054)
Solution: According to the problem, we can set the 4 numbers to be $10x$, $10y$, $10a+2$, and $10b+2$, where $0 < x \leq y$, $0 \leq a \leq b$ ($x, y, a, b \in \mathbf{N}_{+}$).
From the given information, we have
$$
\begin{array}{l}
100 x^{2} + 100 y^{2} + (10 a + 2)^{2} + (10 b + 2)^{2} \\
= 2008.
\end{array}
$$

Simplifying, we get
$$
5\left(x^{2} + y^{2} + a^{2} + b^{2}\right) + 2(a + b) = 100.
$$

From equation (1), we can deduce that
$$
x^{2} + y^{2} + a^{2} + b^{2} \leq 20.
$$

Thus, $0 < x \leq y \leq 4$, $0 \leq a \leq b \leq 4$.
From equation (1), it is easy to see that $51(a + b)$, i.e., $a + b = 5k$.
From equation (2), $k$ can only be $0$ or $1$.
(1) When $k = 0$, $a + b = 0$, i.e., $a = b = 0$. In this case, from equation (1), we get
$$
5\left(x^{2} + y^{2}\right) = 100.
$$

Thus, $x = 2$, $y = 4$.
Therefore, the 4 numbers are $20$, $40$, $2$, $2$.
(2) When $k = 1$, $a + b = 5$. From equation (2), it can only be $a = 2$, $b = 3$ or $a = 1$, $b = 4$.
(i) $a = 2$, $b = 3$, substituting into equation (1) we get
$$
5\left(x^{2} + y^{2}\right) + 5 \times 13 + 2 \times 5 = 100,
$$

i.e., $x^{2} + y^{2} = 5$.
Solving, we get $x = 1$, $y = 2$.
Therefore, the 4 numbers are $10$, $20$, $22$, $32$.
(ii) $a = 1$, $b = 4$, substituting into equation (1) we get
$$
5\left(x^{2} + y^{2}\right) + 5 \times 17 + 2 \times 5 = 100,
$$

i.e., $x^{2} + y^{2} = 1$.
From equation (2), there is no solution.
In summary, the 4 numbers are $20$, $40$, $2$, $2$ or $10$, $20$, $22$, $32$.