To prove that $ab + cd$ is composite in the 42nd (2001) IMO Problem 6. In fact, $ab + cd$, $ac + bd$, and $ad + bc$ are all composite, and the number of prime factors (counting multiplicities) of each is at least 3, 3, and 2, respectively (the proof of this conclusion is left to the reader).

Let
$$\alpha = b + d + a - c, \quad \beta = b + d - a + c. \text{ From (3), we have }$$
$$\alpha \beta = a(b + d + a - \alpha) + bd$$
$$= a^2 + (b + d)a + bd - a\alpha$$
$$= (a + b)(a + d) - a\alpha.$$

Thus, $\alpha \mid (a + b)(a + d)$. Since $\alpha > a + d$, $\alpha$ and $a + b$ are not coprime (otherwise, it would imply $\alpha \mid (a + d)$, which is a contradiction), so they must share at least one common prime factor $p_1$. In this case, from $a + b - \alpha = c - d$, we know $p_1 \mid (c - d)$, hence $p_1 \leq c - d$.

Notice:
$$ad + bc = d(a + b) + b(c - d),$$

so $p_1 \mid (ad + bc)$. Combining $p_1 \leq a + b$ and $p_1 \leq c - d$, we get $p_1 \leq \frac{a + b}{2}$. Combining $\alpha \mid (a + b)(a + d)$, we know $(\alpha, a + d) > 1$. Let $p_2 \mid (a, a + d)$ be a prime, then from $b - c = -(a + d) + \alpha$, we get $p_2 \mid (b - c)$, thus $p_2 \mid (ab + cd)$ (since $ab + cd = b(a + d) - d(b - c))$. Using $p_2 \leq b - c < a + b > b + d + a - c = \alpha,$

Therefore, $\beta$ is a positive integer greater than 1, so $ac + bd$ is composite.