4.28 Let \( m \) and \( n \) be natural numbers. Determine the minimum number of distinct prime divisors of the natural number
\[
m(n+9)\left(m+2 n^{2}+3\right)
\]

(53rd Moscow Mathematical Olympiad, 1990)
[Solution] (1) If $m>1$, then $m$ and $m+2 n^{2}+3$ have different parities. In the product $m(n+9)\left(m+2 n^{2}+3\right)$, there must be the even prime number 2 as a divisor. Additionally, there is at least one odd prime number as a divisor.
Therefore, when $m>1$, the product $m(n+9)\left(m+2 n^{2}+3\right)$ has at least two different prime divisors.
(2) If $m=1$, the product becomes
$$
2(n+9)\left(n^{2}+2\right) \text {. }
$$

It is clear that the even prime number 2 is a divisor.
If $n$ is odd, then $n^{2}+2$ is odd, so there must be an odd prime number as a divisor.
If $n$ is even, then $n+9$ is odd.
Thus, when $m=1$, the product $m(n+9)\left(m+2 n^{2}+3\right)$ has at least two different prime divisors.
(3) If we take $m=27, n=7$, then we have
$$
m(n+9)\left(m+2 n^{2}+3\right)=3^{3} \cdot 2^{11} \text {. }
$$

This number has exactly two different prime divisors, 2 and 3.
From the above, for all natural numbers $m$ and $n$, the number
$$
m(n+9)\left(m+2 n^{2}+3\right)
$$

has at least two different prime divisors.