14.3. 29 * Let $m$ and $n$ be natural numbers, how many different prime factors can the natural number
$$
m(n+9)\left(m+2 n^{2}+3\right)
$$

have at least?
The answer is 2.
If $m>1$, then $m$ and $m+2 n^{2}+3$ have different parity, their product has a prime factor 2, and at least one odd prime factor. If $m=1$, then $m(n+9)\left(m+2 n^{2}+3\right)=2(n+9)\left(n^{2}+2\right), n+9$ and $n^{2}+2$ have at least one odd number greater than 1. Their product still has a prime factor 2 and an odd prime factor.

Take $m=27, n=7, m(n+9)\left(m+2 n^{2}+3\right)=3^{3} \cdot 2^{11}$, which contains only two distinct prime factors 2 and 3.