$3 \cdot 10$ Find all odd natural numbers $n$ such that $(n-1)!$ is not divisible by $n^2$.

Find all odd natural numbers $n$ such that $(n-1)!$ cannot be divided by $n^2$.
[Solution] If $n$ can be expressed as $n=ab$, where $a \geqslant 3, b \geqslant 3, a \neq b$, then $a$, $2a$, and $b$, $2b$ are factors of the product $1 \cdot 2 \cdot \cdots \cdot (n-1)$, so $(n-1)!$ can be divided by $a^2 b^2 = n^2$.

If $n=a^2, a \geqslant 3$, then when $a^2-1 \geqslant 4a$, $a, 2a, 3a$, and $4a$ are factors of the product $\left(a^2-1\right)! = 1 \cdot 2 \cdot 3 \cdots \left(a^2-1\right)$, so $\left(a^2-1\right)!$ can be divided by $a^4 = n^2$ (when $a \geqslant 5$, the inequality $a^2-4a-1 \geqslant 0$ holds).

Since $n=1$ clearly does not meet the requirements of this problem, only two possibilities remain: $n$ is a prime number and $n=9$.

It is easy to verify that in these two cases, $(n-1)!$ cannot be divided by $n^2$. Therefore, the solution to this problem is that $n$ is a prime number or 9.