48. Find all odd natural $n$ for which $(n-1)!$ does not divide $n^2$.

 untranslated part:
( $n-1$ )! does not divide $n^2$.
48. Answer: $n$ is a prime number and $n=9$.

If $n$ can be represented in the form $n=ab$, where $a \geqslant 3, b \geqslant 3$, and $a$ is, then the product $1.2 \cdot \ldots \cdot(n-1)$ includes the factors $a, 2a$ and $b, 2b$, so $(n-1)$ is divisible by $a^2 b^2=n^2$.

If $n \Rightarrow p^2$, where $p$ is a prime number, then for $p^2-1 \geqslant 4p$ in the product $\left(p^2-1\right)!=1 \cdot 2 \cdot 8 \cdot \ldots \cdot\left(p^2-1\right)$, the numbers $p$, $2p$, $8p$, and $4p$ are included and it is divisible by $p^4=n^2$. (The inequality $p^3-4p-1 \geqslant 0$ is satisfied for all prime numbers $p \geqslant 5$.)

Two possibilities remain: $n$ is a prime number (in this case, $(n-1)!$ is not even divisible by $n$) and $n=9$ (in this case, 81 is divisible by 9, but not by 81).

$\nabla$ According to the well-known Wilson's theorem, for a prime $p$, the number $(p-1)!$ always gives a remainder of $(p-1)$ when divided by $p$.