Let $n$ be a positive integer. Given an integer coefficient polynomial $f(x)$, define its signature modulo $n$ to be the (ordered) sequence $f(1), \ldots, f(n)$ modulo $n$. Of the $n^n$ such $n$-term sequences of integers modulo $n$, we need to determine how many are the signature of some polynomial $f(x)$ under the following conditions:

a) $n$ is a positive integer not divisible by the square of a prime.

b) $n$ is a positive integer not divisible by the cube of a prime.

### Solution:
Part (a):

Let $n = p_1 p_2 \cdots p_k$, where $p_i$ are distinct primes. We need to find the number of signatures modulo $n$.

Lemma 1: There is a polynomial for all signatures modulo $p$ for any prime $p$. Thus, there are $p^p$ possible signatures modulo $p$.

Lemma 2: If we have a polynomial $f(x)$ modulo $m$ and a polynomial $g(x)$ modulo $n$, where $m$ and $n$ are coprime, then we can find an $h(x)$ modulo $mn$ such that $h(x) \equiv f(x) \pmod{m}$ and $h(x) \equiv g(x) \pmod{n}$.

Using Lemma 2 repeatedly, we can combine the signatures modulo each $p_i$ to get a signature modulo $n$. Therefore, the number of signatures modulo $n$ is:
$$\prod_{i=1}^k p_i^{p_i}.$$

Part (b):

Let $n = p_1 p_2 \cdots p_k q_1^2 q_2^2 \cdots q_l^2$, where $p_i$ are distinct primes and $q_i$ are distinct primes not divisible by the cube of a prime.

Lemma 6: For any sequence of $p$ numbers, $a_0, a_1, \cdots, a_{p-1}$, we can find a polynomial $f(x)$ that has its signature starting with $a_0, a_1, \cdots, a_{p-1} \pmod{p^2}$. There are $p^{3p}$ signatures modulo $p^2$.

Using similar reasoning as in part (a), the number of signatures modulo $n$ is:
$$\prod_{i=1}^k p_i^{p_i} \prod_{j=1}^l q_j^{3q_j}.$$

The answer is:
$$\boxed{\prod_{i=1}^k p_i^{p_i} \text{ for part (a)}}$$
$$\boxed{\prod_{i=1}^k p_i^{p_i} \prod_{j=1}^l q_j^{3q_j} \text{ for part (b)}}$$