Problem 1607

Given a prime number $p > 3$, we need to analyze the sets $S \subseteq \{1, 2, \ldots, p-1\}$ and the derived sets $S_a$ for $a \in \mathbb{N}$. The set $S_a$ is defined as:
$$S_a = \{ x \in \{0, 1, 2, \ldots, p-1\} \mid (\exists s \in S) \, x \equiv_p a \cdot s \}$$

### Part (a)
We need to determine how many sets $S \subseteq \{1, 2, \ldots, p-1\}$ exist such that the sequence $S_1, S_2, \ldots, S_{p-1}$ contains exactly two distinct terms.

1. Primitive Root Representation:
Suppose $g$ is a primitive root of $p$. Any element $a_i \in S$ can be written as $g^{b_i} \pmod{p}$. Thus, $S = \{g^{b_1}, g^{b_2}, \ldots, g^{b_t}\}$.

2. Transformation of Sets:
For each $i$, the set $S_i$ can be written as:
$$S_i = \{i \cdot a_1, i \cdot a_2, \ldots, i \cdot a_t\} \pmod{p}$$
Using the primitive root, this becomes:
$$S_i = \{g^{b_1+j}, g^{b_2+j}, \ldots, g^{b_t+j}\}$$
where $g^j \equiv i \pmod{p}$.

3. Distinct Sets Analysis:
We need $S_1, S_2, \ldots, S_{p-1}$ to contain exactly two distinct sets. This implies that the sets $M_j = \{b_1+j, b_2+j, \ldots, b_t+j\}$ (considered modulo $p-1$) must form exactly two distinct sets.

4. Conditions for Two Distinct Sets:
- If $t = p-1$, then $S = \{1, 2, \ldots, p-1\}$, leading to $S_1 = S_2 = \ldots = S_{p-1}$, which contradicts the requirement of having exactly two distinct sets.
- Therefore, $t \neq p-1$.

5. Structure of Sets:
- We need $M_1, M_2, \ldots, M_{p-1}$ to form exactly two distinct sets. This can be achieved if $M_1 = M_3 = \ldots = M_{p-2}$ and $M_2 = M_4 = \ldots = M_{p-1}$.
- The sets $M_1$ and $M_2$ must be such that they repeat every two steps.

6. **Determining $t$**:
- The sets $M_1$ and $M_2$ must be of the form $\{x, x+2, x+4, \ldots, x+2(t-1)\}$ modulo $p-1$.
- This implies $t = \frac{p-1}{2}$.

7. Conclusion:
- The sets $S$ that satisfy the condition are those where $t = \frac{p-1}{2}$.
- There are exactly 2 such sets: $\{1, 3, 5, \ldots, p-2\}$ and $\{2, 4, 6, \ldots, p-1\}$.

### Part (b)
We need to determine all numbers $k \in \mathbb{N}$ for which there exists a set $S \subseteq \{1, 2, \ldots, p-1\}$ such that the sequence $S_1, S_2, \ldots, S_{p-1}$ contains exactly $k$ distinct terms.

1. **Partitioning into $k$ Groups**:
- We partition $M_1, M_2, \ldots, M_{p-1}$ into $k$ groups $G_1, G_2, \ldots, G_k$ such that each group contains identical sets.

2. Injective Mapping:
- Let $c$ be a number such that $\gcd(c, p-1) = 1$. Define a map $f: M_{i_1} \rightarrow M_{i_1} + c \cdot r$ for $r = 1, 2, \ldots, p-2$.
- Since $\gcd(c, p-1) = 1$, $c \cdot r$ will cover all residues modulo $p-1$.

3. Equal Group Sizes:
- The map $f$ ensures that each group $G_i$ has the same size.
- Therefore, $|G_1| = |G_2| = \ldots = |G_k|$.

4. Divisibility Condition:
- Since the total number of sets is $p-1$, we have $k \mid (p-1)$.

5. Conclusion:
- For any $k$ that divides $p-1$, there exists a set $S$ such that the sequence $S_1, S_2, \ldots, S_{p-1}$ contains exactly $k$ distinct terms.

The final answer is: