Problem 1505

### Part (a)
1. Pigeonhole Principle Application:
- Consider the pairs $(F_i, F_{i+1}) \mod m$ for $i = 0, 1, 2, \ldots, m^2$.
- There are $m^2 + 1$ such pairs but only $m^2$ possible distinct pairs modulo $m$.
- By the Pigeonhole Principle, at least two of these pairs must be the same.
- Therefore, there exist integers $0 \le i < j \le m^2$ such that:
$$(F_i, F_{i+1}) \equiv (F_j, F_{j+1}) \pmod{m}$$

### Part (b)
1. Using Result from Part (a):
- From part (a), we have $0 \le i < j \le m^2$ such that:
$$F_i \equiv F_j \pmod{m} \quad \text{and} \quad F_{i+1} \equiv F_{j+1} \pmod{m}$$
- Let $k = j - i$. Then:
$$F_{n+k} \equiv F_n \pmod{m} \quad \text{for all natural numbers } n$$

### Part (c)
1. **Definition of $k(m)$**:
- By definition, $k(m)$ is the smallest positive integer such that:
$$F_{n+k(m)} \equiv F_n \pmod{m} \quad \text{for all natural numbers } n$$
2. Properties of Fibonacci Sequence:
- Consider the sequence starting from $F_0 = 0$ and $F_1 = 1$.
- The smallest period $k(m)$ must satisfy:
$$F_{k(m)} \equiv 0 \pmod{m} \quad \text{and} \quad F_{k(m)+1} \equiv 1 \pmod{m}$$
- This is because the Fibonacci sequence modulo $m$ must reset to the initial conditions $F_0 \equiv 0 \pmod{m}$ and $F_1 \equiv 1 \pmod{m}$ to repeat.

### Part (d)
1. **If $k \vdots k(m)$**:
- If $k$ is a multiple of $k(m)$, then:
$$k = t \cdot k(m) \quad \text{for some integer } t$$
- Therefore:
$$F_{n+k} = F_{n + t \cdot k(m)} \equiv F_n \pmod{m} \quad \text{for all natural numbers } n$$
2. **If $k$ is not a multiple of $k(m)$**:
- Suppose $k$ is not a multiple of $k(m)$.
- Then there exists a smaller period in the Fibonacci sequence modulo $m$, which contradicts the minimality of $k(m)$.
- Hence, $k$ must be a multiple of $k(m)$.

$\blacksquare$