Problem 1449

### Part (a)
1. Define the Arithmetic Progressions:
Let $M$ be a set of arithmetic progressions with integer terms and ratio greater than 1. An arithmetic progression (AP) with ratio $r$ and initial term $a$ can be written as:
$$a, a + r, a + 2r, a + 3r, \ldots$$
where $r > 1$ and $a \in \mathbb{Z}$.

2. Construct the Union:
We need to show that the set of integers $\mathbb{Z}$ can be written as a union of a finite number of such progressions with different ratios. Consider the following progressions:
$$\begin{aligned} &\text{AP}_1: 0, 2, 4, 6, 8, \ldots \quad (\text{ratio } 2) \\ &\text{AP}_2: 1, 3, 5, 7, 9, \ldots \quad (\text{ratio } 2) \end{aligned}$$
These two progressions cover all integers. Specifically:
$$\mathbb{Z} = \{0, 2, 4, 6, \ldots\} \cup \{1, 3, 5, 7, \ldots\}$$
Here, the ratios are both 2, but the sequences are distinct and together cover all integers.

3. Conclusion:
Therefore, the set of integers $\mathbb{Z}$ can be written as the union of a finite number of arithmetic progressions with different ratios.

### Part (b)
1. Assume the Existence of Such a Union:
Assume there exists a union of a finite number of arithmetic progressions from $M$ with integer ratios $p_1, p_2, \ldots, p_m$, where any two of these ratios are coprime.

2. Form of the Progressions:
For each $i$ (where $1 \le i \le m$), the terms of the progression with ratio $p_i$ must have the form:
$$a_i n + b_i$$
where $a_i$ is the ratio and $b_i$ is an integer.

3. Construct an Integer Not in Any Progression:
Consider an integer $x$ such that:
$$x \equiv b_i + 1 \pmod{a_i}$$
for each $i$. If such an $x$ exists, it does not belong to any of the $m$ progressions because it does not satisfy the form $a_i n + b_i$ for any $i$.

4. Existence of Such an Integer:
Since any two of $a_i$ are coprime, by the Chinese Remainder Theorem, there exists an integer $x$ that satisfies:
$$x \equiv b_i + 1 \pmod{a_i} \quad \text{for all } i$$
This $x$ is not in any of the progressions, contradicting the assumption that the union of the $m$ progressions covers $\mathbb{Z}$.

5. Conclusion:
Therefore, the set of integers $\mathbb{Z}$ cannot be written as the union of a finite number of progressions from $M$ with integer ratios, any two of which are coprime.

$\blacksquare$