Problem 1714

To solve this problem, we need to construct two nonempty subsets $A$ and $B$ of $\mathbb{N}$ that satisfy the given conditions. Let's go through each condition step-by-step.

1. **Condition 1: $A \cap B = \{1\}$**

We need to ensure that the only common element between $A$ and $B$ is $1$.

2. **Condition 2: Every number in $\mathbb{N}$ can be expressed as the product of a number in $A$ and a number in $B$**

This means for any $n \in \mathbb{N}$, there exist $a \in A$ and $b \in B$ such that $n = a \cdot b$.

3. **Condition 3: Each prime number is a divisor of some number in $A$ and also some number in $B$**

For any prime $p$, there should be some $a \in A$ and $b \in B$ such that $p$ divides $a$ and $p$ divides $b$.

4. **Condition 4: One of the sets $A$ and $B$ has the property that if the numbers in this set are written as $x_1 < x_2 < x_3 < \cdots$, then for any given positive integer $M$, there exists $k \in \mathbb{N}$ such that $x_{k+1} - x_k \ge M$

This means that one of the sets has arbitrarily large gaps between consecutive elements.

5. Condition 5: Each set has infinitely many composite numbers**

Both $A$ and $B$ should contain infinitely many composite numbers.

Let's construct the sets $A$ and $B$ as follows:

- Let $A = \{1, 2, 2^2 - 1, 2^3 - 1, 2^4 - 1, \ldots\}$
- Let $B = \{1\} \cup (\mathbb{N} \setminus A)$

Now, let's verify each condition:

1. **Condition 1: $A \cap B = \{1\}$**

By construction, $A$ contains $1$ and other specific numbers, while $B$ contains $1$ and all natural numbers not in $A$. Therefore, $A \cap B = \{1\}$.

2. **Condition 2: Every number in $\mathbb{N}$ can be expressed as the product of a number in $A$ and a number in $B$**

For any $n \in \mathbb{N}$, we can take $a = 1 \in A$ and $b = n \in B$. Thus, $n = a \cdot b$.

3. **Condition 3: Each prime number is a divisor of some number in $A$ and also some number in $B$**

- For $p = 2$, $2 \in A$.
- For any odd prime $p$, $p \mid (2^{p-1} - 1) \in A$ (by Fermat's Little Theorem), and $2p \in B$.

4. **Condition 4: One of the sets $A$ and $B$ has the property that if the numbers in this set are written as $x_1 < x_2 < x_3 < \cdots$, then for any given positive integer $M$, there exists $k \in \mathbb{N}$ such that $x_{k+1} - x_k \ge M$**

The set $A$ has this property because the numbers $2^n - 1$ grow exponentially, so the gaps between consecutive elements increase exponentially.

5. Condition 5: Each set has infinitely many composite numbers

- For $B$, we can take powers of $2$ (e.g., $4, 8, 16, \ldots$).
- For $A$, note that $2^{2k} - 1$ is composite for all $k > 1$.

Thus, all conditions are satisfied.

The final answer is $\boxed{ A = \{1, 2, 2^2 - 1, 2^3 - 1, 2^4 - 1, \ldots\} }$ and $B = \{1\} \cup (\mathbb{N} \setminus A)$.