Problem 1256

1. We need to determine if there exists a function $f: \mathbb{N} \to \mathbb{N}$ such that $f(f(n)) = 2n$ for all $n \in \mathbb{N}$.

2. Let's start by analyzing the given condition $f(f(n)) = 2n$. This implies that applying the function $f$ twice to any natural number $n$ results in $2n$.

3. To explore this further, let's consider the implications of this condition. If we apply $f$ to both sides of the equation $f(f(n)) = 2n$, we get:
$$f(f(f(n))) = f(2n)$$
Since $f(f(n)) = 2n$, we can substitute $2n$ for $f(f(n))$ in the equation above:
$$f(2n) = f(f(f(n))) = 2f(n)$$
This shows that $f(2n) = 2f(n)$ for all $n \in \mathbb{N}$.

4. We can use induction to generalize this result. Suppose $n = 2^m q$, where $m \ge 0$ and $q$ is an odd natural number. By repeatedly applying the result $f(2n) = 2f(n)$, we get:
$$f(2^m q) = 2^m f(q)$$
This means that if we can define $f$ for all odd natural numbers, then the values of $f$ for even natural numbers are determined.

5. Now, we need to define $f$ for all odd natural numbers. Let's consider the following definition:
$$f(2^m q) = \begin{cases} 2^m (4t + 3) & \text{if } q = 4t + 1 \\ 2^{m+1} (4t + 1) & \text{if } q = 4t + 3 \end{cases}$$
where $t$ and $m$ are non-negative integers.

6. We need to verify that this function satisfies the condition $f(f(n)) = 2n$. Let's check this for both cases:
- If $q = 4t + 1$, then:
$$f(q) = 4t + 3$$
and
$$f(f(q)) = f(4t + 3) = 2(4t + 1) = 2q$$
- If $q = 4t + 3$, then:
$$f(q) = 2(4t + 1)$$
and
$$f(f(q)) = f(2(4t + 1)) = 2(4t + 3) = 2q$$

7. Therefore, the function $f$ defined above satisfies the condition $f(f(n)) = 2n$ for all $n \in \mathbb{N}$.

The final answer is \( f(2^m q) = \begin{cases}