Monday, January 9th

Problem
Determine if there exists $f:\mathbb{Z}^+\to \mathbb{Z}^+$ such that $f(f(n))=2n$ for all positive integers n.

Solution
As long as $f$ does not have a fixed point there is no problem in the definition of $f$. The key is to analyze the powers of 2, as it appears on the defining property of $f$.

We have that $f\left(2^p q\right)=2^p f(q)$ for $q$ an odd integer, hence $f$ is determined by the values it takes on the odd integers. Since the powers of 2 are not affected by $f$, it suffices that $f$ map $2^r a$ to $2^s b$, where $a,b$ are odd integers and $r,s$ are non-negative integers and the only restriction is that $a\neq b$.