Can anybody explain this solution in more detail?

Question

I have come across this problem, but I cannot understand the solution given here.

This is the link:

Functional equation $f(f(x))=2x$ on $\mathbb{Z}_{>0}$

Answer 1

I personally think that it is slightly unclear. I shall give my own solution to this problem. If a function $f(x)$ satisfies $f(f(x)) = 2x$ it must be injective. I assume you got there already. This is a consequence of the fact that if $f(x) = f(y)$, then $2y = f(f(y)) = f(f(x)) = 2x$ which implies that $2y=2x$ and hence, $x = y$. We now proceed by the method of trippling an involution. In this case, we do the change of variable $x = f(y)$ which gives $f(f(f(x))) = 2f(x)$. In other words, in general, $f(2^ix) = 2^if(x)$. Now we define $f(1)=k$. We have that $f(2^i) = 2^ik$. For some k such that $f(k) = 2$. We have no information about $f(k)$, however, implying that there are at least infinitely many. Turns out that this functional equation leaves out all of the odd numbers and specifies nothing about them. Hence, the odd numbers mean that the function can be very ugly.