Two possible equivalent statements regarding iterations of a map on $\mathbb{Z}_+\times\mathbb{Z}_+$ Consider the map $f:X\to X$ where $X=\mathbb{Z}_+\times\mathbb{Z}_+$ and $\mathbb{Z}_+$ denotes the set of positive integers and
$$
f(x,y) := 
\begin{cases}
(2x,y-x)& \text{if $x<y$},\\
(x-y,2y)& \text{if $x>y$},\\
(x,y)   & \text{if $x=y$}.
\end{cases}\;
$$
Question.
Let $(a,b)\in X$. Are the following two statements equivalent?

*

*The ratio $\displaystyle\frac{a+b}{\gcd(a,b)}$ is some (positive) power of $2$, i.e.,
$$
\log_2\left(\frac{a+b}{\gcd(a,b)}\right)\in \mathbb{Z}_+\tag{0}
$$


*There exists a positive integer $n$ such that
$$
f^{n}(a,b) = (c,c)\tag{1}
$$
where $c:=(a+b)/2$, [added: and $f^n$ means function compositions].

Background.
This question is closely related to a recent one I asked on MathOverflow (MO). Here, the question focuses on a specific condition ($0$), which is inspired by several exchanges of comments under the linked question on MO.
Thanks to some observations of the map $f$ below, one can write a program with any given $(a,b)$ to simulate iterations of $f$ to see if $n$ in ($1$) exists. All the cases I have tested so far say yes to the above question. The statement is particularly true for two simple cases, $(3,13)$ and $(3,9)$, which were used in some unsuccessful attempts mentioned on MO.
Here are some observations of the map; some have been mentioned on MO:

*

*The sum of the two components of $f^{n}(x,y)$ is fixed for all $n$.


*Since the sum is fixed, by the pigeonhole principle, we must have
$$
f^{M}(x,y) \in \{f^{k}(x,y)\mid k = 1,2,\cdots, M-1\}\;.
$$


*If ($1$) is ever true, then we must have $2\mid (x+y)$.


*The map $f$ is homogeneous: $f(kx,ky) = kf(x,y)$ for any positive integer $k$.
 A: The key insight to this problem is that, at any step of the process of repeating applying the function, the truth value of statement $(1)$ won't change if we divide both $a,b$ by $gcd(a,b)$ and the expected result $c$ by the same factor immediately .
In other word, for co-prime $a,b$ and arbitrary $k$, $(ak,bk)$ will result in $({ak+bk\over2}, {ak+bk\over2})$ after some iteration if and only if $(a,b)$ will result in $({a+b\over 2}, {a+b\over 2})$ after the same number of iteration.
So the procedure can be safely modified to, each step containing applying $f(x,y)$ and then dividing both numbers by its $gcd$ instead of just applying $f(x,y)$. We denote this function $g$ consisting of $f$ and cancelling out $gcd$, and let the statement $(2)$ be "applying $g$ repeatedly resulting in both side being equal after finite number of steps". It follows that statements $(1)$ and $(2)$ are equivalent.
The problem thus becomes proving $(0)$ and $(2)$ are equivalent.
Now consider the cases:
$(i)$ $x, y$ are even and odd. It is clearly both statements $(0)$ and $(2)$ are false.
$(ii)$ $x, y$ are both odd or both even. Now after applying one $f(x,y)$ both numbers becomes even, and therefore we immediately will divide them by a number greater or equal to $2$ so the new $x,y$ will become a strictly different pair than $x,y$ where the sum is strictly less than $x+y$. So in less than $x+y$ steps, this will eventually reach a point where one is even and the other is odd, which is the $(i)$ case, or both are $1$s. If both are $1$s then both statements $(0)$ and $(2)$ are true.
Edit: it is also necessary to show the truth value of $(0)$ won't change when dividing both number by its $gcd$, which is trivial.
A: I believe the answer is yes. Rewrite the map $f$ as $f(x; s)$ where $s=x+y$. Then $f(x;s) = 2x$ if $x\lt \frac s2$, $s-2x$ if $x\gt \frac s2$, and if $x=\frac s2$ then we have our fixed point. Now, let $s=2^kt$ where $t$ is odd, and consider the behavior of $f(\cdot;s)\mod t$; the fixed point will have $f^n(x;s)\equiv 0\bmod t$. But it's easy to see that $f^n(x;s)\equiv (-1)^i2^nx\mod t$ for some $i$, so this can only happen if $x\equiv 0\bmod t$. This is equivalent to the stated condition on the GCD, so that condition is necessary.
OTOH, if we have $x\equiv 0\bmod t$, then we can consider the behavior of the iteration $\mod 2^k$; again, $f^n(x;s)\equiv (-1)^i2^nx$ for some $i$. But if we let $m=x\bmod 2^k$ then it's clear that there's some minimal $a$ such that $2^am\equiv 0\bmod 2^k$, and then $2^{a-1}m$ will be $\equiv 2^{k-1}\bmod 2^k$. Then by the Chinese Remainder Theorem, $f^{a-1}(x;s)$ will equal $s/2$. This should establish sufficiency.
