There is a checker at $(1, 1)$ of the lattice $(x, y)$ with $x, y$ positive integers. It moves as follows. At any move it may double one coordinate, or it may subtract the smaller coordinate from the larger. Which points of the lattice can the checker reach?
(Source: Problem Solving Strategies by Engel, Ch. 1 Problem 15)
The book gives the solution that $(x,y)$ is reachable if and only if $\gcd(x,y)=2^n$ for some $n\in\mathbb{N}$. It explains that with any move, $\gcd(x,y)$ doubles or remains the same, so necessarily $\gcd(x,y)=2^n$ has to be true for $(x,y)$ to be reachable.
However, I am stuck on the sufficient part of the condition (to which the book provides no answer). Let's say that $\gcd(x,y)=2^n$ is true. Then, is $(x, y)$ always reachable? I tried to prove that we could go from any point to another with the operations as long as the gcd is the same, but to no avail. I also considered that Bezout's Identity could be useful, but couldn't utilize it in a useful way.
Why is $\gcd(x,y)=2^n$ a sufficient condition for $(x, y)$ to be reachable?
We first show that any $(x,y)$ with $x,y$ positive integers and $\gcd(x,y)=1$ is reachable.
Assume that
$$S:=\{(x,y) : \gcd(x,y)=1 , (x,y)\text{ is not reachable}\}\not=\emptyset.$$ Then there is an element $(x,y)$ in $S$ with the smallest value of $x^2+y^2$. $x$ and $y$ are both odd otherwise we can divide the even element by $2$ and find another element of $S$ with a smaller value of $x^2+y^2$. Since $\gcd(x,y)=1$ then $x\not=y$ because $(1,1)$ is not in $S$. If $x>y$ then $(x+y,y)\in S$ and $((x+y)/2,y)\in S$ (note that $x+y$ is even). It follows that $$\frac{(x+y)^2}{4}+y^2<x^2+y^2$$ getting a contradiction. So $S$ has to be empty.
Finally, if $\gcd(x,y)=2^n$ then $x=2^nX$, $y=2^nY$ for some positive integers $X,Y$ such that $\gcd(X,Y)=1$. Therefore $(X,Y)$ is reachable which implies that $(x,y)= (2^nX,2^nY)$ is reachable too.