The problem: Given $2^n$ positive integers around a circle. In a step, we write down the positive difference of any consecutive numbers, and delete the previous numbersa (so there will be the same number of numbers on the circle after each step). Prove that after finite many steps, each number will be 0.
I have a half proof, but something is missing. How can I prove that after some steps each number on the circle will be even? And how can I find counterexamplea for the problem if there are $k\neq 2^n$ numbers?
This is a very cool problem!
To show that the numbers eventually become even, notice that $|x-y|$ has the same parity as $x+y$. Therefore, we can consider a different process, where you replace the $k^\text{th}$ number in the circle $a_k$ with $a_k + a_{k+1}$, where the indices wrap around $\pmod{2^n}$. As long as we can show these numbers eventually become even, it follows the original number will as well.
Let $a_k^{(t)}$ be the value of $k^\text{th}$ number in the circle after $t$ steps in the modified process. It is easy to prove by induction on $t$ the following formula for $a_k^{(t)}$: $$ a_k^{(t)}=\sum_{i=0}^t \binom{t}{i} a_{k+i} $$ Now, let $t=2^n$. You can show that the coefficient $\binom{2^n}{i}$ is even whenever $0<i<2^n$. Therefore, modulo $2$, we can throw out all but the first and last terms of the above sum without affecting the parity, so $$ a_{k}^{(2^n)}=\sum_{i=0}^{2^n} \binom{2^n}{i} a_{k+i}\equiv a_k+a_{k+2^n}=a_{k}+a_{k}\equiv 0\pmod 2 $$ That is, after $2^n$ steps, every number is even. Note that $a_{k+2^n}=a_k$ because the indices wrap around mod $2^n$.
When the number $m$ of numbers in the circle is not a power of $2$, suppose that one of the number is odd, say, $a_0$, while the rest are odd. Then by the same logic, $$ a_{0}^{(2^n)}\equiv a_0+a_{2^n}\equiv 1+0=1 \pmod 2 $$ where $a_{2^n}\equiv 0\pmod2$ because $2^n$ is not a multiple of $m$.