I recently learned about a puzzle called a "Diffy". You take four numbers positioned at the vertices of a square. On the midpoint of each side, you write the difference of the two numbers on the vertices of that side. Once you have done this for all four sides, you connect the four new numbers with a square and repeat.
My conjecture is that it always comes to four zeroes, if we repeat to infinity. I have not yet been able to prove this, though I tried with variables $w, x, y, z$ as the corners of the square. Unfortunately, I ended up with math mush. Can someone help me prove or disprove this?
First, you can show that all four numbers will eventually become even. This lets you conclude the result by induction on the largest of the four numbers; this maximum never increases, and when all four numbers are even, you can cut them all in half and consider a smaller problem.
The parities of any numbers will become even in at most $4$ steps, as shown below. Here, $0$ represents any even number, $1$ any odd number.
\begin{array}{ccccccc} \begin{matrix}0&0\\0&1\end{matrix} & \to & \begin{matrix}0&1\\0&1\end{matrix} & \to & \begin{matrix}0&1\\1&0\end{matrix} & \to & \begin{matrix}1&1\\1&1\end{matrix} & \to & \begin{matrix}0&0\\0&0\end{matrix} \\ &&\uparrow \\ && \begin{matrix}1&1\\1&0\end{matrix} \end{array}
In general, if you do the same process with numbers on the vertices of an $n$-gon, then the numbers will always all become zero if and only if $n$ is a power of $2$.