Start with a sequence $S =(a, b, c, d)$ of positive integers and find the derived sequence $$S_1 = T(S) = (|a −b|, |b−c|, |c−d|, |d −a|).$$
Does the sequence $S, S_1, S_2 = T(S_1), S_3 = T(S_2), \ldots$ always end up with $(0, 0, 0, 0)$?
Observations:
$(0, 3, 10, 13) \rightarrow (3, 7, 3, 13) → (4, 4, 10, 10) →(0, 6, 0, 6) → (6, 6, 6, 6) → (0, 0, 0, 0)$
$(8, 17, 3, 107) → (9, 14, 104, 99) → (5, 90, 5, 90) → (85, 85, 85, 85) → (0, 0, 0, 0)$
$(91, 108, 95, 294) → (17, 13, 99, 203) → (4, 86, 104, 186) → (82, 18, 82, 182) → (64, 64, 100, 100) → (0, 36, 0, 36) → (36, 36, 36, 36) → (0, 0, 0, 0)$
(This is Ross Millikan's answer, but in more detail.)
Consider your recursion scheme mod $2$. Then $$(0,1,0,0),\ (1,0,1,1)\to(1,1,0,0)\to(0,1,0,1)\to(1,1,1,1)\to(0,0,0,0)\to(0,0,0,0).$$ Since (modulo rotation) each of the $16$ possible bitstrings of length $4$ occurs in this sequence, and all of them are transformed into $(0,0,0,0)$ after at most $4$ steps we know that after $\leq4$ steps the resulting quadruple consists of even numbers only. At this point we can divide all four numbers by $2$ without harm done concerning the length of the procedure. After at most $4$ further steps we again have arrived at an even quadruple and can again divide by $2$, and so on. Noting that the $\max$ of the $a_i$ can only decrease in one step it follows that all is over after about $4\lceil\log_2(m)\rceil$ steps, where $m$ is the largest of the numbers $a_i$ given at the beginning.