A proof for this challenging math problem?!?

I have a proof. It is quite ugly and I am sure that a better proof exists. (It seems like there might be a proof involving an analogue of Taylor's theorem and some ring theory, but I'm not seeing it right now.)

---

I'm going to re-cast your problem slightly so I can make my argument more easily.

Let T be the set of all 4-tuples (a, b, c, d) of nonnegative integers. We will not distinguish between cyclic permutations of 4-tuples; for example, (1, 2, 3, 4) = (2, 3, 4, 1). (Elements of T correspond to possible numberings of four vertices of a square by nonnegative integers.)

Define a function f: T -> T by f(a, b, c, d) = (|a - b|, |b - c|, |c - d|, |d - a|). (Application of the function f corresponds to one iteration of the process you have described.)

Note that f is well-defined under the convention that cyclic permutations of 4-tuples are equal, because applying f to a cyclic permutation of a 4-tuple is a cyclic permutation of f applied to the original 4-tuple. (For example, f(1, 2, 5, 10) = (1, 3, 5, 9), and f(2, 5, 10, 1) = (3, 5, 9, 1) = (1, 3, 5, 9) also.)

Let S_0 be the set of 4-tuples (a, b, c, d), where a, b, c, and d are positive integers. (This corresponds to possible initial numberings of the square.)

Let S_1 = f(S_0); let S_2 = f(S_1); ...; let S_{k + 1} = f(S_k). (Then S_k corresponds to possible numberings of the square after k iterations of the process from a legal starting point.)

Let S be the union of S_k over all nonnegative integers k. (Then S corresponds to all possible numberings on the square after some number of iterations of your process from a legal starting point. It turns out that S does NOT equal all of T; it's a proper subset.)

Let f^k(s) denote application of f, k times. So, for example, f^3(s) = f(f(f(s))).

We wish to show that, for every s in S_0, there exists k such that f^k(s) = (0, 0, 0, 0). (This is just a recasting of your problem in my notation.)

To this end, define a "norm" N on the set S by N(a, b, c, d) = max(a, b, c, d). For example, N(1, 3, 8, 5) = 8. Note that the norm is always an integer.

---

LEMMA. For each nonzero s in S, there exists k such that N(f^k(s)) < N(s).

(If you don't see the point of this lemma immediately, then skip to the next section after the proof, and read that first.)

PROOF OF LEMMA:

Case 1: s = (a, b, c, d), where a, b, c, d are all nonzero. Then

|a - b| < max(a, b) < max(a, b, c, d);

|b - c| < max(b, c) < max(a, b, c, d);

|c - d| < max(c, d) < max(a, b, c, d);

|d - a| < max(d, a) < max(a, b, c, d).

Thus, N(f(s)) < N(s), which establishes this case.

Case 2: s = (0, a, 0, b), where a and b are nonzero. Then

f(0, a, 0, b) = (a, a, b, b), which now reduces to case 1.

Case 3: s = (0, 0, a, b), where a and b are nonzero. Note that s is not in S_0, so s must be in S_{k + 1} for some nonnegative integer k. Thus, s is the image under f of S_k; that is, there exists (x, y, z, w) such that f(x, y, z, w) = (0, 0, a, b). This means that |x - y| = 0, |y - z| = 0, |z - w| = a, and |w - x| = b, which means x = y = z. But since x = z, then |z - w| = |x - w| = |w - x|, which means a = b.

We then have

f(0, 0, a, b) = f(0, 0, a, a)

f(0, 0, a, a) = (0, a, 0, a), which reduces to case 2.

Case 4: s = (0, a, b, c), where a, b, c are nonzero, a ≠ b, and b ≠ c. Then:

f(0, a, b, c) = (a, |a - b|, |b - c|, c), which reduces to case 1 by the assumptions of this case.

Case 5: s = (0, a, a, a/2), where a is nonzero even. Then:

f(0, a, a, a/2) = (a, 0, a/2, a/2);

f(a, 0, a/2, a/2) = (a, a/2, 0, a/2) = (0, a/2, a, a/2), which reduces to case 4.

Case 6: s = (0, a, a, b), where a and b are distinct and nonzero, and b ≠ a/2. Then:

f(0, a, a, b) = (a, 0, |a - b|, b) = (0, |a - b|, b, a). Note that |a - b| ≠ b, because if so then a - b = b or b - a = b, so either a = 2b or a = 0, either of which is a contradiction of the assumptions for this case. Thus, this reduces to case 4.

Case 7: s = (0, a, 2a, 2a), where a is nonzero. Then:

f(0, a, 2a, 2a) = (a, a, 0, 2a)

f(a, a, 0, 2a) = (0, a, 2a, a), which reduces to case 4.

Case 8: s = (0, a, b, b), where a and b are distinct and nonzero, and b ≠ 2a. Then:

f(0, a, b, b) = (a, |a - b|, 0, b) = (0, b, a, |a - b|). Note that a ≠ |a - b|, because if so then a = a - b or a = b - a, so either b = 0 or b = 2a, either of which is a contadiction of the assumptions for this case. Thus, this reduces to case 4.

Case 9: s = (0, a, a, a), where a is nonzero. Then

f(0, a, a, a) = (a, 0, 0, a) = (0, 0, a, a), which reduces to case 3.

Case 10: s = (0, 0, 0, a), where a is nonzero. Note that s is not in S_0, so s is in S_{k + 1} for some nonnegative k. Thus, there exists (x, y, z, w) such that f(x, y, z, w) = (0, 0, 0, a). Then |x - y| = 0, |y - z| = 0, and |z - w| = 0, so x = y = z = w, so |w - x| = 0. This contradicts the fact that a was nonzero. So this case cannot happen (that is, there are no elements of S of the form (0, 0, 0, a) where a is nonzero, so we do not have to consider this case).

This completes the proof of the lemma. (Note that all cases are covered, because any nonzero element of S is of one of the forms (*, *, *, *), (0, *, *, *), (0, 0, *, *), (0, *, 0, *), or (0, 0, 0, *), as I am not viewing cyclic permutations as distinct--so, for example, the 4-tuple (2, 0, 3, 5) equals (0, 3, 5, 2), so is handled in the (0, a, b, c) case.)

---

This lemma directly implies what we want, because if we start with a 4-tuple (a, b, c, d), where a, b, c, and d are nonnegative, then

* Since (a, b, c, d) is in S, then there exists k such that N(f^k(a, b, c, d)) < N(a, b, c, d). If N(f^k(a, b, c, d)) = 0, then we are done. Otherwise:

* Since f^k(a, b, c, d) is in S, then there exists p such that N(f^(k + p)(a, b, c, d)) < N(f^k(a, b, c, d)). If N(f^(k + p)(a, b, c, d)) = 0, then we are done. Otherwise...

...and so on. This process cannot continue indefinitely, because the norm is always an integer, and thus N(a, b, c, d) can be decreased only finitely many times before it eventually becomes zero.

So, for large enough j, N(f^j(a, b, c, d)) = 0, which means f^j(a, b, c, d) = (0, 0, 0, 0).

once you're making the radius a million and the periods a million aside, they wont cover the airplane. The circumferences of the circles will touch, yet there will be gaps. once you're making your circles a minimum of sqrt2 cases the size of the periods, then they are going to cover the airplane. i assume you could enable the circles to overlap.