KGS math club/solution 11 21

It is assumed that the initial square is the unit square with coordinates (0,0), (0,1), (1,0) and (1,1). Note that jumping always yields points with integer coordinates.

1) There can never be a larger square.

For, if there is a jumping sequence forming a square of side length 3, say, from the unit square, then there is also a jumping sequence from this larger square back to the unit square (the sequence that reverses each step of the original sequence). This "inverse" sequence may be applied to the unit square itself, which would result in a _smaller_ square of side length 1/3. But this is not possible, since jumping always yields integer coordinates.

Note: this not only applies to squares, but to any configuration.

2) Three points can never be collinear.

Consider a jump from A over B. Note that this is realized by adding 2*(B-A) to A. Therefore, the difference vector 2*(B-A) has _even_ coordinates. Hence, modulo 2, the points don't change: in each configuration ever achieved, modulo 2 the points are equal to the corners of the unit square. In particular, modulo 2 the points are all different.

Claim: If three points are collinear, then two of them are equal modulo 2.

Let A, B, C three collinear points, B being the inner point. First, let AB = BC. Then C = A+2*(B-A). Hence A and C are equal modulo 2, as claimed. Now, w.l.o.g., assume AB < BC. Then perform a jump from A over B, thus making A the new inner point. Repeating this process finally yields the first case, where the inner point is equidistant to the outer points.

Summing up: The corners of the unit square are pairwise distinct modulo 2, hence the same is true in all possible configurations. As shown in the claim, in the case of three collinear points, two of them are equal modulo 2. Hence, such a configuration can never be achieved.

Greetings from "Georg" on KGS