Four Pegs That Form a Square

The applet bellow allows one to experiment with the following problem:

All the small dots are draggable and, when dragged, snap into legal positions. As a variant, you can select to work with five dots, in which case you can, at the outset, move the extra dot to any location you wish. After that, it behaves as a regular dot. An extra control modifies the size of the square relative to the grid.

The problem with 4 dots has a beautiful proof, that is presented in an intuitive and a formalized forms. The answer to the question is negative.

With an additional peg available, the grid could be modified to make jumps shorter than the side length of the square. May this matter? You may want to think this over before checking the answer.

Now consider the case with an auxiliary peg. Suppose the pegs are initially at (0, 0), (0, n), (n, 0), (n, n) and (x, y). The minimal lattice A of this set is {(an + kx, bn + ky): a, b, k in Z}, Z being the set of whole numbers. Suppose for a contradiction that this lattice contains some square of side m < n with sides parallel to those of the original. Then (0, m) is in A (since A is a lattice) and so x is a rational multiple of n. Similarly so is y. Thus by a suitable rescaling we may assume that m, n, x and y are all integers.

Now let l = gcd(m, n). Then since l is a linear combination of m and n with integer coefficients, (0, 0), (0, l) and (l, 0) are all in A, and so the standard square lattice L which is the minimal lattice of these points is a sublattice of A. Let p be the least positive integer such that px and py are both divisible by l. Then we have L = {(an + kpx, bn + kpy): a, b, k in Z}. Indeed, we may include the restriction 0 k < n/l since n/l · (px, py) = (un, vn) for some u, v in Z. But then it follows that the intersection of L with the square [0, n)×[0, n) contains at most n/l points. Since it actually contains (n/l)² points we have n/l (n/l)² and so n/l = 1 and so n|m, which is the desired contradiction.