# Four Pegs That Form a Square

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

Four pegs on a square grid form a square, i.e., the pegs are located at the corners of a grid square. The pegs are allowed to jump - taking turns - over each other so as to land on a grid point across their original position (at the same distance from the jumped over peg.) The jumped over pegs remain in place. Is it possible that after a series of such moves the four pegs form a square larger than the original?

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.

What if applet does not run? |

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.

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Copyright © 1996-2018 Alexander Bogomolny(By Nathan Bowler.)

Now consider the case with an auxiliary peg. Suppose the pegs are initially at **Z**}, **Z** being the set of whole numbers. Suppose for a contradiction that this lattice contains some square of side

Now let *l* = gcd(m, n).*l* is a linear combination of m and n with integer coefficients, *l*)*l*, 0)*l*. Then we have **Z**}.*l**l* · (px, py) = (un, vn)**Z**. But then it follows that the intersection of L with the square *l* points. Since it actually contains (n/*l*)^{2} points we have *l* ≥ (n/*l*)^{2}*l* = 1

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Copyright © 1996-2018 Alexander Bogomolny71769459