Newsgroups: rec.puzzles
From: hoey@ai.etl.army.mil (Dan Hoey)
Date: 20 Dec 89 20:01:01 GMT
Subject: Re: Old chestnut (spoiler)

a...@myrias.com (Alfred Nykolyn) writes:

>Find 3 points with integer coordinates in the XY plane that
>form an equilateral triangle.

There are no such three points.  For if there were, we could translate
the figure so that one of the points is at the origin.  (We could
*not* necessarily rotate the figure so that a second point is on the
X-axis, as Randall Rathbun suggested.  Rotation could move a point
from integer coordinates to non-integer coordinates.)  Also, if the
four nonzero coordinates have a common factor, we may divide through
by their GCD to form a minimal equilateral triangle.

Let the endpoints be O=(0,0), A=(Xa,Ya), and B=(Xb,Yb).  To have an
equilateral triangle we must have |A|^2 = |B|^2 = |A-B|^2.  Now |A|^2
mod 4 is 0, 1, or 2 according as Xa and Ya are both even, mixed,
or both odd.  Since |B|^2 is equal, Xb and Yb must have the same
population of even coordinates.  Similarly, Xa-Xb and Ya-Yb must also
have the same population of even coordinates.

If Xa and Ya are both odd then Xb and Yb must both be odd, so Xa-Xb
and Ya-Yb will both be even, so the triangle is not equilateral.

If Xa and Ya have mixed parity, then Xb and Yb must have mixed parity.
This will make Xa-Xb and Ya-Yb both even (if Xa and Xb have the same
parity) or both odd (if not).  In either case the triangle is not
equilateral.

If Xa and Ya are both even, then Xb and Yb must both be even,
contradicting the minimality of the triangle.

Therefore no such triangle exists.

Dan