Rope-stretching a right corner

A colleague asked today if one could find positive integers a, b, c, and d so that a² + b² + c² = d², or if that was an open problem.  He added that he'd heard that Egyptians stretched ropes to create 3-4-5 triangles in order to form right angles, and was wondering about the possibility of a three-dimensional analog.

I mentioned the Google Group investigating the harder problem of finding a rectangular box with integer sides, integer diagonals, and integer main diagonal.  (See http://groups.google.com/group/theperfectcuboid?lnk=iggc.)

But only while driving home did it occur to me that it's straightforward to produce lots of examples of my colleague's easier problem.

Start with your favorite primitive Pythagorean triple (a, b, c).  (See my earlier post about Pythagorean triples:  http://byoshiwara.blogspot.com/2009/12/blog-post.html.)

Then c is odd, so c = 2n + 1, and

a² + b² + [2n(n + 1)]² = [2n(n + 1) + 1]²

For example, 3² + 4² + 12² = 13².

Generating Pythagorean Triples

The 5 millennia old clay tablet designated Plimpton 322 contains a trig table. The second and third columns represent a leg and hypotenuse of a right triangle with positive integer sides. The rows are arranged in approximately equal steps of angle.

The existence of such a table suggests that the Babylonians were adept at producing Pythagorean triples (integers a, b, and c satisfying a² + b² = c²), a trick which is also useful to many algebra, geometry, and trig teachers attempting to create exercises with nice values.

Every positive Pythagorean triple can be generated by choosing positive integers u and v with u > v and setting a = 2uv, b = u² - v², and c = u² + v² (or by scaling such a triple by a positive integer). We'll derive that fact below. (Pythagorean triples with no common factor are called primitive Pythagorean triples, and all the primitive Pythagorean triples are generated when u and v are relatively prime with exactly one of them being odd.)

It's straightforward to verify that the a, b, and c so defined do form a Pythagorean triple. And conversely, if a, b, and c form a Pythagorean triple, then (a/c, b/c) is a point on the unit circle , so the positive Pythagorean triples can be mapped onto the rational points of the unit circle that lie in the first quadrant.

The line y = 1 + mx will intersect the unit circle at (0,1) and also at a point in the first quadrant when the slope m is between -1 and 0. In fact, we can find the x-coordinate of the second intersection point by solving the equation x² + (1 + mx)² = 1--we find that , so .

Thus the second intersection point is a rational point if m is rational. Of course the slope between (0,1) and any rational point on the unit circle is rational, so we have a 1-1 correspondence between positive rational points on the unit circle and rational slopes between -1 and 0.

We now assume that m is a rational number between -1 and 0, so we can write m = -v/u, where u and v are positive integers with u > v. Then the second intersection point we found above has the form

.

Thus every rational point on the unit circle can be written in this form. In particular, every primitive Pythagorean triple a, b, and c can be expressed as above in terms of u and v.