Arctangent Diophantine Equations

April 9, 2012

The natural numbers, which are the counting numbers (one, two, three... and first, second, third...), are really useful things. The number, 0.25, can be expressed as 1/4; or, 2/10 + 5/100. You can see that any real number can be expressed in this way, although some numbers will be harder to express than others, and pi will take a little time.

The mathematician, Leopold Kronecker was a fan of the natural numbers. Kronecker said, "God made the integers, all the rest is the work of man." This is a mathematical version of Einstein's statement that God doesn't play dice.

There are equations called Diophantine equations that require integer solutions. I mentioned Diophantine equations in a previous article (The Lonely Runner Conjecture, August 11, 2011). The need for integer solutions to certain problems is illustrated by the cattle problem, a Diophantine problem supposedly communicated by Archimedes to his friend, Eratosthenes. You wouldn't expect a fractional cow to be grazing on the plain. A spherical cow might be a special case.

A recent article on the arXiv Preprint Server examines some trigonometric Diophantine equations.[1] This paper, by Konstantine Zelator, a mathematician at Bloomsburg University of Pennsylvania, examines equations similar to the following:

(Eq. 1)

There's apparently a geometrical proof of this particular equation in a published paper, but the paper is not freely accessible online.[2] As a firm believer of the "information wants to be free" principle, I've done my own proof. The geometrical construction for the proof is shown below.

Geometrical construction to prove that arctan(1/3) + arctan(1/7) = arctan(1/2). (Rendered by the author using Inkscape).

The arctan(1/7) and arctan(1/3) triangles are ABC and ADE. Since you need a triangle that's the sum of the angles, you obviously join the triangles together, as shown. To get the tangent of the summed angle, you need to construct another right triangle, AFG, as shown. The triangle ADE is similar to the triangle ABF, and the triangle CFG is similar to triangle ABC. That knowledge, application of the Pythagorean theorem, and a lot of calculation, proves the equation.

Although a high school student would disagree, I find geometry to be fun. What I don't like is calculation, so I resort to computer models, instead. The general form of Eq. 1, shown as Eq. 2, has many integer solutions. I've written a computer program to find them, and some of these are shown in the table. The source code can be found here. The one caveat, however, is that my equations are certain only within the limits of the arithmetical precision. Since I checked the results using Gnumeric, this is about one part in 10¹⁶, but it isn't mathematical certainty.

(Eq. 2)

α	β	γ	_	α	β	γ
2	3	1	_	15	98	13
3	7	2	_	17	41	12
4	13	3	_	18	47	13
5	8	3	_	22	75	17
5	21	4	_	23	30	13
6	31	5	_	23	83	18
7	18	5	_	27	46	17
7	43	6	_	31	43	18
8	57	7	_	33	76	23
9	32	7	_	34	55	21
9	73	8	_	38	47	21
10	91	9	_	47	83	30
11	50	9	_	57	68	31
12	17	7	_	57	73	32
13	4	3	_	70	99	41
13	21	8	_	80	93	43
13	72	11	_	83	23	18

The fractions, 1/α, 1/β and 1/γ, needn't have just 1 as a numerator. Zelator offers Eq. 3 as an example. My computer offers Eq. 4, and many others.

	(Eq. 3)
	(Eq. 4)

References:

1. Konstantine Zelator, "The Diophantine Equation arctan(1/x)+arctan(m/y)= arctan(1/k)," arXiv Preprint Server, March 28,2012
2. Hasan Unal, "Proof without words: an arctangent equality," Mathematics and Computer Education, vol. 45, no. 3 (Fall 2011), p 197.