From: voloch@fireant.ma.utexas.edu (Felipe Voloch)
Subject: Re: A trig equation
Date: 12 Mar 1999 20:13:48 GMT
Newsgroups: sci.math.research
Keywords: linear relations between roots of unity
Thomas Mattman (mattman@math.mcgill.ca) wrote:
: I'm interested in the following equation:
:
: cos( Pi*j/p) + cos( Pi * k/q) = 1
: where j,k,p,q are positive integers 0 < j < p/2 and 0 < k < q/2.
:
: It seems that the only solutions are j/p = k/q = 1/3. However
: I haven't been able to prove it. Any suggestions would be
: welcome. I think I have an argument that p = q. So ideas
: for that case would also be helpful.
:
: Thomas Mattman
: Mattman@math.mcgill.ca
Let z = exp(Pi*j/p), w=exp(Pi * k/q), which are complex roots
of unity. Your equation is (z+z^{-1})/2 + (w+w^{-1})/2) =1
which I'll rewrite as z+z^{-1} + w+w^{-1} -1-1 = 0.
A theorem of Mann classifies all vanishing sums of roots of unity
and the rewritten equation is just that. I'm sure Mann's theorem
implies what you want, i.e., j/p = k/q = 1/3.
The reference is:
H. B. Mann, {\it On linear relations between roots of unity},
Mathematika {\bf 12} (1965) 107-117.
Felipe
==============================================================================
From: Thomas Mattman
Subject: Re: A trig equation
Date: Mon, 15 Mar 1999 15:36:39 GMT
Newsgroups: sci.math.research,sci.math,sci.math.symbolic
Your right, Mann's theorem does the job. The equation is
"irreducible" (see the paper for a definition) and Mann's
theorem implies that either -1, w and z are at the corner's
of an equilateral triangle or else they are all 30th roots of
unity. In either case, we deduce that j/p = k/q = 1/3.
Thanks very much for your help. Thanks also to David
Reiss for suggesting the Mathematica notebook.
Thomas
Felipe Voloch wrote:
[see above -- djr]