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