rjs (06/23/82)
The answer to the pizza question was instructive (and in fact showed that you cannot divide a circle into 7 equal parts with a straight edge and compass). However, it does not show how to accomplish divisions it claims are possible. My questions are: I can see how to construct the roots of x^pq - 1 = 0 from the roots of x^p - 1 = 0 and x^q - 1 = 0 if p and q are mutually prime (all products of roots, one from each set are roots). But what if they are not mutually prime? Then there are roots (other than 1) in common and multiplying roots doesn't generate enough roots. I.e. how do you get the roots of x^9 - 1 = 0 from the roots of x^3 - 1 = 0? Once you try to get the roots for a Fermat prime > 3 you have the problem of finding roots of a polynomial of degree > 2. This is usually quite difficult, but all polynomials of interest are of the form 1 + x + x^2 + x^3 + ... + x^2^2^n. Is there an easy (or even known) method for finding these roots? Thanks. Robert Snyder, III (harpo!floyd!rjs)
G:shallit (06/23/82)
[1] You hit the nail on the head. You CAN'T construct the 9-th roots, since there is no way to get them from cube roots. The only roots you can construct are products of a power of 2 and a set of Fermat primes. That set includes at most 1 of each prime. In particular, you can't construct the roots corresponding to powers of Fermat primes. [2] With respect to actually finding the roots of the polynomials in question, this can be done. The procedure is not terrible easy to describe, but I can give a reference: Jacobson's Basic Algebra I, pp. 215-218. I must admit this is rough going. But really one needs to be conversant with Galois theory to REALLY understand what is going on here. Herstein's Topics in Algebra is a good intro to Galois theory. /Jeff Shallit, Department of Mathematics, University of California, Berkeley
rwhaas (06/23/82)
If your problem is simply to find the n roots of unity, i.e. the solution to x^n - 1 = 0, then note that x^n=1=exp(2*pi*i) so that all the roots are of the form root[k] = cos(2*pi*k/n) + i*sin(2*pi*k/n), k=0,...,n-1 . As to construction of these roots, the previous articles apply. Roy Haas Bell Labs, Indian Hill