Ah, Now I understand what Erich is talking about (sorry for being so dense). So here's a way of explaining what's going on: Suppose that f(x) is a monic integral polynomial (in Erich's case it's (x^n-1)/(x-1) for various n, and we want to find those integer m (or more generally rational number m) such that f(x) - m factors non-trivially over the rational numbers (i.e. it's not irreducible and it doesn't factor as a linear times an irreducible). So, if n = degree(f), and k is an integer 2 <= k<= n/2 we're interested in a factorization as a rational polynomial of degree k and one of degree n-k. Make the coefficients (other than the leading one, which is 1) of the factor of degree k be unknowns, and go through the division algorithm of dividing f(x) by that polynomial. We desire the remainder to be a constant. What this does is to give k-1 equations in the k unknown coefficients of the factor of degree k, which should give a space curve that the coefficients must satisfy. If this is irreducible (which it will undoubtedly be) and is of genus >1, by Falting's theorem it will have only a finite number of rational points. If it's of genus 1 it might have an infinite number of rational points, and if it's of genus 0 it definitely does. So I've been doing a few calculations For n=5 (i.e. f(x) = (x^5-1)/(x-1)) only k=2 is possible: the curve is of genus 0, so there are an infinite number of factorizations of (x^5-1)/(x-1) - m into a product of two quadratics (here m must be rational). for n=6 only k=2 is possible and the curve is of genus 1 (though not an elliptic curve in standard Weierstrass form). I don't know yet whether or not it has an infinite number of rational points. for n=7, k=2 and k=3 are possible. The genera of the curves are 2 and 4 respectively, so there are only a finite number of rational m. for n=8, k=2 and 3 are possible. The genera of the curves are 4 and 11 respectively, so again there are only a finite number of rational m. Victor On Mon, Aug 3, 2009 at 11:39 AM, victor miller <victorsmiller@gmail.com>wrote:

Erich, Perhaps I'm not understanding what you're doing. I just calculated the factorizations

of (x^6-1)/(x-1) -m over the rationals for 1<=m<1000 and found the following statistics:

995 of them were irreducible 1 factored into a product of a quadratic and a cubic 4 factored into a linear factor and a quartic 1 factored into a linear factor and a product of two quadratics

for n=8

995 were irreducible 1 factored into a product of a cubic and quartic 3 factored into a linear and a sextic 1 factored into a linear a quadratic and a quartic.

I would expect by some argument using the Hilbert Irreducibility theorem that the asymptotic proportion of m for which there is a linear factor is 0.

Victor

On Fri, Jul 31, 2009 at 4:54 PM, Erich Friedman <efriedma@stetson.edu>wrote:

...

for integers m and n, i've been factoring (over the integers) the polynomials:

P = (x^n-1) / (x-1) - m

obviously when m = (k^n-1)/(k-1) for some integer k, there is a linear factor.

i'm interested in the cases when P factors without a linear factor. i've only found 6 cases:

(x^8-1)/(x-1) - 3 = (x^3 + x – 1)(x^4 + x^3 + x + 2)

(x^14-1)/(x-1) - 4 = (x^3 + x^2 – 1)(x^10 + x^8 + x^7 + 2x^5 + x3 + 2x^2 – x + 3)

(x^5-1)/(x-1) + 11 = (x^2 + 3x + 4)(x^2 – 2x + 3)

(x^9-1)/(x-1) + 19 = (x^4 – x^3 + x^2 – 3x + 4)(x^4 + 2x^3 + 2x^2 + 4x + 5)

(x^11-1)/(x-1) - 23 = (x^2 + x + 2)(x^8 – x^6 + 2x^5 + x^4 – 4x^3 + 3x^2 + 6x – 11)

(x^6-1)/(x-1) - 56 = (x^2 – x + 5)(x^3 + 2x^3 – 2x – 11)

is there some rhyme or reason for these, or are they just random?

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Put x = 1/18 and x = 17/90 in the identity: gamma(x)*gamma(x+1/3)*gamma(x+2/3) = 2*pi*3^(1/2-3x)*gamma(3x), and multiply the results together. Warut On Thu, Aug 6, 2009 at 6:22 PM, <rwg@sdf.lonestar.org> wrote:

can be brutal! Can someone or something do

gamma(1/18)*gamma(17/90)*gamma(7/18)*gamma(47/90)*gamma(13/18)*gamma(77/90)/(gamma(1/6)*gamma(17/30)) = 4*3^(4/15)*%pi^2

1 17 7 47 13 77 gamma(--) gamma(--) gamma(--) gamma(--) gamma(--) gamma(--) 18 90 18 90 18 90 4/15 2 ----------------------------------------------------------- = 4 3 pi 1 17 gamma(-) gamma(--) 6 30 ? --rwg

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...

Put x = 1/18 and x = 17/90 in the identity:

gamma(x)*gamma(x+1/3)*gamma(x+2/3) = 2*pi*3^(1/2-3x)*gamma(3x),

and multiply the results together.

Warut

[Also caught by Dick Askey]

[And Rich, whose message was delayed somehow.]

Gack! Macsyma's and Mathematica's Gamma simplifiers are supposed to try that, so there are bugs in both! --rwg I was lazy to trust them.

Rich pointed out http://arxiv.org/abs/0907.4384, which gives a simple formula for Product Gamma(k/n) 1<=k<=n (k,n)=1 This seems to generalize to /===\ /===\ | | k sqrt(2 pi) phi(n) | | mu(d) n z | | Gamma(z + -) = (-----------------) | | Gamma (---), | | n 1 | | d 1<=k<=n /===\ ----- d | s(n) (k,n)=1 | | p - 1 z (n | | p ) | | p prime p | n where s(n) is the squarefree part of n, phi is Euler's totient, and mu is Moebius's trivalue. For (sort of) legible, see http:\\gosper.org\gamprd.html. (Vertical derangement not in .nb). Just two Gammas for n a prime power, p - 1 ----- 2 m - 1 (2 pi) p m sqrt(p) (------------------) Gamma(p z) /===\ (m (p - 1) + 1) z | | k p | | Gamma(z + --) = ----------------------------------------------, | | m m - 1 m p Gamma(p z) 1<=k<=p (k,p)=1 and for powers of 2, just one: /===\ n - 2 n - 1 | | k 2 - 1/2 2 (1 - n) z n - 1 1 | | gamma(z + --) = (2 pi) 2 gamma(2 z + -) | | n 2 n 2 1<=k<=2 k odd The paper also gives prod Gamma(Farey sequence), which presumably generalizes as above. And both identities should have prod|sum trig analogs, e.g. sum(tan(x+k/n)), (k,n)=1. --rwg