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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