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    University of Minnesota

    (March 16, 2004) Factoring x n ? 1: cyclotomic and Aurifeuillian polynomials

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    Polynomials of the form x 2 ? 1, x 3 ? 1, x 4 ? 1 have at least one systematic factorization x n ? 1 = (x ? 1)(x n?1 + x n?2 +... + x 2 + x + 1) Equivalently, polynomials like x 2 ? y 2, x 3 ? y 3, and x 4 ? y 4 have factorizations For odd n, replacing y by ?y gives a variant x n ? y n = (x ? y)(x n?1 + x n?2 y +... + xy n?2 + y n?1) x n + y n = (x + y)(x n?1 ? x n?2 y +... ? xy n?2 + y n?1) For composite exponent n one obtains several different factorizations x 30 ? 1 = (x 15) 2 ? 1 = (x 15 ? 1)(x 15 + 1) x 30 ? 1 = (x 10) 3 ? 1 = (x 10 ? 1)(x 20 + x 10 + 1) x 30 ? 1 = (x 6) 5 ? 1 = (x 6 ? 1)((x 6) 4 +... + 1) Such algebraic factorizations yield numerical partial factorizations of some special large numbers, such as 2 33 ? 1 = (2 11) 3 ? 1 = (2 11 ? 1)(2 22 + 2 11 + 1) 2 33 ? 1 = (2 3) 11 ? 1 = (2 3 ? 1)(2 30 +... + 1) Thus, 2 33 ?1 has factors 2 3 ?1 = 7 and 2 11 ?1 = 23·89. It is then easier to complete the prime factorization 2 33 ? 1 = 7 · 23 · 89 · 599479 But that largish number 599479 might be awkward to understand. How do we verify that a number such as N = 599479 is prime? That is, how do we show that N is not evenly divisible by any integer D in the range 1 < D < N? One could divide N by all integers D between 1 and N, but this is needlessly slow, since if D evenly divides N and D> ? N then N/D is an integer and N/D < ? N. That is, we need only do trial divisions by D for D ? ? N. And, after dividing by 2, we need only divide by odd numbers D thereafter. Also, we need only divide by primes, if convenient. For example, since N = 101 is not divisible by the primes D = 2, 3, 5, 7 no larger than ? 101 ? 10, we see that 101 is prime.

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    Title : (March 16, 2004) Factoring x n ? 1: cyclotomic and Aurifeuillian polynomials
    Abstract : Polynomials of the form x 2 ? 1, x 3 ? 1, x 4 ? 1 have at least one systematic factorization x n ? 1 = (x ? 1)(x n?1 + x n?2 +... + x 2 + x + 1) Equivalently, polynomials like x 2 ? y 2, x 3 ? y 3, and x 4 ? y 4 have factorizations For odd n, replacing y by ?y gives a variant x n ? y n = (x ? y)(x n?1 + x n?2 y +... + xy n?2 + y n?1) x n + y n = (x + y)(x n?1 ? x n?2 y +... ? xy n?2 + y n?1) For composite exponent n one obtains several different factorizations x 30 ? 1 = (x 15) 2 ? 1 = (x 15 ? 1)(x 15 + 1) x 30 ? 1 = (x 10) 3 ? 1 = (x 10 ? 1)(x 20 + x 10 + 1) x 30 ? 1 = (x 6) 5 ? 1 = (x 6 ? 1)((x 6) 4 +... + 1) Such algebraic factorizations yield numerical partial factorizations of some special large numbers, such as 2 33 ? 1 = (2 11) 3 ? 1 = (2 11 ? 1)(2 22 + 2 11 + 1) 2 33 ? 1 = (2 3) 11 ? 1 = (2 3 ? 1)(2 30 +... + 1) Thus, 2 33 ?1 has factors 2 3 ?1 = 7 and 2 11 ?1 = 23·89. It is then easier to complete the prime factorization 2 33 ? 1 = 7 · 23 · 89 · 599479 But that largish number 599479 might be awkward to understand. How do we verify that a number such as N = 599479 is prime? That is, how do we show that N is not evenly divisible by any integer D in the range 1 < D < N? One could divide N by all integers D between 1 and N, but this is needlessly slow, since if D evenly divides N and D> ? N then N/D is an integer and N/D < ? N. That is, we need only do trial divisions by D for D ? ? N. And, after dividing by 2, we need only divide by odd numbers D thereafter. Also, we need only divide by primes, if convenient. For example, since N = 101 is not divisible by the primes D = 2, 3, 5, 7 no larger than ? 101 ? 10, we see that 101 is prime.
    Subject : unspecified
    Area : Mathematics
    Language : English
    Affiliations : 
    Url : http://www.math.umn.edu/~garrett/m/number_theory/aurifeuillian.pdf
    Doi : 10.1.1.121.3592

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