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Thread: Factorising x^n-1

  1. #1
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    Factorising x^n-1

    I have been playing around with factorising x^n-1 for positive integers n.
    For example, x^2-1 = (x-1)(x+1)
    X^3-1 = (x-1) ( x^2+x+1).
    I have been using Wolfram to see what happens as n gets large: x^105-1
    (x - 1) (x^2 + x + 1) (x^4 + x^3 + x^2 + x + 1) (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1) (x^12 - x^11 + x^9 - x^8 + x^6 - x^4 + x^3 - x + 1) (x^24 - x^23 + x^19 - x^18 + x^17 - x^16 + x^14 - x^13 + x^12 - x^11 + x^10 - x^8 + x^7 - x^6 + x^5 - x + 1) (x^48 + x^47 + x^46 - x^43 - x^42 - 2 x^41 - x^40 - x^39 + x^36 + x^35 + x^34 + x^33 + x^32 + x^31 - x^28 - x^26 - x^24 - x^22 - x^20 + x^17 + x^16 + x^15 + x^14 + x^13 + x^12 - x^9 - x^8 - 2 x^7 - x^6 - x^5 + x^2 + x + 1)

    I noticed that this is the first time to coeff of the terms are not -1, 0 or 1 . The above in bold has a '2'
    I can't see why 105 would do this but not others?

    Any ideas?
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  2. #2
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    Re: Factorising x^n-1

    It doesn't.

    The geometric series $\displaystyle \begin{align*} 1 + x + x^2 + x^3 + \dots + x^{n - 1} \end{align*}$ has as its closed form $\displaystyle \begin{align*} \frac{1 \left( x^n - 1 \right) }{x - 1} \end{align*}$, so

    $\displaystyle \begin{align*} 1 + x + x^2 + x^3 + \dots + x^{n - 1} &= \frac{x^n - 1}{x - 1} \\ x^n - 1 &= \left( x - 1 \right) \left( 1 + x + x^2 + x^3 + \dots + x^{n - 1} \right) \end{align*}$
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    Re: Factorising x^n-1

    Errmm.. so whats going on with wolfram expansion?
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    Re: Factorising x^n-1

    I think if you go on to fully factorise the expression then the 2 occurs
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    Re: Factorising x^n-1

    Polynomials, like integers, can have multiple factorizations. But, up to ordering, the factorization by irreducible polynomials is unique. Wolframalpha showed you the factorization by irreducible polynomials. If you were to multiply out everything except the first term, you would get:
    (x-1)(x^(104)+x^(103)+...+x+1)

    But, that second term is reducible (it can be factored further).
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    Re: Factorising x^n-1

    Thanks from rodders
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    Re: Factorising x^n-1

    Thanks. Opened my eyes.
    I guess the question is why the case n=105 produces an reducible polynomial over the integers which has coefficients other than -1, 0 or 1.
    The wikipedia link has this:

    "The case of the 105th cyclotomic polynomial is interesting because 105 is the lowest integer that is the product of three distinct odd prime numbers and this polynomial is the first one that has a coefficient other than 1, 0, or −1:"

    But doesn't explain why?
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  8. #8
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    Re: Factorising x^n-1

    Quote Originally Posted by rodders View Post
    Thanks. Opened my eyes.
    I guess the question is why the case n=105 produces an reducible polynomial over the integers which has coefficients other than -1, 0 or 1.
    The wikipedia link has this:

    "The case of the 105th cyclotomic polynomial is interesting because 105 is the lowest integer that is the product of three distinct odd prime numbers and this polynomial is the first one that has a coefficient other than 1, 0, or −1:"

    But doesn't explain why?
    Probably because it is not known why. It has been observed to be true, but an explanation is not yet forthcoming. If you really want to know, follow the research to its current conclusion, then look for some form of math that may be able to extend it further until you are satisfied by the answer. This is how new math is created.
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