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Math Help - Polynomials which factors but have no roots?

  1. #1
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    Polynomials which factors but have no roots?

    Hello all,

    Can anyone think of an example of a polynomial f(g) with integer coefficients which factors (poly mod n) but has no roots, e.g. for which there are no such integers g where f(g)=0(mod n) ?

    What are appropriate values of n, the coefficients of f(g), and how can I show that it has no roots?

    Any help is appreciated!
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by 1337h4x View Post
    Hello all,

    Can anyone think of an example of a polynomial f(g) with integer coefficients which factors (poly mod n) but has no roots, e.g. for which there are no such integers g where f(g)=0(mod n) ?

    What are appropriate values of n, the coefficients of f(g), and how can I show that it has no roots?

    Any help is appreciated!
     x^4+x^2+1=(x^2+x+1)(x^2-x+1)

    Now consider this polynomial modulo  2 .
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    Quote Originally Posted by chiph588@ View Post
     x^4+x^2+1=(x^2+x+1)(x^2-x+1)

    Now consider this polynomial modulo  2 .
    Thank you ! How can I prove that this has no roots though? I can't just say it can't be factored any further with integers, or can I? Do I just find the quadratic roots and show that they aren't integers? Aren't those numbers still roots? Or how does this have no roots still?

    I'm a little confused about this as you can tell...
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by 1337h4x View Post
    Thank you ! How can I prove that this has no roots though? I can't just say it can't be factored any further with integers, or can I? Do I just find the quadratic roots and show that they aren't integers? Aren't those numbers still roots? Or how does this have no roots still?

    I'm a little confused about this as you can tell...
    Just plug in  0 and  1 into the polynomial.
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    Quote Originally Posted by chiph588@ View Post
    Just plug in  0 and  1 into the polynomial.
    If you plug in 0 you get 2, if you plug in 1 you get three. What is your point?
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by 1337h4x View Post
    If you plug in 0 you get 2, if you plug in 1 you get three. What is your point?
    No when you plug in  0 you get  1 .

    Anyway those are all the elements modulo 2 and none of them gave you a root, but you can still factor your polynomial.
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    Answer

    All polynomials have roots. The question is determining those roots.
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by wonderboy1953 View Post
    All polynomials have roots. The question is determining those roots.
    Over the complex numbers yes, but not an arbitrary field like \mathbb{Z}/2\mathbb{Z}.
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    Quote Originally Posted by chiph588@ View Post
    No when you plug in  0 you get  1 .

    Anyway those are all the elements modulo 2 and none of them gave you a root, but you can still factor your polynomial.
    Okay, so when I plug in 0 I get 1, when I plug in 1 I get 3, so how are the results of 1 and 3 relate to modulo 2 ? Is it because 2|(3-1) ? Is that what you mean?

    And from what you and wonderboy are discussing, I take it that when we say there are no roots, we mean that there are no REAL roots? no integer roots? Both?
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  10. #10
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by 1337h4x View Post
    Okay, so when I plug in 0 I get 1, when I plug in 1 I get 3, so how are the results of 1 and 3 relate to modulo 2 ? Is it because 2|(3-1) ? Is that what you mean?

    And from what you and wonderboy are discussing, I take it that when we say there are no roots, we mean that there are no REAL roots? no integer roots? Both?
    Modulo 2 there are no roots what so ever.

    Let  f(x)=x^4+x^2+1 .

     f(0)=1\not\equiv0\bmod{2}
     f(1)=3\equiv1\not\equiv0\bmod{2}

    Thus there are no roots modulo  2 .
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    Quote Originally Posted by chiph588@ View Post
    Modulo 2 there are no roots what so ever.

    Let  f(x)=x^4+x^2+1 .

     f(0)=1\not\equiv0\bmod{2}
     f(1)=3\equiv1\not\equiv0\bmod{2}

    Thus there are no roots modulo  2 .
    So how do you consider the system "modulo 2" then? I was looking back at your work, and then you stated that it was modulo 2. How ?
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  12. #12
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by 1337h4x View Post
    So how do you consider the system "modulo 2" then? I was looking back at your work, and then you stated that it was modulo 2. How ?
    Read your original question. I showed you a polynomial f(x) that can be factored but there is no g such that  f(g)\equiv0 \bmod{2} .
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  13. #13
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    Ah, I see what you're saying. Thanks for making the connection. I lost track of what I was originally asking apparently!
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