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Math Help - A polynomial that factors (poly mod n) but has no roots?

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    A polynomial that factors (poly mod n) but has no roots?

    How is it possible for a polynomial to factor (poly mod n) but have no roots? That is no integers x such that f(x)\equiv0\bmod{n}.

    Give an example of a polynomial that factors (poly mod n) and prove that it has no roots.

    I'm completely stumped.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Pn0yS0ld13r View Post
    How is it possible for a polynomial to factor (poly mod n) but have no roots? That is no integers x such that f(x)\equiv0\bmod{n}.

    Give an example of a polynomial that factors (poly mod n) and prove that it has no roots.

    I'm completely stumped.
    Here are some examples :
    If n=2, in \bmod 2, x^2+x+1 has no root. (and it's the only one if n=2)

    So you can take as an example : f(x)=(x^2+x+1)^2=x^4+2x^3+3x^2+2x+1 \equiv x^4+x^2+1 \bmod 2 this factors into (x^2+x+1) \cdot (x^2+x+1) but it has no root \bmod 2.

    Check it out : x \equiv 0~,~1~,~2 \bmod 2 doesn't give any zero for x^2+x+1
    Last edited by Moo; July 29th 2008 at 01:29 PM. Reason: simplified a little more
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  3. #3
    Moo
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    Let's see for n=3.

    x^2+1 has no root \bmod 3 (once again, check it out )

    x^4+1 also has no root \bmod 3.

    Thus f(x)=(x^2+1) \cdot (x^4+1)=x^6+x^4+x^2+1 can be factored, but has no root \bmod 3

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