Results 1 to 3 of 3

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

  1. #1
    Member
    Joined
    Jul 2008
    Posts
    78

    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 $\displaystyle f(x)\equiv0\bmod{n}$.

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

    I'm completely stumped.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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 $\displaystyle f(x)\equiv0\bmod{n}$.

    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 $\displaystyle \bmod 2$, $\displaystyle x^2+x+1$ has no root. (and it's the only one if n=2)

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

    Check it out : $\displaystyle x \equiv 0~,~1~,~2 \bmod 2$ doesn't give any zero for $\displaystyle x^2+x+1$
    Last edited by Moo; Jul 29th 2008 at 01:29 PM. Reason: simplified a little more
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Let's see for $\displaystyle n=3$.

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

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

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

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Mar 4th 2012, 11:41 PM
  2. Polynomials which factors but have no roots?
    Posted in the Number Theory Forum
    Replies: 12
    Last Post: May 28th 2010, 09:37 AM
  3. factors of a polynomial
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Jan 15th 2009, 06:34 AM
  4. Linear factors of polynomial
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Feb 17th 2008, 11:03 PM
  5. Replies: 10
    Last Post: Sep 11th 2006, 05:50 AM

Search Tags


/mathhelpforum @mathhelpforum