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Math Help - solutions of a biquadratic equation.

  1. #1
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    Lightbulb solutions of a biquadratic equation.

    Consider the biquadratic equation, x^4 + x^3+x^2 +x +1 =0 in Z _11

    I find that the equation has no solutions from a brute force method of trying all numbers from 1 to 10 for x.

    I would like to know if there is another method to actually prove that there is no solution for this equation.

    Thanks in advance,
    MAX
    Last edited by MAX09; May 4th 2011 at 06:44 AM. Reason: Latex error
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by MAX09 View Post
    Consider the biquadratic equation, x^4 + x^3+x^2 +x +1 =0 in Z _11

    I find that the equation has no solutions from a brute force method of trying all numbers from 1 to 10 for x.

    I would like to know if there is another method to actually prove that there is no solution for this equation.

    Thanks in advance,
    MAX
    Surely x=3, 4, 5 and 9 are solutions.

    Note that x^4+x^3+x^2+x+1\equiv 0\ (mod\ 11)\Leftrightarrow \frac{x^5-1}{x-1}\equiv 0\ (mod\ 11)

    x^{10}\equiv 1\ (mod\ 11) (Fermat's little theorem)

    \Rightarrow (x^5-1)(x^5+1)\equiv 0\ (mod\ 11)

    \Rightarrow x^5\equiv \pm1\ (mod\ 11)

    etc.
    Last edited by alexmahone; May 4th 2011 at 07:22 AM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by alexmahone View Post
    Surely x=3, 4, 5 and 9 are solutions.

    Note that x^4+x^3+x^2+x+1\equiv 0\ (mod\ 11)\Leftrightarrow \frac{x^5-1}{x-1}\equiv 0\ (mod\ 11)

    x^{10}\equiv 1\ (mod\ 11) (Fermat's little theorem)

    \Rightarrow (x^5-1)(x^5+1)\equiv 0\ (mod\ 11)

    \Rightarrow x^5\equiv \pm1\ (mod\ 11)

    etc.
    I can't fault your logic, but I'm getting the solution to x^5 \equiv \pm 1~\text{(mod 11)} to be all numbers x in Z_11! Obviously all of these cannot solve the original equation. I don't see how this gets us anywhere. How do you whittle them down to the solutions to the original problem?

    -Dan

    Edit: Oh Heavens I'm being silly. I got it. (I hope you aren't responding to this yet.)
    Last edited by topsquark; May 4th 2011 at 09:01 AM.
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    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by topsquark View Post
    I can't fault your logic, but I'm getting the solution to x^5 \equiv \pm 1~\text{(mod 11)} to be all numbers x in Z_11! Obviously all of these cannot solve the original equation. I don't see how this gets us anywhere. How do you whittle them down to the solutions to the original problem?

    -Dan
    The four solutions (3, 4 ,5 and 9) satisfy x^5\equiv 1\ (mod\ 11)

    On the other hand, the numbers 2, 6, 7, 8 and 10 satisfy x^5\equiv -1\ (mod\ 11)
    Last edited by alexmahone; May 4th 2011 at 09:17 AM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by alexmahone View Post
    The four solutions (3, 4 ,5 and 9) satisfy x^5\equiv 1\ (mod\ 11)

    On the other hand, the numbers 2, 6, 7 and 8 satisfy x^5\equiv -1\ (mod\ 11)
    Yup. I got it (see my last edit), but apparently you were responding before you got the edit. Sorry about that!

    -Dan
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