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Simple algebra problem (fields)

  1. Apr 16, 2008 #1
    [SOLVED] Simple algebra problem (fields)

    1. The problem statement, all variables and given/known data
    Solve x²+x+41=0 in the fields Z43 and Z167

    2. The attempt at a solution
    Well Z43={0,1,...,42} so any y that is n=0 mod 43 works...

    x²+x+41=0=43 -> x²+x=2 -> x=1

    Is this how it's done or am I doing it incorrectly?

    If the solution above is right, then what about the same equation in the field Z167? I can't find any x that is in that field and solves the problem.
  2. jcsd
  3. Apr 16, 2008 #2


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    Don't be intimidated by the fact you're working in a finite field! It's just a quadratic equation, and you've known how to solve them for a long time!

    Huh? y? n?

    The first implication is correct (and it's an "if-and-only-if"), but you have the second one backwards. x=1 is one solution to that quadratic, but you have not proven it is the only solution.
  4. Apr 16, 2008 #3


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    How would you normally solve x2+ x- 2= 0? What are its roots? What do negative numbers correspond to "mod 43"?
  5. Apr 16, 2008 #4
    Thanks for your answers.

    But I still don't get it for Z167:


    [tex]x=\frac{-1\pm\sqrt{1-4\cdot 1\cdot(-126)}}{2}[/tex]

    The only way I get a reasonable result from that is if (under the square root):
    or: -126=41 => 4*1*41=164 => 1-4*1*(-126)=1-164=-163=4
    or: -4*1*(-126)=504=3 => 1-4*1*(-126)=1+3=4

    [tex]x=\frac{-1\pm\sqrt{4}}{2}=\frac{-1\pm 2}{2}[/tex]

    But then neither of the possible values of x is in Z167.
    Last edited: Apr 16, 2008
  6. Apr 16, 2008 #5


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    You're working in Z167, so shouldn't you use its division operator, rather than the division operator for rational numbers?
  7. Apr 16, 2008 #6
    Yes, right.

    I tried that but got a result that didn't solve the equation so I thought that my approach was wrong.

    I tried it once again and this time got a result that works.

    Ok, I've got it.

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