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Solving equation in finite field

  1. Nov 8, 2006 #1
    Hi all,

    I encountered a problem I don't know how to solve: In [itex]\mathbb{Z}_{21}[/itex], I want to find numbers x such that [itex]x^2 = x[/itex].

    I don't know what to do with that, I just wrote that (since 21 divides [itex]x^2 - x[/itex])

    [tex]
    21y = x^2 - x
    [/tex]

    for some integer y, but it didn't seem to help me much.

    Could you give me please some advice?


    And I have one more (probably trivial, but I can't find it nowhere) question regarding [itex]Z_{n}[/itex]: These fields consist of elements [itex]\{0, 1, ... , n - 1\}[/itex]. But what if I am given something like -7 or [itex]\frac{2}{5}[/itex], how should I represent it in this field?

    Thank you.
     
  2. jcsd
  3. Nov 8, 2006 #2

    Hurkyl

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    Z_21 isn't a finite field. If it was, solving this problem would be trivial. (You'd do exactly the same thing you did in your precalculus classes)

    But you can factor Z_21 into the product of two finite fields...


    -7 should be easy. (Remember that Z_n is just working modulo n)

    As for 2/5, that would simply require solving the equation 5x = 2. The extended Euclidean algorithm will be useful here. (run it on (5, n))
     
    Last edited: Nov 8, 2006
  4. Nov 8, 2006 #3

    mathwonk

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    solve it as usual, i.e. x(x-1) = 0, so either x = 0, or x-1 = 0, or x(x-1) is a multiple of 3(7). e.g. x=7 works.

    but as i noticed on my midterm in college when i had not studied this topic, it aint too hard to find all solutions of a problem when there are only 21 possible solutions. i.e. try them all.
     
  5. Nov 8, 2006 #4

    Hurkyl

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    Hey, wait a minute! I thought x² - x = (x - 15)(x - 7)!

    (Yes, I know both are true. I just thought it would be fun to say it that way. :wink:)
     
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