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Working with finite fields of form Z_p

  1. May 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Let p be an odd prime. Then Char(Z_p) is nonzero.

    Prove: Not every element of Z_p is the square of some element in Z_p.

    2. Relevant equations

    3. The attempt at a solution

    I first did this, but i was informed by a peer that it was incorrect because I was treating the congruency as an equality:

    Suppose not. Then every element of Z_p is the square of some element in Z_p. Take 1. Since in mod p: 1 = (p-1)2 = 12, it follows that 0 = p2 - 2p, and hence p = 2, a contradiction.

    What can i do to find a correct way of proving this?
  2. jcsd
  3. May 2, 2012 #2


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    Gold Member

    In a finite field, the group of units is cyclic. This gives you a quick way of proving the result.
  4. May 2, 2012 #3


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    Homework Helper

    It's probably even easier if you simply observe that if i^2=k then (-i)^2=k. But since your first proof is so bogus, you probably really don't understand what Z_p is. Try doing a warmup and figuring out what elements of Z_3, Z_5 and Z_7 are squares and which aren't.
  5. May 2, 2012 #4
    Are you allowed to divide by p in Zp?

    For example suppose we're in Z5. Now

    5 * 2 = 10 = 0 (mod 5)

    and 5 * 3 = 15 = 0 (mod 5).

    So since 5*2 = 5*3 we conclude that 2 = 3.

    What's wrong with that proof?
  6. May 2, 2012 #5
    Yes, I suppose that's silly... dividing by p (which is congruent to zero). I have done what dick suggested and discovered that for Z_3, 2 is not a square. For Z_5, neither 2 nor three are squares. I am trying to work off this for now but cant seem to generalize these cases.
    Last edited: May 2, 2012
  7. May 2, 2012 #6


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    Ok, I think you got what Z_p means. Now use counting logic. If every element of Z_p were a square then for every element z of Z_p there is an element k such that k^2=z. If 1^2=1 and (-1)^2=(p-1)^2=1 do you have enough elements of Z_p to give each element a square root?
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