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Primes, pigeon holes, modular arithmetic

  1. Nov 27, 2012 #1
    1. The problem statement, all variables and given/known data

    q.png


    3. The attempt at a solution

    Don't have a clue how to even start this one, sorry.
     
  2. jcsd
  3. Nov 27, 2012 #2

    Zondrina

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    I don't have much time to help with this one. Do you recall what the pigeonhole principle states?

    The n elements of a set get mapped to n-1 elements of another set, so no matter what, there are elements ai and aj which get mapped to the same element or the same 'hole'.
     
  4. Nov 27, 2012 #3
    Yeah I'm happy with the pigeon hole principle, although I can't quite see how it applies as a can be any natural number or 0, so surely the size of set A is infinite?
     
  5. Nov 27, 2012 #4
    That can't be meant, because the pigeonhole principle can only be used if ##\mathcal A## is finite. So ##0\leq a,b<\sqrt p## probable means ##(0\leq a<\sqrt p)## and ##(0\leq b<\sqrt p)##. This would give you ##|\mathcal A| < (\sqrt p+1)^2 = p+2\sqrt p+1##.

    Now I'd look at the function ##f(x,y)=x^2+2y^2## for all pairs ##(x,y)\in\mathcal A##.
     
  6. Nov 27, 2012 #5
    Ah right yeah I thought they were too separate inequalities which really messed me up. Quite simple now.

    Got down to this..

    (b-b')^2 + 2(a-a')^2 = pk

    for some integer k.

    I'm having a little struggle getting rid of the k (so to speak).

    a, b, a', b' are all < sqrt(p)

    so

    (b-b')^2 + 2(a-a')^2 < 3p

    so k < 3

    if k = 1 we're fine, no worries.

    but what about the k = 2 case? I feel like I should return to the x^2 = -2 (mod p) to get some fact about p I could use...?

    Thanks for the help.
     
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