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Number theory modular arith.

  1. Sep 4, 2009 #1
    b>1, p an odd prime

    b2 = -1 mod p

    if p|b2 + 1
    show p = 1 mod 4

    I know that the order of b is 4 (mod p)
     
    Last edited: Sep 4, 2009
  2. jcsd
  3. Sep 4, 2009 #2
    That's a strange approach to use...
    I would do it this way: Since p is odd, and p|a^2 + 1, a must be an even number. a^2 thus = 0 (mod 4), and consequently a^2 + 1, and p are both congruent to 1 (mod 4)
     
  4. Sep 4, 2009 #3
    haha, oh man it's so obvious, I can't believe I missed it.
     
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