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Sum of 3 squares

  1. Apr 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that if a prime number is a sum of three squares of different primes, then one of the primes must be equal to 3.

    2. Relevant equations

    3. The attempt at a solution

    I really have no idea where to start this one.
  2. jcsd
  3. Apr 22, 2010 #2


    Staff: Mentor

    Start with an equation that represents the given part of what you're trying to prove.
  4. Apr 23, 2010 #3
    So something like:

    Let p, q, r, and s be prime. Then if s = p^2 + q^2 + r^2, either p, q, or r must = 3.

    The only theorem I have on 3 squares is that N >=1 is a sum of three squares if and only if N <> 4^n(8m+7), for some m, n >= 0.
  5. Apr 23, 2010 #4


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    Homework Helper

    If p is a prime different from 3, what is p^2 mod 3?
  6. Apr 23, 2010 #5


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    A couple of hints:

    Try writing your primes as p = 3k + r, r = 0, 1, 2. (Note if k != 1, r cannot be zero, then p isn't prime)

    Consider values mod 3
  7. Apr 26, 2010 #6
    So p would have to be 1(mod 3) ==> a^2+b^2+c^2==0(mod 3) ==> 3|p which is a contradiction, right?
  8. Apr 26, 2010 #7


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    Sure, unless one of the primes is 3.
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