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(Number theory) Sum of three squares solution proof

  1. Jan 26, 2017 #1
    1. The problem statement, all variables and given/known data

    Find all integer solutions to x2 + y2 + z2 = 51. Use "without loss of generality."

    2. Relevant equations

    3. The attempt at a solution

    My informal proof attempt:

    Let x, y, z be some integers such that x, y, z = (0 or 1 or 2 or 3) mod 4
    Then x2, y2, y2 = (0 or 1) mod 4
    So x2 + y2 + z2 = [ (0 or 1) + (0 or 1) + (0 or 1) ] mod 4
    Since 51 = 3 (mod 4) = x2 + y2 + z2, then x, y, z = (1 or 3) mod 4

    It is obvious that the solution is the permutations of ##\pm1, \pm1, \pm7##.

    It's my first proof course and I'm a little shaky. Is my logic correct? I feel like I took a leap from "Since 51..." to the solution, is there a more formal way to write that? I'm also not sure how to use wlog here. Thanks.
     
  2. jcsd
  3. Jan 26, 2017 #2

    fresh_42

    Staff: Mentor

    I have a additional solution.
     
  4. Jan 26, 2017 #3
    If my logic is correct, then x, y, z can take on values of 1, 3, 5, 7. I gave it some thought and I'm not seeing it :nb)
     
  5. Jan 26, 2017 #4

    fresh_42

    Staff: Mentor

    ##2\cdot 25 + 1##
    The remark "use w.l.o.g." probably refers to the assumption ##x \geq y \geq z \geq 0## which you could make.
     
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