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Sum of squares

  1. Jun 24, 2008 #1
    Hi everybody :smile:
    I'm currently reading Burton's Elementary Number Theory (almost done!) and in the chapter about Lagrange's Theorem about the sum of four squares, there is a supposedly easy question which I can't solve for some reason :blushing:. I'd really appreciate a hint or two...

    Prove that at least one of any four consecutive natural numbers is not a sum of two squares [that is, can't be represented as the sum of two squares of whole numbers]

    Thank you all! :smile:
     
  2. jcsd
  3. Jun 24, 2008 #2
    Well I think this one works?

    2 + 3 + 4 + 5 = 14

    1^2 = 1
    2^2 = 4
    3^2 = 9
    4^2 = 16

    It has to be the SUM of TWO squares... 9 + 4 = 13. None of the others work.
     
  4. Jun 24, 2008 #3

    CRGreathouse

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    Consider four consecutive numbers mod 4, then consider squares mod 4. The result follows immediately.

    iwin2000: the problem was to show the result for all {n, n + 1, n + 2, n + 3}, not just for one such instance.
     
  5. Jun 24, 2008 #4
    The square of any natural number mod 4 has to be 0 or 1. Therefore, the sum of two such squares mod 4 has to be 0, 1, or 2. However, out of four consecutive natural numbers mod 4, one has to be 3. Contradiction. Is that right?

    Great hint! Thank-you very much :biggrin:
     
  6. Jun 27, 2008 #5

    CRGreathouse

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    You got it.

    I like minimal hints.
     
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