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Proof involving the sum of squared integers

  1. May 8, 2013 #1

    Entropee

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    1. The problem statement, all variables and given/known data

    Theorem: the numbers in the set {99, 999, 9999, ... } cannot be written as two squared integers, but at least one can be expressed as the sum of 3 squared integers.


    2. Relevant equations

    Well there are a lot of examples but lets go with 32 + 32 + 92 = 99

    We may have to use Euclid's division algorithm as well.


    3. The attempt at a solution

    If we call the first part P and the second part Q then we can assume [itex]\neg[/itex]P [itex]\vee[/itex] [itex]\neg[/itex]Q, and try to show a contradiction here.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 8, 2013 #2

    mfb

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    What is P, what is Q?

    I would consider remainders here.
     
  4. May 8, 2013 #3

    Entropee

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    Sorry I'm on an ipad and was being lazy haha. Let p stand for "the numbers in that set cannot be written as two squared integers" and Q stand for "at least one number in the set can be represented as three integers squared." How should remainders be used here?
     
  5. May 8, 2013 #4

    mfb

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    Staff: Mentor

    Just show both of them separately, there is no need to start with logical statements. You found an example how at least one of those numbers can be written as sum of three squares. Fine, this part is done. Now you have to show that none of those numbers is the sum of two squares.

    As always. Test if an equality can hold mod some number - if not, it cannot be true at all.
     
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