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Prove two squares and a cube equal an integer

  1. Mar 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Disprove or prove the statement that every positive integer is the sum of at most two squares and a cube of non-negative integers.


    2. The attempt at a solution
    I'll call the numbers that can be squares a and b. C will be the cube.

    The easiest way to disprove something is to find a counter example. I figured 8 would work if one doesn't include 0 in the allowed values of a, b, and c. However, since the problem says "non-negative integers" instead of "positive" integers, I figure this is in err. (Edit: also, now thinking about it, it says "at most"-- so, 0 must be an allowed state for a, b, and/or c)

    Just from glancing at it, I see that we have a situation where, since a can equal b, we can have 2a + 0 or 2a + 1 (depending on c)... So, that would seem to cover all even/odd integers. The only thing that's bugging me is the spacing. Since a and b are perfect squares, I can't simply say that they will cover all of the possible integers.... I'm not sure how to go about proving/disproving this. Does anyone have any thoughts, or any other thoughts in general as to how to approach this problem?

    Thanks!

    -Max
     
  2. jcsd
  3. Mar 20, 2010 #2
    If I was trying to solve this problem, I would start by working it out for small positive integers:

    [itex]1 = 1^2[/itex]

    [itex]2 = 1^2 + 1^2[/itex]

    [itex]3 = 1^2 + 1^2 +1^3[/itex]

    and so on. If you're lucky, you'll find a relatively small counter example. Otherwise, you might see a pattern that suggests a proof. Try that and see what you find.

    Petek
     
  4. Mar 20, 2010 #3

    Dick

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    Is 8 the only number you tried as a counter example??
     
  5. Mar 20, 2010 #4
    Ah. By brute force, it appears 15 won't work. Somehow, prior to eating dinner, I must've skipped over that one :P I had thought I checked up until 50.

    So, I guess the moral of the story for these kinds of problems is never let elegance get in the way of plain ol' plug and chug.

    Thanks for the insight, guys :)
     
  6. Mar 20, 2010 #5

    Dick

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    7 doesn't work either. You must have missed it.
     
  7. Mar 20, 2010 #6
    Ah. You're entirely right! For some reason, I had written down the squares as 1, 2, 4, 9, 16...

    Better moral of the story: be sure to have full stomach when doing discrete math :)
     
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