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A variation on a classic problem

  1. May 21, 2012 #1
    Most number theorists will be familiar with the result conjectured in the 19th century and proved in the 20th century that the only square pyramidal numbers that are square numbers are 1 and 4900 (the sum of the squares from 1^2 to 24^2 = 70^2).

    While discussing this, it was pointed out to me that the sum of the squares 1^2 up to 48^2 is 1 short of a perfect square. A little investigation found the same was true of the sum up to 47^2, but I did not find any other small examples.

    This seems intriguing, especially as 48 is double 24. My best guess is that someone must have noticed this 100 years ago, but I have not confirmed this.

    The question is are there any solutions of the diophantine equation:

    1^2 + 2^2 + ... N^2 = M^2 - 1

    other than N=47 and N=48?
     
    Last edited: May 21, 2012
  2. jcsd
  3. May 25, 2012 #2

    What about 1 more than a perfect square (N=2)? Or 4 short (N=7,N=9,N=191,N=192,N=994)?
    Wouldn't they be considered 'small' examples?
     
  4. May 26, 2012 #3
    Well, they are small answers, but to different questions!

    Call S(n) = 1^2 + 2^2 + ... + n^2 for convenience.

    The question as posted was looking for natural numbers n and m such that

    S(n) + 1 = m^2

    I have as yet failed to find any solutions other than the first two.

    I did find a nice sequence of examples of adjacent pairs of numbers where the sum is n^2 short of a perfect square. Some of these are in your examples:

    Look at n = 48*k^2 and 48*k^2-1

    What I wonder is whether the lack of other solutions to the original problem is a theorem, and whether this is related to the old cannonball problem, where the sum has to be an exact square (and the only solutions are N=1 and N=24).
     
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