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Square Root of Sum of Squares

  1. May 3, 2005 #1
    {{A_n}={\sqrt{\sum _{z=1}^{n}{z^2}}} } \\
    {{A_1}=1 } \\

    Is there a proof that only for n =1 or n=24 that An is an integer quantity?
  2. jcsd
  3. May 3, 2005 #2


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    [tex]A_n=\sqrt{\frac{n(n+1)(2n+1)}{6}}[/tex], and for n=24 you have 24/6 * 25 * 49.
  4. May 3, 2005 #3
    OK that is 4*25*49 all being squares proves [tex]A_{24}=2*5*7[/tex] and confirms what I have already said but that is no proof that another n could or does not exist such that [tex]A_n[/tex] is a integer quantity.
    Last edited: May 3, 2005
  5. May 3, 2005 #4
    I think there is a proof. I would check Dickson's History of the Theory of Numbers, if you are near a library that has it.
  6. May 3, 2005 #5
    I got the answer:

    The only numbers that are simultaneously square and pyramidal (the cannonball problem) are 1 and 4900, .........(Dickson 1952, p. 25; Ball and Coxeter 1987, p. 59; Ogilvy 1988), as conjectured by Lucas (1875, 1876) and proved by Watson (1918). The cannonball problem is equivalent to solving the Diophantine equation:

    [tex]y^2 = \frac{(x)(x+1)(2x+1)}{6}[/tex]

    Last edited: May 3, 2005
  7. May 4, 2005 #6
    Robert - thanks but is the fact that finding just one solution proof? it is just one answer. I have to find Watson's proof.

    Interesting to note is that

    [tex]\begin{array}{c}{{B_n}={\sqrt{\sum_{y=1}^{n}{\sum _{z=1}^{y}{z^2}}}}=\sqrt{\frac{n(n+1)^2(2n+1)}{6}} } \\{{B_1}=1 } \\{{B_{6}}=14\\{{B_{25}}=95}\\{etc}\\{B_{19785515999613069781581367686}=113006685964487096197915421007515390839216297500491081054\end{array}\[/tex]

    or solving the Diophantine equation

    [tex]y^2 = \frac{(x)(x+1)^2(2x+1)}{6}[/tex]

    has infinite number of integer solutions (the above shows the 1st, 2nd, 3rd and 50th solution) and are easy to generate numerically
    Last edited: May 4, 2005
  8. May 5, 2005 #7
    Sorry I made a mistake:
    [tex]{{B_n}={\sqrt{\sum_{y=1}^{n}{\sum _{z=1}^{y}{z^2}}}}=\sqrt{\frac{n(n+1)^2(n+2)}{12}} }[/tex]

    [tex]y^2 = \frac{(x)(x+1)^2(x+2)}{12}[/tex]

    The one factor is (n+2) and not (2n+1) and the divisor is 12 not 6

    Careless cut and paste on my part
    Last edited: May 5, 2005
  9. May 12, 2005 #8
    This has got me thinking. Is it possible to to put together squares with sides 1 to 24 to make a square of side 70. It looks unlikely as it is difficult to see how the small squares would fit in - for instance the 1 square has to be surrounded by larger squares. I wondered whether anyone has any more information on fitting squares together in this way.
  10. May 13, 2005 #9
    I doubt if you can fit 1 to 24 in a 70 square, very unlikely

    but 24 different integers squares can be packed into a square with 175 sides, solution in the attched file

    Attached Files:

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