Square Root of Sum of Squares

  • Thread starter AntonVrba
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[tex]
\begin{array}{c}
{{A_n}={\sqrt{\sum _{z=1}^{n}{z^2}}} } \\
{{A_1}=1 } \\
{{A_{24}}=70}\end{array}\
[/tex]

Is there a proof that only for n =1 or n=24 that An is an integer quantity?
 

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  • #2
CRGreathouse
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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.
 
  • #3
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CRGreathouse said:
[tex]A_n=\sqrt{\frac{n(n+1)(2n+1)}{6}}[/tex], and for n=24 you have 24/6 * 25 * 49.
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.
 
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  • #4
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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.
 
  • #5
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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]

http://mathworld.wolfram.com/SquareNumber.html
 
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  • #6
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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
 
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  • #7
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AntonVrba said:
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
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
 
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  • #8
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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.
 
  • #9
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chronon said:
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.
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
 

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