# Square Root of Sum of Squares

## Main Question or Discussion Point

$$\begin{array}{c} {{A_n}={\sqrt{\sum _{z=1}^{n}{z^2}}} } \\ {{A_1}=1 } \\ {{A_{24}}=70}\end{array}\$$

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

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CRGreathouse
Homework Helper
$$A_n=\sqrt{\frac{n(n+1)(2n+1)}{6}}$$, and for n=24 you have 24/6 * 25 * 49.

CRGreathouse said:
$$A_n=\sqrt{\frac{n(n+1)(2n+1)}{6}}$$, and for n=24 you have 24/6 * 25 * 49.
OK that is 4*25*49 all being squares proves $$A_{24}=2*5*7$$ and confirms what I have already said but that is no proof that another n could or does not exist such that $$A_n$$ is a integer quantity.

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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.

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:

$$y^2 = \frac{(x)(x+1)(2x+1)}{6}$$

http://mathworld.wolfram.com/SquareNumber.html

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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

$$\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}\$$

or solving the Diophantine equation

$$y^2 = \frac{(x)(x+1)^2(2x+1)}{6}$$

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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AntonVrba said:
Interesting to note is that

$$\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}\$$

or solving the Diophantine equation

$$y^2 = \frac{(x)(x+1)^2(2x+1)}{6}$$

has infinite number of integer solutions (the above shows the 1st, 2nd, 3rd and 50th solution) and are easy to generate numerically
$${{B_n}={\sqrt{\sum_{y=1}^{n}{\sum _{z=1}^{y}{z^2}}}}=\sqrt{\frac{n(n+1)^2(n+2)}{12}} }$$

$$y^2 = \frac{(x)(x+1)^2(x+2)}{12}$$

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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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.

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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