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What is the smallest integer $n$ greater than $1$ such that the root mean square of the first $n$ integers is an integer?

The root mean square is defined as:

$$\sqrt{\left(\frac{a_1^2 + a_2^2 + a_3^2+\dots+a_n^2}{n}\right)}$$

I started off by the relation:

$$\sum^n_{i = 1}i^2 = \frac{n(n+1)(2n+1)}{6}$$

We need an $n$ such that:

$$\sqrt{\frac{n(n+1)(2n+1)}{6n}} = k$$

$$\implies (n+1)(2n+1) = 6k^2$$

$$\implies 2n^2 + 3n + (1-6k^2) = 0$$

$$\implies n = \frac{-3 \pm \sqrt{9 - 8(1-6k^2)}}{4} = \frac{-3 \pm \sqrt{1 + 48k^2}}{4}$$

For the equation to have rational solutions $1 + 48k^2$ must be a perfect square. So, the problem boils down to solving the smallest $k$ such that $1 + 48k^2$ is a perfect square.

I have no idea how to proceed.

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  • $\begingroup$ Observe that $2(n+1)-(2n+1)=1\implies (n+1,2n+1)=1$ $\endgroup$
    – lab bhattacharjee
    Oct 16, 2013 at 16:05
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    $\begingroup$ The denominator in the third equation should be $6n$. $\endgroup$
    – Michael Hoppe
    Oct 16, 2013 at 16:05
  • $\begingroup$ Let's procede by trying $k=0$ (trivial), $k=1$, $k=2$ and so on. Just remember that you missed the $6$ in the denominator so you are actually trying $1+48k^2$. $\endgroup$
    – Carlos Eugenio Thompson Pinzón
    Oct 16, 2013 at 16:09
  • $\begingroup$ artofproblemsolving.com/Wiki/index.php/1986_USAMO_Problems $\endgroup$
    – lab bhattacharjee
    Oct 16, 2013 at 16:11
  • $\begingroup$ @MichaelHoppe: Thanks for pointing out the mistake. Edited. $\endgroup$
    – Gerard
    Oct 16, 2013 at 16:14

2 Answers 2

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It involves solving the equation $x^2-48y^2=1$ in integers $x,y$ since it has a smallest solution $(7,1)$ we get infinitely many solutions $(x_n,y_n)$ from the relations $(x_n+y_n\sqrt{48})=(7+\sqrt{8})^n$.And if you don't know how I reached this relation then visit the following link: http://en.wikipedia.org/wiki/Pell%27s_equation

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Starting with $1+48k^2= m^2$ or $m^2-48k^2 = 1$, it makes sense to try convergents to $\sqrt{48}$, which are $\left\{6,7,\frac{90}{13},\frac{97}{14},\frac{1254}{181},\frac{1351}{195},\frac{17466}{2521}, ...\right\}$.

We spot solutions $(m, k) = (7, 1), (97, 14), (1351, 195), \;\dots $ corresponding to $n = 1, \frac{47}2, 337, ...$

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