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What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square?

Initially, this seemed like one could work it out with $AM-GM$, but it doesn't seem so.

From $AM-GM$ one gets that $$\frac{1^2+2^2+ \dots+n^2}{n} \geqslant \sqrt[\leftroot{-1}\uproot{2}n]{1^2\cdot2^2\dots\cdot n^2} $$

is this of any help here?

Remark. Thanks to Favst, the source of the problem is Problem 1 of 1994 British Mathematical Olympiad Round 2

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  • $\begingroup$ No., AM-GM seems completely useless here. As a hint... you can simplify $\sum\limits_{k=1}^n k^2$. If you don't know how, see here, but I encourage you to try to find it on your own if you can. $\endgroup$ – JMoravitz Jul 23 '20 at 15:42
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    $\begingroup$ Does this answer your question? Number Theory and Square Number Problem $\endgroup$ – Favst Jul 23 '20 at 16:05
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The mean of the squares $1^2, \ldots, n^2$ is $$ f(n) = \frac{1}{n} \sum_{i=1}^n i^2 = \frac{2n^2+3n+1}{6}$$ It is an integer if and only if $n \equiv 1$ or $5 \mod 6$. The first $n > 1$ for which it is a square is $337$, where $f(337) = 38025 = 195^2$. There are infinitely many. See OEIS sequence A084231.

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This question has already been posted on this website. See my solution via Pell's equation here, where I wrote that the answer is $337.$ The problem appeared on the 1994 British Mathematical Olympiad Round 2

Edit: as suggested by Batominovski, I am copying my old solution here:

A while ago, I found this problem as the 1994 British Mathematical Olympiad - Round 2, Problem 1, but the solution is mine. Here it is.

The equation is $$m^2=\frac{1}{n}\sum_{k=1}^{n}{k^2}=\frac{1}{n}\cdot \frac{n(n+1)(2n+1)}{6}=\frac{(n+1)(2n+1)}{6}.$$

With some manipulation, this is equivalent to $$(4n+3)^2-48m^2=1,$$ which can be solved by Pell's equation. The fundamental solution for $D=48$ in Pell's equation $x^2-Dy^2=1$ is $(x,y)=(7,1),$ so all solutions are parameterized by $$x_t + y_t \sqrt{48}=(7+\sqrt{48})^t.$$ We want to find the first solution $t>1$ for which $x_t\equiv 3\pmod{4}.$ While $t=2$ does not work, $t=3$ yields $$1351+195\sqrt{48}.$$ Since $1351=337\cdot 4+3,$ the answer is $337.$

We can check that $$\frac{(337+1)(2\cdot 337+1)}{6}=3^2\cdot 5^2\cdot 13^2.$$

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  • $\begingroup$ I suggest that you copy your solution here. It is likely that the old thread will be deleted. Your answer should be preserved here. I think this one has a better chance of surviving deletion because the OP show some work. $\endgroup$ – Batominovski Jul 23 '20 at 16:20
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    $\begingroup$ @Batominovski sure, I've pasted it above here now. $\endgroup$ – Favst Jul 23 '20 at 16:43
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Hint

Start from $$1^2+2^2+\cdots +n^2={n(n+1)(2n+1)\over 6}$$

Note that the number ${1^2+2^2+\cdots +n^2\over n}$ becomes a non-integer rational, so perfect squares can mean to be square of a quotient.

Remark

As @Batominovsky stated, the noted number cannot be a non-integer perfect square as no elimination of prime factors of $6$ can lead to a perfect square in the denominator.

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    $\begingroup$ If this number $\dfrac{1^2+2^2+\ldots+n^2}{n}=\dfrac{(n+1)(2n+1)}{6}$ is to be a perfect square, it can only be a perfect square of an integer. This is because the denominator $6$ is squarefree. $\endgroup$ – Batominovski Jul 23 '20 at 15:47
  • $\begingroup$ Nice catch @Batominovski ! $\endgroup$ – Mostafa Ayaz Jul 23 '20 at 15:47
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Following a hint by Mostafa Ayaz, we have $(n+1)(2n+1)=6k^2$ for some integer $k$. That is, $$(4n+3)^2-3(4k)^2=1\,.$$ Hence, $(4n+3)+(4k)\sqrt{3}=(2+\sqrt{3})^m$ for some nonnegative integer $m$. Therefore, $$4n+3=\sum_{r=0}^{\left\lfloor\frac{m}{2}\right\rfloor}\,\binom{m}{2r}\,2^{m-2r}\,3^r\,.$$ If $m$ is odd, then $$4n+3\equiv 2m\cdot 3^{\frac{m-1}{2}}\pmod{4}\,.$$ If $m$ is even, then $$4n+3\equiv 3^{\frac{m}{2}}\pmod{4}\,.$$ Since $$4n+3\equiv 3\pmod{4},$$ we need $$m\equiv 2\pmod{4}\,.$$
That is, $m=4s+2$ for some nonnegative integer $s$ $$4n+3+(4k)\sqrt{3}=(7+4\sqrt{3})\,(97+56\sqrt{3})^s\,.$$ That is, $n=a_s$ and $k=b_s$, where $$a_s:=\frac{(7-4\sqrt{3})\,(97+56\sqrt{3})^s+(7-4\sqrt{3})\,(97-56\sqrt{3})^s-6}{8}$$ and $$b_s:=\frac{(7-4\sqrt{3})\,(97+56\sqrt{3})^s-(7-4\sqrt{3})\,(97-56\sqrt{3})^s}{8\sqrt{3}}\,.$$ Note that $a_0=1$, $a_1=337$, and $$a_s=194\,a_{s-1}-a_{s-2}+144\text{ for }s=2,3,4,\ldots\,.$$ Furthermore, $b_0=1$, $b_1=195$, and $$b_s=194\,b_{s-1}-b_{s-2}\text{ for }s=2,3,4,\ldots\,.$$ Therefore, the next smallest pair $(n,k)$ apart from the one given by Robert Israel is $$(n,k)=(a_2,b_2)=(65521,37829)\,.$$

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