Find the least positive integer $n >1$ such that the arithmetic mean of the first $n$ non zero perfect squares is again a perfect square. Please help. Hope it gives me an idea of equations with integer solutions.

  • $\begingroup$ You should show some effort to avoid votes to close. Do you know how to compute the arithmetic mean of the first $n$ non-zero perfect squares? $\endgroup$ – J. W. Tanner May 31 at 16:39
  • $\begingroup$ Please don't shout in all caps. $\endgroup$ – joriki May 31 at 16:52
  • $\begingroup$ This is a beautiful problem. The cannonball problem is not equivalent to it because the problem posted takes the arithmetic mean, which the cannonball problem does not. $\endgroup$ – Favst May 31 at 17:11

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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  • $\begingroup$ Thanks for the source and the solution. $\endgroup$ – saibal May 31 at 17:17
  • $\begingroup$ @saibal If you believe that the solution is correct and better than other answers, kindly vote for it and accept it as the preferred answer. $\endgroup$ – Favst May 31 at 17:19

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