What A Mean Square

We are trying to find integers $n > 1$ such that $\tfrac{1}{n}\sum_{j=1}^n j^2 \,=\, \tfrac16(n+1)(2n+1) = m^2$ is a perfect square. This condition rearranges to read $(4n+3)^2 - 48m^2 \; = \; 1 \;,$ and so we are interested in the solutions of Pell's equation $x^2 -48y^2 = 1$ . Considering the continued fraction expansion of $\sqrt{48} = [6,\overline{1,12}]$ , the first few solutions in positive integers are $(x,y) = (7,1)\,,\,(97,14)\,,\, (1351,195)$ . Note that these can be obtained as $x + y\sqrt{48} = (7 + \sqrt{48})^n$ for $n = 1,2,3$ . Since $97$ is not congruent to $3$ modulo $4$ , we are not interested in that solution, and so (since $n > 1$ ) we want $4n+3 = 1351$ , and hence $n = \boxed{337}$ .

Our objective is to find the least integer $n$ greater than $1$ which satisfies the Diophantine equation $(n+1)(2n+1)=2.3m^2$ Since $2n+1$ is odd, this immediately implies that $2|(n+1)$ , i.e. $n$ is odd. Writing $n=2k-1$ , for some positive integer $k>1$ , we have $k(4k-1)=3m^2$ Now note that $\text{gcd}(k, 4k-1)=1$ and $4k-1$ can never be a perfect square. Hence the solution of the above equation is always of the form $k=m_1^2, 4k-1=3m_2^2$ For some positive integers $m_1,m_2>1$ . Taking these two conditions together, we have $4m_1^2=1+3m_2^2$ This immediately give us a necessary condition $m_2^2=1 \mod(4)$ , i.e., $m_2=1,3 \mod(4)$ . Since we are looking for the smallest solution $m_2>1$ , a simple search over the first few positive integer reveals that $m_2=15$ is the smallest integer satisfying the above Diophantine equation and hence $m_1=13$ . Thus $n=2k-1=2m_1^2-1=337$

It is well known that $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}6.$ Divide this by $n$ to find the average; equate this to the square of a number $a$ . $(n+1)(2n+1) = 6a^2.$ Both sides of the equation must be even, but $2n+1$ is odd. Therefore $n+1$ must be even. This allows us to substitute $n = 2m-1\ \ \therefore\ \ m(4m-1) = 3a^2.$ Assuming $m > 1$ (easy to check), the two factors $m$ , $4m-1$ have no common factors. Moreover, $4m-1$ cannot be a perfect square (because it $\equiv 3$ mod 4), so it must be 3 times a perfect square. Thus we split the square: $a = bc\ \ \therefore\ \ m = b^2,\ \ 4m-1 = 3c^2 \\ 3c^2 = 4b^2 - 1\ \ \therefore\ \ 3c^2 = (2b+1)(2b-1).$ Since $b > 1$ the two factors on the right have no common factors. Again, split the square: $c = pq\ \ \therefore\ \ b = \tfrac12(p^2 \pm 1) = \tfrac12(3q^2 \mp 1). \\ p^2 - 3q^2 = \mp 2.$ It is easy to see that $p$ and $q$ must both be odd. Substitute and divide by 4: $p = 2s-1, q = 2t-1 \ \ \therefore\ \ s(s-1) - 3 t(t-1) = 0\ \text{or}\ 1.$ The smallest solution of this equation is $s = 3$ and $t = 2$ . Then, working backward $p = 2\cdot 3 - 1 = 5;\ \ q = 2\cdot 2 -1 = 3; \\ c = 5\cdot 3 = 15;\ \ b = \tfrac12(5^2+1) = \tfrac12(3\cdot 3^2 - 1) = 13; \\ a = 13\cdot 15 = 195;\ \ m = 13^2 = 169;\ \ (\text{check:}\ 4\cdot 169 - 1 = 675 = 3\cdot 15^2); \\ n = 2\cdot 169 - 1 = \boxed{337}.$