what is the smallest greater than $1$ , for which the root of mean square of the first $n$ positive integers is an integer?
In other words,
$$\sqrt{1^2+2^2+3^2+\dots +n^2} $$ is an integer.
my first attempt was $1^2+2^2+3^2+\dots +n^2=\frac{n(n+1)(2n+1)}{6}$
$n(n+1)(2n+1)$ is a multiple of $6$. product of any two consecutive integers is even.
i.e,$3\mid n(n+1)(2n+1)$ and $\frac{n(n+1)(2n+1)}{3}$ is perfect square is sufficient. how to I continue? is there exists another solution? is there exists infinitely many $n$ that satisfies this condition?
The smallest answer is $24$. Actually, it's the only answer (greater than $1$). This is the cannonball problem, which was solved one century ago.