Sum of Squares

Number Theory Level pending

Let $x,n>1$ be positive integers such that the following relation is satisfied :

${\large \sum_{i=1}^n \left(\frac{i}{x}\right)^2 = 1}$ .

Find the minimum value of $n$ .

1 solution

Janardhanan Sivaramakrishnan
Jan 19, 2015

The problem is equivalent to $\sum_{i=1}^n i^2 = x^2$ for some positive integer $x$ .

Discounting the trivial case of $n=x=1$ , simple calculations would show that $\sum_{i=1}^{24} i^2 = 4900 = 70^2$ and there is no smaller solution.

Hence, the required solution is $\boxed{24}$ .

It is easy to show that there is no solution for $1<n<24$ . But, there is no easy method to find a third solution (or to comment on its existence).