Find the smallest positive integer $n$ so that $\sqrt{\frac{1^2+2^2+...+n^2}{n}}$ is an integer.
Problem
Source: Singapore Open Math Olympiad 2017 2nd Round p3 SMO
Tags: Integer, number theory, Sum, Sum of Squares
Alphatrion
26.03.2020 20:13
parmenides51 wrote: Find the smallest positive integer $n$ so that $$\sqrt{\frac{(n+1)(2n+1)}{6}}$$is an integer.
golema11
26.03.2020 20:17
$n=1$ satisfy
usernameyourself
26.03.2020 20:18
parmenides51 wrote: Find the smallest positive integer $n$ so that $\sqrt{\frac{1^2+2^2+...+n^2}{n}}$ is an integer. [see also here for more unsolved Singapore Contests] Is it not just 1?
parmenides51
26.03.2020 20:24
by the time is has written 2 in the sum it means more than 2, that was the exact wording anyway
a 3-digit number
i3435
26.03.2020 20:46
337 works. If you call the value of the square root k, then square everything and quadratic formula it, the discriminant is 1+48k^2, which must equal m^2 for some m. We already have a solution for this, k=1 and m=7. We can take these as (7-1*sqrt(48)), or (7-4sqrt3), and repeatedly multiply this to itself to get more values of k and m. The next value is k=14 and m=97, but plugging this into the rest of the quadratic formula gives n as a fraction. The next value is k=195 and m=1351. Upon putting this into the quad.formula, we get n=337. This is also the smallest because all possible values of k and m result from this procedure.
Gryphos
26.03.2020 21:59
The exact same problem has been posted yesterday, see here.