Let
S(3)= 13+23+…+(2n)3
and
S(2)= 12+22+…+n2
For what integer values of n will the ratio r=S(3)/S(2) have an integer result?
List all existing answers.
Source: Russian Math Olympiad
S(3)= 1^3+2^3+.....+ (2n)^3
=[ {2n(2n+1)}^2}]/4
= (n^2)(2n+1)^2
S(2) = 1^2 + 2^2 + .....+n^2
= n(n+1)(2n+1)/6
6n(2n+1)
So, S(3)/S(2) = ---------------- .......#
n+1
Now, gcd(n,n+1) = 1
Also, 2(n+1)-(2n+1) =1
Hence, gcd(n+1, 2n+1) =1
Accordingly, from #, we must have:
o n+1 divides 6
Since n is a positive integer, and 1,2,3, and 6 constitute the positive integer factors of 6, it follows that:
n+1= 1,2,3,6
=> n= 1,2,5, disregarding n=0, which is inadmissible.
1, 2, and 5 seems to be the required values of n in consonance with the given conditions.
AS A CHECK
For n=1, we see that: r = S(3)/S(2) = 9/1 =9
For n =2, we see that: r = S(3)/S(2) = 100/5= 20
For n=5, we see that: r = S(3)/S(2) = 55^2/55 = 55
This CONFIRMS that n=1, 2, and 5 are INDEED the desired values in consonance with all the provisions governing the puzzle under reference.
Edited on May 19, 2022, 10:21 am
Puzzle Solution
