# CMO 1984 Problem 1

Problem 1 of the Canadian Mathematical Olympiad 1984

Prove that the sum of the squares of 1984 consecutive positive integers cannot be the square of an integer.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to "Red Bean" Đỗ Đẩu)

Lets write the sum as follows S = n² + (n + 1)² + (n + 2)² + . . . + (n + 1983)² Now expand the squares, we have

S = 1984n² + 2n (1 + 2 + 3 + . . . + 1983) + 1² + 2² + 3² + . . . + 1983²

According to the Faulhabers formulas

1 + 2 + 3 + . . . + n = n (n + 1)/2 and
1² + 2² + 3² + . . . + n² = n(n + 1)(2n + 1)/6

S now becomes

S = 1984n² + 1983 × 1984n + 1983 × 1984 × 3967 / 6
= 992( 2n² + 1983 × 2n + 661 × 3967 ).

It is straightforward to obtain the prime number decomposition of 992:

992 = 2^531.

For 992( 2n² + 1983 × 2n + 661 × 3967 ) to be a perfect square we have to have

2n² + 1983 x 2n + 661 x 3967 = 62m², (*)

for some integer m.

Note that the product 661 × 3967 is an odd number, and the sum on the left is odd whereas on the right 62m² is even. Thus we can not find an integer m to satisfy (*).

Therefore, the sum of the squares of 1984 consecutive positive integers cannot be the square of an integer.

Note: This problem works for year 2010 even though the solution is different and much more difficult.