**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

According to the Faulhaber’s formulas

S now becomes

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

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

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.