This problem was used for year 1984, and it is very easy to prove if the number is 1984. I replaced 1984 with 2010 and the problem becomes much more difficult. Here is the problem:
Problem 1 of Canadian Mathematical Olympiad 1984
Prove that the sum of the squares of 2010 consecutive positive integers cannot be the square of an integer.
Solution by Vo Duc Dien
Let’s write the sum as follows S = n² + (n + 1)² + (n + 2)² + . . . + (n + 2009)²
Now expand the squares, we have
S = 2010n² + 2n (1 + 2 + 3 + . . . + 2009) + 1² + 2² + 3² + . . . + 2009²
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 = 2010n² + 2009 x 2010n + 2009 x 2010 x 4019 / 6 =
335(6n² + 2009 x 6n + 2009 x 4019) for this to be the square of an integer we have to have 6n² + 2009 x6n + 2009 x 4019 = 335m²
6n² + 2009 x6n + 7 x 1153453 = 335m²
6n² + 12054n + 8074171 = 335m² (*)
but we note that the sum on the left is an odd number so m has to be an odd number, and 335m² will always have the units digit of 5.
When the units digit of n is
1 or 6, units digits of 6n² and 12054n are 6 and 4 and units digit of 6n² + 12054n + 8074171 is 1 which is not 5
2 or 7, units digits of 6n² and 12054n are 4 and 8 and units digit of 6n² + 12054n + 8074171 is 3 which is not 5
3 or 8, units digits of 6n² and 12054n are 4 and 2 and units digit of 6n² + 12054n + 8074171 is 7 which is not 5
4 or 9, units digits of 6n² and 12054n are 6 and 6 and units digit of 6n² + 12054n + 8074171 is 3 which is not 5
5, units digits of 6n² and 12054n are 0 and 0 and units digit of 6n² + 12054n + 8074171 is 1 which is not 5
Thus we can not find an integer m to satisfy (*). Therefore, the sum of the squares of 2010 consecutive positive integers cannot be the square of an integer.