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# Bài giải toán Xuân năm 2010

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:

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.

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