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Summing Squares Sequentially! (Posted on 2004-10-01) Difficulty: 3 of 5
Show that if you sum 9999 consecutive squares, the result cannot be a perfect power.

See The Solution Submitted by Old Original Oskar!    
Rating: 2.0000 (2 votes)

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Solution Puzzle Solution Comment 10 of 10 |

Let the squares to be summed be denoted by be (y + i)^2, where:
i = -4999, -4998, -4997, ......., -3, -2, -1, 0, 1,
2, 3, ......., 4998, 4999

Then, by conditions of the problem:

y^2 + Sum (j = 1 to 4999) [(y+j)^2 + (y-j)^2] = z^p, where z is an
integer with p> =2
Or, 9999*(y^2) + (4999*5000*9999)/3 = z^p
Or, (3*11*101) [3*y^2 + (4999)*(5000)] = z^p.
This is possible only when z=3a for some integer a.

Accordingly, we have:

(11*101) [3*y^2 + (4999)*(5000)] = 3^(p-1)*(a^p).....(*)

Since p> =2, it follows that the rhs is divisible by 3.

However, neither 4999 nor 5000 is divisible by 3. Also:
(3, 11) = (3, 101) = 1

Thus, the lhs of the relationship (*) is NOT divisible by 3.

This is a contradiction.

Consequently,  the sum of sum 9999 consecutive squares can NEVER correspond to a perfect power.


  Posted by K Sengupta on 2007-05-23 05:26:49
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