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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.

  Submitted by Old Original Oskar!    
Rating: 2.0000 (2 votes)
Solution: (Hide)
We can express the sum as (n-4999)²+(n-4998)²+...+(n+4999)² and we wish to show that it cannot be B to the P for integer B, P (P>1).

The sum is 9999n²+2(1²+2²+...+4999²)=
9999n²+2x4999x5000x9999/6 =
3333(3n²+4999x5000)

If this is B to the P, then 3 must divide B, and thus 3 to the P divides B to the P. However, 3333(3n²+4999x5000) can only be divided by 3, so P=1.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionPuzzle SolutionK Sengupta2007-05-23 05:26:49
Correct Solutionnp_rt2004-10-03 10:49:21
re(4): Solution (need more work)np_rt2004-10-02 10:38:36
re(2): SolutionKen Haley2004-10-02 00:00:37
re: SolutionKen Haley2004-10-01 23:57:21
re(3): Solution (slight correction)Federico Kereki2004-10-01 16:11:28
re(2): Solution (slight correction)np_rt2004-10-01 15:35:04
re: SolutionFederico Kereki2004-10-01 15:24:10
Solutionnp_rt2004-10-01 14:15:05
Some ThoughtsI am really smartRandyOrton2004-10-01 13:56:53
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