Show that if you sum 9999 consecutive squares, the result cannot be a perfect power.

Basically everything in the beginning is the same as what I wrote earlier. It's just that the reasoning at the end is different.

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Because the problem didn't specify, there are actually three cases to consider. The first case is that all of the original integers are non-negative. The second case is that all of them are non-positive, it is the same as the first case due to symmetry. The third case is that some of them are positive and some of them are negative. Although it appears that the problem is only asking for case 1 (since you wouldn't really consider 9, 4, 1, 0, 1, 4, 9 consecutive squares), I'll do it just for completion.

CASES I/II: All integers are non-negative or all are non-positive.

We all know that the sum of the first n squares of positive integers is n(n+1)(2n+1)/6. Let's say that the series is (n+1)^2 + (n+2)^2 + ... + (n+9999)^2. Keep in mind that n could be -1 so that the series starts at 0.

Now the sum that we are seeking is the sum of the first (n+9999) squares minus the sum of the first n squares. Assuming my algebra is correct

S=(n+9999)*(n+10000)*(2n+19999)/6-n(n+1)(2n+1)/6
  =9999n^2+99990000n+333283335000
  =3333(3n^2+33330000n+99995000)

For S to be a perfect power, the quadratic must be some multiple of 3333. So it has to be divisible by 3. However, the quadratic cannot be divisible by 3 because the first two terms are and the last term is not. Therefore S cannot be a perfect power.

CASE III: Some integers are positive and negative

Let's there are n positive integers (starting from 1). Because 0 is included, there will be 9999-n-1 negative integers (starting from -1).

The sum of the positive-integer squares is n(n+1)(2n+1)/6. The sum of the negative-integer squares is (9998-n)(9999-n)(19997-2n)/6.

S=n(n+1)(2n+1)/6+(9998-n)(9999-n)(19997-2n)/6
  =9999n^2-99970002n+33318335500
  =3333(3n^2-29994n+99965003)

Same as Cases I/II, the quadratic needs to be a multiple of 3333. But it's not divisible by 3 again. So S is not a perfect power.

Posted by np_rt on 2004-10-03 10:49:21