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

Because the problem didn't specify, there are actually two 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)

If S is indeed a perfect power (other than 1), the quadratic must contain factors of 33 and 101 (since 3333=33*101), which means that the quadratic must be factorable, which is equivalent to it having those factors as roots. However, the discriminant is actually negative so it does not have any real roots, meaning the quadratic is prime. Hence, 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 have factors of 33 and 101. But it turns out to it's negative again. So the quadratic is prime.

Edited on October 2, 2004, 10:33 am

Posted by np_rt on 2004-10-01 14:15:05