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

(In reply to Solution by np_rt)

Didn't follow your whole proof; I stopped where you said the sum of the first n positive integers is n(n+1)(2n+1)/6.  Not true; it's n(n+1)/2.

I don't know what this does to your proof....

Posted by Ken Haley on 2004-10-01 23:57:21