Can the sum of any 2112 consecutive positive integers be a perfect square?

x + (x+1) + ... + (x+2111) = (2x+2111) * 1056

(2x+2111) is odd (has no factor of 2)

2 appears as a factor in 1056 (and therefore the whole total), five times, which is odd, so the 2's can't be separated equally into two equal factors (that being what a square root is).

Posted by Charlie on 2007-04-05 10:25:45