Sum the integers, get squares? (Posted on 2007-04-05)

	Submitted by K Sengupta
Rating: 2.7500 (4 votes)
Solution:	(Hide)
(p+1) + (p+2) + ... + (p+2112) = (p+2112)(p+2113)/2 - p(p+1)/2 = 2112p + 2112.2113/2 = 1056(2p + 2113). But 1056 = 32.33 and (2p+2113) is odd, so the highest power of 2 dividing 1056(2p+2113) is 32 = 2^5. Hence the sum of 2112 consecutive positive integers cannot be a perfect square.

	Subject	Author	Date
	Solution	Praneeth	2007-09-12 08:19:53
	i like it	Ady TZIDON	2007-04-05 14:27:59
	solution	Charlie	2007-04-05 10:25:45
	Different method	Marc	2007-04-05 10:09:15
	No Subject	Ady TZIDON	2007-04-05 09:24:16