Curious Consecutive Conundrum (Posted on 2008-04-08)

	Submitted by K Sengupta
Rating: 5.0000 (1 votes)
Solution:	(Hide)
By the given conditions, we have: (L+1)3 - L3 = 3L2 + 3L +1 = P2 or, 12L2 + 12L + 4 = 4P2 or, 3(2L+1)2 = (2P+1)(2P-1) Since each of 2P+1 and 2P-1 are odd, while (2P+1)-(2P-1) = 2, it follows that gcd(2P+1, 2P-1) = 1 Thus, for some integers x and y, it follows that: (2P-1, 2P+1) = (3x2, y2), or: (x2, 3y2) If the former case is true, then y2 = 3x2 + 2, so that: y2(mod 3) = 2, which is a contradiction, since y2 (mod 3) must be congruent to either 0 or 1. If the latter case is true, then: x2 = 3y2 - 2 or, x2 (mod 3) = 1, which is indeed valid. Again, x2 = 2P-1, so that x is odd. Thus, x = 2z+1, for some integer z Accordingly, (2z+1)2 = 2P - 1 Hence: 4z2 + 4z + 1 = 2P-1 or, P = 2z2 + 2z + 1 = z2 + (z+1)2 Consequently, P is always expressible as the sum of squares of two consecutive positive integers. ------------------------------------------------- Alternate Method: (L+1)3 - L3 = P2......(i) or, 3(2L+1)2 = (2P+1)(2P-1)......(ii) From (i), it follows that P is odd. Now, gcd(2P-1, 2P+1) = 1, as proved earlier, so in terms of (ii), it follows that at least one of 2P+1 and 2P-1 is a perfect square. Recalling that P is odd, it follows that: (2P+1)(Mod 4) = 3, so that 2P+1 cannot be a perfect square. Thus, (2P-1) is a perfect square. Since 2P-1 is odd, it follows that: 2P - 1 = (2z + 1)2, whenever z is an integer. or, P = z2 + (z+1)2 Consequently, P is always expressible as the sum of squares of two consecutive positive integers.

	Subject	Author	Date
	re: need a hint	K Sengupta	2008-07-23 16:11:47
	need a hint	Praneeth	2008-07-23 03:40:32
	AND the n-th member is:	Ady TZIDON	2008-04-10 17:01:24
	re: What value for L?	Brian Smith	2008-04-10 00:15:01
	re: What value for L?	brianjn	2008-04-09 03:44:49
	What value for L?	brianjn	2008-04-09 03:36:06
	first thoughts	FrankM	2008-04-08 22:58:08