Another 2007 Problem (Posted on 2007-02-19)

	Submitted by K Sengupta
Rating: 3.0000 (1 votes)
Solution:	(Hide)
PART (A) We observe that: S_1(x) = sqrt(x); S_2(x) = sqrt(x+sqrt(x)) = sqrt(x+(S_1(x))) ; S_3(x) = sqrt( x+ sqrt(S_2(x)) S_(i+1)(x) = sqrt( x+ sqrt(S_i(x))); for i = 2,3,…, 2006. Clearly, for any given S_(i+1)(x) to correspond to an integer, it is necessary that S_(i)(x) is an integer. So, each of S_(i)(x) for i = 1,2,....,2007 must be integers. But S_2(x) = sqrt(x+(S_1(x))) ; = sqrt(S_1(x) + S_1(x)^2); and so: S_2(x)^2 = S_1(x)(S_1(x) +1)). Since both S_1(x) and S_2(x) must be integers, this is possible only if: x = S_1(x) = S_2(x) = 0. This gives S_(i)(x) = 0 for all i = 1,...,2007 and y=0 Consequently, (x,y) = (0,0) constitutes the only possible solution to Part A. PART (B): We observe that: S_1(x) = sqrt(2x); S_2(x) = sqrt(x+sqrt(2x)) = sqrt(x+(S_1(x))) ; S_3(x) = sqrt( x+ (S_2(x)) S_(i+1)(x) = sqrt( x+ (S_i(x))); for i = 2,3,…, 2006. Clearly, for any given S_(i+1)(x) to correspond to an integer, it is necessary that S_(i)(x) is an integer. Now, S_1(x) is an integer only when x =2* k^2, for some integer k. Accordingly, S_2(x) = sqrt(2k(k+1)). Clearly, for y to correspond to an integer, it follows that S_2(p) must be an integer; so that: 2k(k+1) = (S_2(x))^2 Solving the underlying Pell’s equation for k<300; we obtain: (k,p,S_1(x), S_2(x), S_3(x)) = (1,2,2,2,2); (8,128, 16, 12, sqrt(140)); (49, 4802, 98, 70, sqrt(4872)); (288, 165888, 576, 408, sqrt(166296)) Now, S_1(x) = S_i(x) = 0, whenever p=0; for i=2,3,…., 2007 S_1(x) = S_i(x) = 2, whenever p=2; for i = 2,3,…,2007 Since S_3(x) does not correspond to an integer for x< 2*300^2 = 180,000; we can safely conclude that the equation (#) admits of the solution: (x,y) = (0,0); (2,2) whenever x<180,000. --------------------------------------------------------------- For an alternative methodology, refer to the solution posted by Gamer in the comments section for Part A and Part B.

	Subject	Author	Date
	Solution for B (using same method)	Gamer	2007-02-20 22:58:02
	re(2): a partial solution to B	Daniel	2007-02-20 03:33:11
	Solution	Art M	2007-02-20 02:59:48
	re: a partial solution to B	Joel	2007-02-20 02:38:05
	a partial solution to B	Daniel	2007-02-20 01:00:08
	Solution to A	Gamer	2007-02-19 19:41:33
	Obvious ones	Federico Kereki	2007-02-19 16:22:58