Sophie Squares (Posted on 2005-07-13)

	Submitted by Gamer
Rating: 3.2500 (4 votes)
Solution:	(Hide)
If two numbers a and b work, then a+b and 2a+b also work, except in this case 2(a+b)²-1=(2a+b)², which is the wrong sign for what we want. So apply this recursively again to switch the sign back. We get ((a+b)+(2a+b)) and (2(a+b)+(2a+b)) which equals (3a+2b) and (4a+3b) as terms. So, the formula (a,b)->((3a+2b),(4a+3b)) works going forward, but we want one that goes backwards, so we need to come up with c and d less than a and b. If (c,d)->(a,b), then a = 3c+2d and b = 4c+3d. Solving for c and d gives c = 3a-2b and d = 3b-4a. If c = 3a-2b and d = 3b-4a, then a+(3a-2b)+(3b-4a)=b which satisfies a+c+d=b. The proof that it satisfies 2c²+1 = d² is as follows: 2a²+1=b² (16a²-24ab+8b²)+2a²+1=(16a²-24ab+8b²)+b² 18a²-24ab+8b²+1=9b²-24ab+16a² 2(3a-2b)²+1=(3b-4a)² 2c²+1=d²

	Subject	Author	Date
	re: True?	gail	2005-07-19 06:08:17
	re(2): Solution	Bractals	2005-07-14 03:24:53
	re: Solution	Bob Smith	2005-07-13 20:05:16
	No Subject	pcbouhid	2005-07-13 18:23:54
	Solution	Bractals	2005-07-13 18:09:21
	re(2): True?	Bob Smith	2005-07-13 17:43:19
	re: True?	Steve Herman	2005-07-13 17:29:07
	True?	e.g.	2005-07-13 17:18:01
	Counter-example	Bob Smith	2005-07-13 16:17:12