A multi-solution diophantine problem (Posted on 2006-09-15)

	Submitted by K Sengupta
Rating: 5.0000 (1 votes)
Solution:	(Hide)
Substituting: x = 2a, y = 2b, and z = 2c, in the original equation, one obtains: 2^2a + 2^5b = 2^3c Considering 2^2a + 2^5b = 2^3c in conjunction with the identity 2^(k-1) + 2^(k-1) = 2^k generates the equation : 2a + 1 = 5b + 1 = 3c Since, the minimum possible positive integral solution to (#) is (a,b,c)=(10,4,7), by way of application of Chinese Remainder Theorem to (#), we obtain the following parametric relationship: (a,b,c) = (15*s + 10, 6*s + 4, 10*s + 7)---------(#), so that (#) generates infinite number of triplets (a,b,c) corresponding to integral s (positive 0r negative) Since, by assumption x=2^a; y=2^b and z = 2^c; it immediately follows that there exists an infinite number of solutions to the equation under reference. For example, for s= 1, 2, 3, 4, 5, 6 and 7 we obtain seven triplets: (x,y,z) = (2^10, 2^4, 2^7), ( 2^25, 2^10, 2^17), (2^40, 2^16, 2^27), (2^55, 2^22, 2^37), ( 2^70, 2^28, 2^47), (2^85, 2^34, 2^57), (2^100, 2^40, 2^67) NOTE: The parametric relationship in (#) may correspond to one of the numerous parametric solutions for the problem under reference. However, the relationship (#) alone is sufficient to prove that the total number of solutions to the problem is indeed infinite.

	Subject	Author	Date
	re: No Subject	K Sengupta	2006-12-31 12:43:13
	No Subject	Ferdinand	2006-12-31 07:27:48
	re: re: no subject	Dennis	2006-09-17 17:59:06
	re: No Subject	John Reid	2006-09-15 15:19:21
	No Subject	Dennis	2006-09-15 14:30:07
	Solution for part A	Dej Mar	2006-09-15 13:15:57
	Many solutions (all of them?)	Old Original Oskar!	2006-09-15 11:49:40