A Powers and Square Puzzle (Posted on 2006-12-18)

	Submitted by K Sengupta
Rating: 4.0000 (1 votes)
Solution:	(Hide)
Let 2^m+3^n = k^2, where k is a positive integer. Hence, -1 ^m = k^2 (Mod 3) and since k^2 is not a multiple of 3, it follows that k^2 = 1( Mod 3). Hence, we observe that m is even and accordingly, m=2p, for some positive integer p. Therefore, 4^p+3^n = k^2; so that, -1^n = k^2 (Mod 4). This implies that k is odd and, accordingly, k^2 = 1(Mod 4), so that n is even. This implies that, n = 2q, for some positive integer q. Hence, (2^p)^2 + (3^q)^2 = k^2, so that: (2^p, 3^q, k) is a Pythagorean triplet. Hence, there exists a, b with g.c.d(a,b) = 1 and a greater than b such that: 2^p = 2ab; 3^q = a^2-b^2; k = a^2+b^2. Accordingly, 2^(p-1) = ab and since a is greater than b with (a,b) = 1, it follows that: A = 2^(p-1) and b=1. Hence, 3^q = 2^(2p-2) -1 = (2^(p-1) -1)(2^(p-1) +1), so that: Exactly one of 2^(p-1) -1) and 2^(p-1) +1 is divisible by 3 while the other must be equal to 1. Now, 2^(p-1) +1 = 1, gives: 2^(p-1) = 0 which is a contradiction. Thus,(2^(p-1) -1 = 1 , giving p=2, so that: 2^(p-1) +1 =3. Consequently, 3^q = 3^1, so that, q=1 But, (p.q) = (2,1) gives (x,y)= (4,2) Thus, (x,y) = (4,2) is the only possible pair satisfying all the conditions of the problem.

	Subject	Author	Date
	re(2): unique solution?	Joel	2006-12-18 23:15:23
	re: unique solution?	Dennis	2006-12-18 22:14:30
	unique solution?	Dennis	2006-12-18 21:37:51