Another Remainder Puzzle (Posted on 2007-01-25)

	Submitted by K Sengupta
Rating: 4.0000 (1 votes)
Solution:	(Hide)
Since, each of a, b and c are positive, reducing both sides of the given equation 3^a + 5^b = 7^c + 1 to Mod 3; we obtain: 2 ^b = 2 ( Mod 3 ); so that b must be odd. Hence, 3^a + 5 = 7^c + 1 ( Mod 8). Since, 3^a = 1, 3 (Mod 8) and 7^c = 1,7 (Mod 8); it follows that : (3^a Mod 8, 7^c Mod 8) = (3,7), so that both a and c must be odd. Otherwise, if at least one of a and c is even, if follows that: (3^a Mod 8, 7^c Mod 8) = (1,7); (1,1), (3,1). None of these three cases satisfy the relationship: 3^a +5 = 7^c + 1 (Mod 8), and accordingly, this is a contradiction. Now, each of a, b and c are positive, and so, the equation 3^a + 5^b = 7^c + 1 gives: 3^a + 5 = 7^c + 1 (Mod 10). Since, both a and c are odd; this is possible if and only if a = 4p-3 and c = 4q -3 for positive integers p and q. Consequently, dividing a and c separately by 4 we obtain the respective remainders as 1 and 1. ------------------------------------------------------------------ NOTE: When each of a, b and c are positive integers the minimum possible solution to the given equation is (a,b,c) = (1,1,1). This is also pointed out by Ady TZIDON in the Comments section.

	Subject	Author	Date
	re: A partial solution? (spoiler) ....what if...	Ady TZIDON	2007-01-30 09:55:05
	A partial solution? (spoiler)	Steve Herman	2007-01-26 09:22:44