A Multi-solution Puzzle (Posted on 2006-08-03)

	Submitted by K Sengupta
Rating: 5.0000 (1 votes)
Solution:	(Hide)
Let U = 3P+2Q and V = 3P – 2Q-------(#) We have, V^2 = (U^2 – V^2)/(U-1) Or, -U(V^2 - U) = 0 So, either, U=0 or, U = V^2 Therefore, solving (#) and substituting U=V^2, we obtain: P = V(V+1)/6 ; Q = V(V -1)/4, where V is an integer Case 1: V=0(Mod 6) and V= 0 (Mod 4) V=0(Mod 12), so that, V = 12S for an integer S, giving: P = 2S(12S + 1); Q = 3S(12S -1) Case 2: V = 5(Mod 6); V = 1 (Mod 4) Here, V = 5 (Mod 12); or, V = 12S+5, for an integer S, giving: P = (2S+1)(12S+5); Q = (3S +1)(12S+5) Case 3: V = 0 (Mod 6) ; V = 3 (Mod 4) This is not possible since, in former instance V is even, while in the latter instance V is odd. Case 4: V = 5(Mod 6); V = 0 (Mod 4) This is not possible since, in latter instance V is even, while in the former instance V is odd. Hence; P = 2S(12S + 1); Q = 3S(12S -1) ; where S is a positive integer......(a) Or, P = (2S+1)(12S+5); Q = (3S +1)(12S+5)....(b) ; where S is a non- negative integer, constitutes all possible solutions to the equation under reference. Substituting S = 1,2,3,.....in (a), we obtain: (P,Q) = (26,33); (100, 138); (222, 315); ....... Substituting S = 0,1,2,3,..... in (b), we obtain: (P,Q) = (5,5);(51,68);(145,203);(287,410);...... Proceeding in this manner, we observe that (a) would generate a positive integer solution for each positive integer value of S, while (b) would generate a positive integer solution for each non-negative integer S. Consequently, the given equation admits of an infinite number of solutions. NOTE: It has been pointed out by Daniel (in this location ) that other parametric solutions exist for this problem. However, relationships (a) and (b) are sufficient to prove that the total number of solutions to the problem is indeed infinite.

	Subject	Author	Date
	re: what mathematica has to say	Richard	2006-08-03 19:14:50
	what mathematica has to say	Daniel	2006-08-03 18:03:12
	Some answers w/o derivations	Richard	2006-08-03 16:41:51