A Constant Puzzle (Posted on 2006-12-02)

	Submitted by K Sengupta
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An infinite number of values of c is possible and the minimum value of c for which this is possible is given by c = 8. This has also been pointed out by Charlie in this area and by Daniel in this location. An analytical proof that c=8 is the minimum x*y^2 - y^2 + x+ y =c Or, P*y^2+ y + 1+ P = c, whenever P = x-1 Now, y=1 gives c = 2P+2; y=2 gives c = 5P+3; y=3 gives c = 10P +4, and so on. Hence, we observe that c is increasing in P for fixed y. Now the minimum value of c such that : 5s+3 = 2t+2 = c. yields a positive integer solution in (s,t) for positive integer c occurs at (s,t,c) = (1,3,8) Hence, (P,y) = (3,1) and (1,2) are two of the non-negative integer solution of P(1+ y^2)+y+1 = 8.....(#) Substituting, P = 0 in (#), we obtain:(P, y) = (0,7) as the remaining non negative integer solution. Recalling that P = x-1, we obtain: (x,y) = (4,1), (2,2) and (1,7) as the only possible positive integer solution to the equation x*y^2 - y^2 + x +y = 8. Consequently, the required minimum value of the constant c is 8. -------------------------------------------------------------------------------- For the cases c =90, 92; refer to the comment posted by Charlie in this location.

	Subject	Author	Date
	re(2): solutions (spoiler)	Daniel	2006-12-02 19:35:42
	re: solutions (spoiler)	Charlie	2006-12-02 14:31:00
	extreme analysis	Charlie	2006-12-02 14:17:02
	solutions (spoiler)	Daniel	2006-12-02 12:38:25