A Real Solution Puzzle (Posted on 2006-07-30)

	Submitted by K Sengupta
Rating: 3.0000 (2 votes)
Solution:	(Hide)
Daniel's METHODOLOGY CULMINATING IN THE SOLUTION IS GIVEN HERE. ALSO REFER TO JLo's SOLUTION. (A Particular Case: If it was required to determine all possible positive real solutions instead of "real solutions" in the first line of the problem, the solution would have been as follows: x^3 + y = 12x + 20 Or, x^3 – 12x – 20 = -y Or, (x-4)*(x+2)^2 = 4-y -----------(1) 2*y^3 +z = 24y +36 or, 2*y^3 – 24y – 36 = -z or, 2(y-4)*(y+2)^2 = 4-z ----------(2) 3*z^3 + x = 36z +52 or, 3*z^3 – 36z -52 = -x or, 3(z-4)(z+2)^2 = 4-x ---------------(3) From (1),(2) and (3) we obtain: 6*(x-4)(y-4)(z-4)[(x+2)^2 + (y+2)^2 + (z+2)^2 + 1/6] = 0 As the last factor is always positive for all real numbers x, y and z; we have (x-4)(y-4)(z-4) = 0 This gives us at least one of x = 4,y = 4,z = 4. In conjunction with (1), (2) and (3); this gives all possible positive real solutions as x = y = z = 4.)

	Subject	Author	Date
	An easy one	Federico Kereki	2006-07-30 16:27:10
	Maple "proof"	JLo	2006-07-30 13:29:10
	Errata	K Sengupta	2006-07-30 11:36:41
	solution with proof :-)	Daniel	2006-07-30 11:36:17
	At least partial solution	atheron	2006-07-30 10:35:20