A 2009 problem (Posted on 2006-08-27)

	Submitted by K Sengupta
Rating: 2.0000 (1 votes)
Solution:	(Hide)
Let q = p + r. Then the equation becomes (27 – 3r)(p^2) + (27r – 3*r^2)p - r^3 + 2009 = 0 : As a quadratic equation in p, its discriminant is (27r – (3*(r^2))^2 - 4(27 – 3r)(-r^3 + 2009) = -3(r - 14)(r - 9)(r^2 + 41*r + 574) Now,r^2 + 41r + 574 = ((2r+41)^2)/4 + 615/4; and so, The factor r^2 + 41r + 574 is always positive for any integer value of r. Therefore the equation has integer solution in p only when: r = 9, 10, 11, 12, 13, 14. When r = 9, the equation becomes -3(p^2) – 30p + 1009 = 0 which has no integer solution in p. Similarly for r = 10,11,12,13, the resulting quadratic equations do not give integer solution in p. When r = 14, the equation becomes -15*(p+7)^2 = 0. Thus p = -7; q = r+p = 14-7 = 7 is a solution to the given equation.

	Subject	Author	Date
	a solution	Dennis	2006-09-13 16:24:09
	Factors and restrictions	Gamer	2006-08-28 00:00:41
	further restrictions	Daniel	2006-08-27 22:14:38
	mathematica code	Daniel	2006-08-27 11:32:59
	Found 1.	Larry	2006-08-27 09:59:23