Subtract The Cubics, Get Product (Posted on 2007-07-22)

	Submitted by K Sengupta
Rating: 3.3333 (3 votes)
Solution:	(Hide)
v^3 – w^3 = vw + 61 > 0, giving: v> w> = 1 Now, v^3 – w^3 = vw + 61 or, (v-w)(v^2 + vw + w^2) = vw + 61 or, (v^2 + w^2)*P + (P-1)vw = 61, where P = v-w Now for P> = 3, we observe that: v> = 3+w, so that: 3(v^2 + w^2) < = P(v^2+w^2)< 61 Or, v^2 + w^2) < = 20. Since v> = w+3, this is only possible iff (v, w) = (4,1), but 4^3 – 1^3 = 63, while 4*3 + 61 = 73. This is a contradiction. Hence P =1, 2 If P = 2, then 5*w^2 + 10*w – 53 = 0, which does not yield any integer root for w. If P =1, then w^2+w – 30 = 0, so that; w=5, ignoring the negative root which is inadmissible. This yields v=6, and consequently (v, w) = (6,5) is the only possible solution to the given problem. .................Q E D ..................... *** Also refer to an alternative methodology given by Brian Smith in this location .

	Subject	Author	Date
	Another solution	Brian Smith	2023-04-15 17:19:31
	Solution	Brian Smith	2007-07-23 10:35:35
	No Subject	xdog	2007-07-23 08:18:04
	re: solution	Federico Kereki	2007-07-23 00:19:46
	solution	xdog	2007-07-22 20:36:32
	(Very) partial spoiler!	Josie Faulkner	2007-07-22 18:56:57