My Math Forum Problem solving an equation

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 June 30th, 2008, 11:39 AM #1 Newbie   Joined: Jun 2008 Posts: 4 Thanks: 0 Problem solving an equation How do I show that the equation y^9 + y^8 - 7y^7 - 5y^6 + 21y^5 + 11y^4 - 27y^3 - 5y^2 + 20y + 4 = 0, where y > 0 has no integers solutions? I think estimating the left term may help to show that the left term is larger than zero for each y > 0. Thanks in advance, frank
 June 30th, 2008, 11:55 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Problem solving an equation You can check that it fails for y = 1 or 2. Otherwise y >= 3. The expression on the left can be rewritten as y^6(y^3 + y^2 - 7y - 5) + y^2(21y^3 + 11y^2 - 27y - 5) + 20y + 4 From here it suffices to notice that each term is positive: y^3 + y^2 >= 12y, so y^3 + y^2 - 7y - 5 >= 5y - 5 >= 10 21y^3 + 11y^2 >= 222y, so 21y^3 + 11y^2 - 27y - 5 >= 195y - 5 >= 580 20y >= 60 4 = 4
 June 30th, 2008, 09:59 PM #3 Newbie   Joined: Jun 2008 Posts: 4 Thanks: 0 Re: Problem solving an equation Thank you for the fast reply! There is another problem: How can I show that this equation has no solutions which are larger than zero?
 July 1st, 2008, 04:05 AM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Problem solving an equation What do you mean by that?
 July 1st, 2008, 04:13 AM #5 Senior Member   Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 Re: Problem solving an equation I think he means $\text{for all}\; x\in \mathbb{Z}^+,\; \text{show that}\; x\;\text{is not a solution to the above equation}$
 July 1st, 2008, 05:55 AM #6 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Problem solving an equation If you mean y, I already showed that. If you mean x, well... I suppose I also showed that.
 July 24th, 2008, 12:22 PM #7 Newbie   Joined: Jun 2008 Posts: 4 Thanks: 0 Re: Problem solving an equation Thank you! I got the solution of the equation.

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