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June 30th, 2008, 11:39 AM   #1
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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
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June 30th, 2008, 11:55 AM   #2
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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
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June 30th, 2008, 09:59 PM   #3
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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?
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July 1st, 2008, 04:05 AM   #4
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Re: Problem solving an equation

What do you mean by that?
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July 1st, 2008, 04:13 AM   #5
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Re: Problem solving an equation

I think he means
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July 1st, 2008, 05:55 AM   #6
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Re: Problem solving an equation

If you mean y, I already showed that. If you mean x, well... I suppose I also showed that.
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July 24th, 2008, 12:22 PM   #7
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Re: Problem solving an equation

Thank you!
I got the solution of the equation.
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