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 February 17th, 2017, 11:08 AM #1 Member   Joined: Oct 2012 Posts: 78 Thanks: 0 Show that Let x be a real number such that : x+1/x=3 Prove that x^n + 1/x^n is a positive integer where n is positive integer (without using mathematical Induction). Last edited by fahad nasir; February 17th, 2017 at 11:10 AM. February 17th, 2017, 11:11 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,035 Thanks: 2272 Use mathematical induction. February 17th, 2017, 11:33 AM   #3
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 Originally Posted by skipjack Use mathematical induction.
without mathematical induction February 17th, 2017, 06:48 PM #4 Newbie   Joined: Aug 2015 From: china Posts: 6 Thanks: 4 By $\displaystyle x+\frac{1}{x}=3$,obtained $\displaystyle x_1=\frac{3+\sqrt{5}}{2},x_2=\frac{3-\sqrt{5}}{2}$ When $\displaystyle x=\frac{3+\sqrt{5}}{2},then \frac{1}{x}=\frac{3-\sqrt{5}}{2}$. When $\displaystyle x=\frac{3-\sqrt{5}}{2},then \frac{1}{x}=\frac{3+\sqrt{5}}{2}$. So that $\displaystyle (\frac{3-\sqrt{5}}{2})^n+(\frac{3+\sqrt{5}}{2})^n$ Thanks from topsquark and JeffM1 February 21st, 2017, 09:46 AM   #5
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 Originally Posted by fengnr By $\displaystyle x+\frac{1}{x}=3$,obtained $\displaystyle x_1=\frac{3+\sqrt{5}}{2},x_2=\frac{3-\sqrt{5}}{2}$ When $\displaystyle x=\frac{3+\sqrt{5}}{2},then \frac{1}{x}=\frac{3-\sqrt{5}}{2}$. When $\displaystyle x=\frac{3-\sqrt{5}}{2},then \frac{1}{x}=\frac{3+\sqrt{5}}{2}$. So that $\displaystyle (\frac{3-\sqrt{5}}{2})^n+(\frac{3+\sqrt{5}}{2})^n$
then what? February 21st, 2017, 11:04 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra Consider the individual terms of $(3-\sqrt5)^n$ and $(3+\sqrt5)^n$. What can you say about the terms containing to an even power of $\sqrt5$? What can you say about the terms containing an odd power of $\sqrt5$? Thanks from fahad nasir February 21st, 2017, 11:37 AM #7 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 552 $\displaystyle \left ( \dfrac{3 - \sqrt{5}}{2} \right )^n = \dfrac{1}{2^n} * (3 - \sqrt{5})^n = \dfrac{1}{2^n} * \left \{ \sum_{i=0}^n (-\ 1)^i * \dbinom{n}{i} * 3^{(n - i)} ( \sqrt{5})^i \right \}.$ $\displaystyle \left ( \dfrac{3 + \sqrt{5}}{2} \right )^n = \dfrac{1}{2^n} * (3 + \sqrt{5})^n = \dfrac{1}{2^n} * \left \{ \sum_{i=0}^n 1^i * \dbinom{n}{i} * 3^{(n - i)} ( \sqrt{5})^i \right \}.$ Now add those sums. Whenever i is odd, the summand will be zero. You only need consider summands where i is even, and in that case the square root of 5 will be replaced by a power of 5. So you now are definitely into the realm of the rationals. That is not the end of the road because a product of powers of 3 and 5 is not divisible by a power of 2. But do not forget the binomial coefficient. I'll let you formalize it. EDIT: I see that Archie was ahead of me. Thanks from fahad nasir February 21st, 2017, 01:39 PM   #8
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 Originally Posted by JeffM1 That is not the end of the road because a product of powers of 3 and 5 is not divisible by a power of 2. But do not forget the binomial coefficient.
I don't think the binomial coefficient is important. We have $$\frac12(\text{integer}) + \frac12(\text{the same integer}) = (\text{the integer})$$ February 21st, 2017, 05:14 PM   #9
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 Originally Posted by v8archie I don't think the binomial coefficient is important. We have $$\frac12(\text{integer}) + \frac12(\text{the same integer}) = (\text{the integer})$$
Perhaps I am missing something, but in the expansion each summand is going to be

$2 * \dfrac{some\ integer}{2^n} = \dfrac{some\ integer}{2^{(n-1)}}.$

If that is correct, the integer is a product of a power of 3, a power of 5, and a binomial coefficient. In that case, you must show that the power of 2 divides into that integer. Because 2, 3, and 5 are all prime, the binomial coefficient must be heavily involved.

I admit that I did not work the proof out to the end and merely did what you had already done about showing that the odd powers of $\sqrt{5}$ cancel out so this is a hand waving argument. February 21st, 2017, 05:37 PM #10 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra I think you are right. I forgot the exponent on the 2. A result that may apply to some extent is that $$\sum_{k=0}^{n} {n \choose k} = 2^n$$ Tags show Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post fahad nasir Trigonometry 2 October 2nd, 2016 08:51 AM jiasyuen Calculus 12 January 25th, 2015 01:47 AM notnaeem Real Analysis 4 August 16th, 2010 12:32 PM naserellid Algebra 2 August 15th, 2010 02:20 AM 450081592 Calculus 2 January 25th, 2010 09:38 AM

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