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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: 20,804 Thanks: 2150 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: 7,671 Thanks: 2651 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: 551 $\displaystyle \left ( \dfrac{3 - \sqrt{5}}{2} \right )^n = \dfrac{1}{2^n} * (3 - \sqrt{5})^n = \dfrac{1}{2^n} * \left \{ \sum_{i=0}^n (-\ 1)^i * \dbinom{n}{i} * 3^{(n - i)} ( \sqrt{5})^i \right \}.$ $\displaystyle \left ( \dfrac{3 + \sqrt{5}}{2} \right )^n = \dfrac{1}{2^n} * (3 + \sqrt{5})^n = \dfrac{1}{2^n} * \left \{ \sum_{i=0}^n 1^i * \dbinom{n}{i} * 3^{(n - i)} ( \sqrt{5})^i \right \}.$ 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: 7,671 Thanks: 2651 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 $$\sum_{k=0}^{n} {n \choose k} = 2^n$$

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