February 17th, 2017, 12:08 PM  #1 
Member Joined: Oct 2012 Posts: 57 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 12:10 PM. 
February 17th, 2017, 12:11 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,250 Thanks: 1439 
Use mathematical induction.

February 17th, 2017, 12:33 PM  #3 
Member Joined: Oct 2012 Posts: 57 Thanks: 0  
February 17th, 2017, 07:48 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 $\displaystyle (\frac{3\sqrt{5}}{2})^n+(\frac{3+\sqrt{5}}{2})^n$ 
February 21st, 2017, 10:46 AM  #5  
Member Joined: Oct 2012 Posts: 57 Thanks: 0  Quote:
 
February 21st, 2017, 12:04 PM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,094 Thanks: 2360 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 $\sqrt5$?

February 21st, 2017, 12:37 PM  #7 
Senior Member Joined: May 2016 From: USA Posts: 862 Thanks: 348 
$\displaystyle \left ( \dfrac{3  \sqrt{5}}{2} \right )^n = \dfrac{1}{2^n} * (3  \sqrt{5})^n = \dfrac{1}{2^n} * \left \{ \sum_{i=0}^n (\ 1)^i * \dbinom{n}{i} * 3^{(n  i)} ( \sqrt{5})^i \right \}.$ $\displaystyle \left ( \dfrac{3 + \sqrt{5}}{2} \right )^n = \dfrac{1}{2^n} * (3 + \sqrt{5})^n = \dfrac{1}{2^n} * \left \{ \sum_{i=0}^n 1^i * \dbinom{n}{i} * 3^{(n  i)} ( \sqrt{5})^i \right \}.$ 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. 
February 21st, 2017, 02:39 PM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,094 Thanks: 2360 Math Focus: Mainly analysis and algebra  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, 06:14 PM  #9  
Senior Member Joined: May 2016 From: USA Posts: 862 Thanks: 348  Quote:
$2 * \dfrac{some\ integer}{2^n} = \dfrac{some\ integer}{2^{(n1)}}.$ 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, 06:37 PM  #10 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,094 Thanks: 2360 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  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Show that  fahad nasir  Trigonometry  2  October 2nd, 2016 09:51 AM 
How to show?  jiasyuen  Calculus  12  January 25th, 2015 02:47 AM 
want to show that show that two infinite summations R equal  notnaeem  Real Analysis  4  August 16th, 2010 01:32 PM 
show that :  naserellid  Algebra  2  August 15th, 2010 03:20 AM 
Show that fog is onetoone  450081592  Calculus  2  January 25th, 2010 10:38 AM 