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February 17th, 2017, 12:08 PM   #1
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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).

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February 17th, 2017, 12:11 PM   #2
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Use mathematical induction.
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February 17th, 2017, 12:33 PM   #3
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Use mathematical induction.
without mathematical induction
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February 17th, 2017, 07:48 PM   #4
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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
$\displaystyle (\frac{3-\sqrt{5}}{2})^n+(\frac{3+\sqrt{5}}{2})^n$
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February 21st, 2017, 10:46 AM   #5
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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?
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February 21st, 2017, 12:04 PM   #6
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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 $\sqrt5$?
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February 21st, 2017, 12:37 PM   #7
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$\displaystyle \left ( \dfrac{3 - \sqrt{5}}{2} \right )^n = \dfrac{1}{2^n} * (3 - \sqrt{5})^n = \dfrac{1}{2^n} * \left \{ \sum_{i=0}^n (-\ 1)^i * \dbinom{n}{i} * 3^{(n - i)} ( \sqrt{5})^i \right \}.$

$\displaystyle \left ( \dfrac{3 + \sqrt{5}}{2} \right )^n = \dfrac{1}{2^n} * (3 + \sqrt{5})^n = \dfrac{1}{2^n} * \left \{ \sum_{i=0}^n 1^i * \dbinom{n}{i} * 3^{(n - i)} ( \sqrt{5})^i \right \}.$

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.
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February 21st, 2017, 02:39 PM   #8
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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 $$\frac12(\text{integer}) + \frac12(\text{the same integer}) = (\text{the integer})$$
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February 21st, 2017, 06:14 PM   #9
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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.
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February 21st, 2017, 06:37 PM   #10
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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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