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If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then

$$x^{2000}+\frac{1}{x^{2000}}=?$$

My try:

$$\left(x^{1000}\right)^2+\left(\frac{1}{x^{1000}}\right)^2=\left(x^{1000}+\frac{1}{x^{1000}}\right)^2-2$$

Continuation ?

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  • $\begingroup$ You have marked this algebra-precalculus. Your original equation $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ has no solutions in real numbers. So is there a tpyo or are you allowed complex numbers. $\endgroup$
    – Anurag A
    Dec 1, 2016 at 18:41
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    $\begingroup$ $\frac{1+\sqrt{5}}{2} = 2\cos\frac{\pi}{5}$. $\endgroup$ Dec 1, 2016 at 22:11
  • $\begingroup$ Why wouldnt we be allowed complex numbers? And the fact that x is complex doesn't effect solving the post. $\endgroup$
    – fleablood
    Dec 1, 2016 at 22:48

7 Answers 7

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Note that $a = \frac{1+\sqrt{5}}{2}$ satisfies the equation $a^2 - a - 1 = 0$

Substituting $a=x+\frac{1}{x}$ gives:

$$0 = \left(x+\frac{1}{x}\right)^2 - \left(x+\frac{1}{x}\right) - 1 = x^2 - x + 1 - \frac{1}{x} + \frac{1}{x^2}$$

$$\iff \quad x^4 - x^3 + x^2 - x + 1 = 0$$

Multiplying by $x+1$ results in $x^5+1=0$, so $x$ is a complex $5^{th}$ root of $-1$ therefore $x^5 = -1$.

Then $x^{2000}+\frac{1}{x^{2000}} = \big(x^{5}\big)^{400} + \cfrac{1}{\big(x^{5}\big)^{400}} = (-1)^{400} + \cfrac{1}{(-1)^{400}} = 1 + 1 = 2$.


P.S. For a heavy-handed "solution" to the tune of "how to crack a nut with a sledgehammer", let Wolfram Alpha do all the work:

resultant[resultant[x^2 - a x + 1, a^2 - a - 1, a ], x^4000 - b x^2000 + 1, x] = (b-2)^4

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  • $\begingroup$ Niice. But, should be $x^5$ and not $x^{-5}$. $\endgroup$ Dec 2, 2016 at 17:29
  • $\begingroup$ @Physicist137 Thanks for catching that. Edited and fixed. $\endgroup$
    – dxiv
    Dec 2, 2016 at 17:41
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If $a_n=x^n+\frac{1}{x^n}$ then $a_n=a_1\cdot a_{n-1}+a_{n-2}=\phi\cdot a_{n-1}-a_{n-2}$ which is a second-order linear recurrence, where $\phi=\frac{1+\sqrt 5}{2}$. The initial conditions are $a_1=\phi$ and $a_2=a_1^2-2=\phi^2-2=\phi-1$, since $\phi^2=\phi+1$

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    $\begingroup$ Helps to note that the $a_n$ are periodic ($a_6=-a_1$ and $a_{11}=a_1$). $\endgroup$
    – lulu
    Dec 1, 2016 at 18:51
  • $\begingroup$ Yes, depending on how much an elementary solution OP wants, it would work just to use the recurrence and calculate $a_1, a_2, ...$ while using $\phi^2=\phi+1$ $\endgroup$
    – Momo
    Dec 1, 2016 at 18:57
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Edit: Found another solution, removed old answer (it was incorrect anyway)

You have $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$. By simple algebraic manipulation you can get the $$x^4-x^3+x^2-x+1 = 0$$ Now notice that $x^4 = x^3-x^2+x-1$ and multiplying both sides by $x$ you get $x^5 = x^4-x^3+x^2-x=-1$.

Therefore $$x^{2000}+\frac{1}{x^{2000}} = ({x^{5}})^{400}+\frac{1}{(x^{5})^{400}} = (-1)^{400}++\frac{1}{(-1)^{400}} = 1+1 = 2$$

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Here's another approach for the record.

  1. The equation $x+ \frac{1}{x}=\alpha$ where $\alpha \in [-2,2]$ can be solved as follows. Identify $\alpha$ is $2\cos (\theta)$, and observe that letting $z=e^{i\theta}$,from the definition of $\cos (\theta)$, we have $$z+ \frac{1}{z}=2\cos(\theta)=\alpha.$$

  2. Also, from the definition of $\cos$,

$$z^k + \frac{1}{z^k} = 2\cos (k \theta).$$

  1. Now back to our question. Let $\alpha = \frac{1+\sqrt{5}}{2}$. Then $\cos(\theta) = \frac{1+\sqrt{5}}{4}$. This gives us $\theta = \pm \frac{\pi}{5} \mod 2\pi$. Therefore the answer to the problem is $2\cos (400\pi)=2$.
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  • $\begingroup$ That was a wonderful argument. $\endgroup$ Dec 2, 2016 at 18:01
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Let $\psi = \frac{1+\sqrt{5}}{2} $ $$x + \frac{1}{x} = \psi$$ $$x^2 + 1 = \psi x $$ $$x^2 - \psi x + 1 = 0$$ Use quadratic formula here to solve for $x$. Then plug that into the given expression.

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  • $\begingroup$ Superior answer $\endgroup$
    – Decaf-Math
    Dec 1, 2016 at 18:34
  • $\begingroup$ $x + \frac{1}{x} = \psi$ has no real solutions. $\endgroup$
    – Anurag A
    Dec 1, 2016 at 18:41
  • $\begingroup$ @AnuragA Go for complex..? $\endgroup$
    – Frank
    Dec 1, 2016 at 18:53
  • $\begingroup$ That might be very dificult to simplify. $(a+b\sqrt{c})^{2000} + \frac 1{(a+b\sqrt{c})^{2000}}$ might not be an acceptable answer. (Then again maybe it will work out nice.) $\endgroup$
    – fleablood
    Dec 1, 2016 at 23:00
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Nice try of yours. You can further continue...: $$ x^{2000} + \frac{1}{x^{2000}} = \left(x^{1000} + \frac{1}{x^{1000}}\right)^2 - 2 \\ x^{1000} + \frac{1}{x^{1000}} = \left(x^{500} + \frac{1}{x^{500}}\right)^2 - 2 \\ x^{500} + \frac{1}{x^{500}} = \left(x^{250} + \frac{1}{x^{250}}\right)^2 - 2 \\ \vdots \\ x^{n} + \frac{1}{x^n} = \left(x^{n/2} + \frac{1}{x^{n/2}}\right)^2 - 2 \\ $$

In the case of $2000 = 2^4 \cdot 5^3$. That is, when you are done with all the $2$s, you will be left with $5$: $$ x^{125} + \frac{1}{x^{125}} = \left[\left(x^{25}\right)^5 + \left(\frac{1}{x^{25}}\right)^5\right] = $$

We do basically the same you did: $$ (a+b)^5 = {{5}\choose{0}}a^5 + {{5}\choose{1}}a^4 b + \cdots \\ (a+a^{-1})^5 = {{5}\choose{0}}a^5 + {{5}\choose{5}}b^{-5} + \cdots \\ $$

Notice the intermediate terms: $$ {{5}\choose{1}}a^{4} a^{-1} + {{5}\choose{2}}a^{3} a^{-2} + {{5}\choose{3}}a^{2} a^{-3} + {{5}\choose{4}}a^{1} a^{-4} = {{5}\choose{1}}a^{3} + {{5}\choose{2}}a + {{5}\choose{3}}a^{-1} + {{5}\choose{4}}a^{-3} $$

That is: $$ \left[x^{5} + \frac{1}{x^{5}}\right]^5 = \left[x^{5} + \frac{1}{x^{5}}\right] + 10\left[\left(x^{5}\right)^3 + \left(\frac{1}{x^{5}}\right)^3\right] + 5\left[\left(x^{5}\right)^5 + \left(\frac{1}{x^{5}}\right)^5\right] $$


You can keep doing so and simplifying to smaller and smaller terms. You can do this because: $$ {{n}\choose{k}} = {{n}\choose{n-k}} = \frac{n!}{k!(n-k)!} $$

That is, $(a + a^{-1})^n$ will have the terms $a^{-k} a^{n-k}$ and $a^k a^{-(n-k)}$ with the same coeficients. And this term forms: $$ a^k a^{-n+k} = a^{k-n+k} = a^{2k-n} \\ a^{-k} a^{n-k} = a^{-k+n-k} = a^{-2k+n} = a^{-(2k-n)} $$

So, the coeficients opposite terms will be equal, then you will always be able to build terms like $x^k + 1/x^k$. Thus you can keep doing it until you are left only with the $x + 1/x$. It might take a while..... =D.

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  • $\begingroup$ @egreg Oh. Thanks. I'll fix. $\endgroup$ Dec 2, 2016 at 0:54
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Just giving another way. $$x+\frac 1x=\frac{1+\sqrt2}{2}\\ x^2+\frac{1}{x^2}=\frac{-1+\sqrt5}{2}\\ x^4+\frac{1}{x^4}=-\frac{1+\sqrt5}{2}\\ x^8+\frac{1}{x^8}=\frac{-1+\sqrt5}{2}$$ Hence $$\left(x^8+\frac{1}{x^8}\right)\left(x^2+\frac{1}{x^2}\right)=\left(\frac{-1+\sqrt5}{2}\right)^2\\ x^{10}+\frac{1}{x^{10}}=\left(\frac{-1+\sqrt5}{2}\right)^2-\left(-\frac{1+\sqrt5}{2}\right)=2\iff(x^{10}-1)^2=0$$ Thus $x^{10}=1$ from which the result $2$.

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    $\begingroup$ Top equation: Replace 2 with 5 $\endgroup$
    – Improve
    Dec 2, 2016 at 1:58

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