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I was trying to solve this problem on my textboox

Given $x + \frac{1}{x}$ = $\frac{1+\sqrt{5}}{2}$ find the value of $x^{2000} + \frac{1}{x^{2000}}$

After doing a bit of exploration i have noticed that the value of $x^n + \frac{1}{x^n}$ doesn't diverges, however i am unable to find a clear pattern for it and i did some long algebra process to get the answer.

Although i got the value for $x^{2000} + \frac{1}{x^{2000}}$ i still am not very satisfied with my method of solving it, i've done some more exploration but i'm still unable to notice a pattern within the first few terms.

Any help or hints would be appreciated.

Gary
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3 Answers3

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For $x+\dfrac1x=2\cos A$

$x=\cos A\pm i\sin A$

For integer $n,$ using Proof for de Moivre's Formula, $$x^n=\cos nA\pm i\sin nA$$

$\implies x^n+\dfrac1{x^n}=2\cos nA$

Now we can show, $\dfrac{\sqrt5+1}4=\cos36^\circ$

See also: Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer.

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Hint:

$$x^2+\dfrac1{x^2}=\cdots=\dfrac{\sqrt5-1}2$$

$$\implies x^2+\dfrac1{x^2}-\left(x+\dfrac1x\right)=-1$$

As $x\ne0,$ multiply out $x^2$ to find $$x^4-x^3+x^2-x+1=0$$

$$\implies x^5+1=(x+1)(x^4-x^3+x^2-x+1)=0\implies x^5=?$$

Can you take it from here?

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Letting $a_n = x^n + x^{-n},$ we can easily express $a_{2n}, a_{5n}$ in terms of $a_n.$ Repeatedly doing this while using the factorization of $2000$ will give you the answer.

The algebra is still heavy, so you might prefer the alternative of using a recursion for $a_{n+1}$ in terms of $a_n, a_{n-1}.$ Hint: consider $a_1 a_n.$

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