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I played with the solution for the problem $$\text{if }\; x + \frac{1}{x} = a$$ what is $$x^5 + \frac{1}{x^5}$$ I tried different exponents other than 5 and tried finding the solution to it. I defined $f(x) = a^x + \frac{1}{a^x}$. I got $f(x + y)=f(x)f(y) - f(x-y).$ I tried reversing the equation I got to get $f(x)$ but I only got these: $$f(0) = 2$$ by substituting $b=0$ $$f(x)=f(-x)$$ $$(f(a)^2 - 4)(f(b)^2 - 4) \geq 0$$ $$f'(0)=0$$ Can this be solved using the given information? Is $f(x) = a^x + \frac{1}{a^x}$ the only solution? Thanks in advance!

Edit: I already got the solution for $x^5 + \frac{1}{x^5}$, I'm asking if how can I get the family of functions $f(x)$ from $f(x + y)=f(x)f(y) - f(x-y)$, sorry for the unclear question

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Hint:

$$\left(x^3+\dfrac1{x^3}\right)\left(x^2+\dfrac1{x^2}\right)=x^5+\dfrac1{x^5}+x+\dfrac1x$$

Now $x^3+\dfrac1{x^3}=\left(x+\dfrac1{x}\right)^3-3\left(x+\dfrac1{x}\right)$

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Hint: Expand $$ \left(x+\frac{1}{x} \right)^5 $$ and the problem will suddenly become super easy.

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    $\begingroup$ I wouldn't call it super easy. You still have at least one more step of actual thinking to do. $\endgroup$ – Arthur Nov 28 '19 at 10:09
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    $\begingroup$ Well, you're right. But when you have the expanded form, you basically immediately see what you have to do. $\endgroup$ – Matti P. Nov 28 '19 at 10:10

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