# Finding the family of functions given an equation

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

$$\left(x^3+\dfrac1{x^3}\right)\left(x^2+\dfrac1{x^2}\right)=x^5+\dfrac1{x^5}+x+\dfrac1x$$
Now $$x^3+\dfrac1{x^3}=\left(x+\dfrac1{x}\right)^3-3\left(x+\dfrac1{x}\right)$$
Hint: Expand $$\left(x+\frac{1}{x} \right)^5$$ and the problem will suddenly become super easy.