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

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