Find a Polynomial in $x-\frac1x$ Given that $x^n - (1/x^n)$ is expressible as a polynomial in $x - (1/x)$ with real coefficients only if $n$ is an odd positive integer, find $P(z)$ so that $P(x-(1/x)) = x^5 - (1/x)^5.$
To start, I factored, giving:
$P(x-\frac{1}{x}) = (x-\frac{1}{x})(x^4+\frac{1}{x^4}+x^2+\frac{1}{x^2}+1).$
However, I cannot find a way to connect this to $x-\frac{1}{x}$. What should I do?

$$\left(x^m-\dfrac1{x^m}\right)\left(x^2+\dfrac1{x^2}\right)=x^{m+2}-\dfrac1{x^{m+2}}+x^{m-2}-\dfrac1{x^{m-2}}$$
$$\implies F_{m+2}=F_m\left(x^2+\dfrac1{x^2}\right)+F_{m-2}$$  where $F_r=x^r-\dfrac1{x^r}$
Now  $x^2+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2=F_1^2+2$
$$F_3=\left(x-\dfrac1x\right)^3+3\left(x-\dfrac1x\right)$$
$$F_5=\left(\left(x-\dfrac1x\right)^2+2\right)F_3+F_1$$
