Given that $x^n - \frac1{x^n}$ is expressible as a polynomial in $x - \frac1x$ with real coefficients only if $n$ is an odd positive integer, find $P(z)$ so that $P\left(x-\frac1x\right) = x^5 - \frac1{x^5}.$
Let's substitute $z$ for $x - \frac{1}{x}$. Then the equation becomes $P(z) = x^5 - \frac{1}{x^5}$.
Notice that $x^5 - \frac{1}{x^5} = (x - \frac{1}{x})(x^4 + 1 + \frac{1}{x^4})$. Since we have $z = x - \frac{1}{x}$, we can express $x^4 + 1 + \frac{1}{x^4}$ as a polynomial in $z$.
To do this, we can calculate $(x^4 + 1 + \frac{1}{x^4})(x-\frac{1}{x})$ and observe that it is equal to $x^5 - \frac{1}{x^5}$. Let's perform this multiplication:
$(x^4 + 1 + \frac{1}{x^4})(x-\frac{1}{x}) = x^5 - \frac{1}{x^5} + x - \frac{1}{x} = x^5 + x - \frac{1}{x^5}- \frac{1}{x} = x^5 + x - \frac{1}{x^5} - \frac{1}{x}$
We can see that $x^5 - \frac{1}{x^5} + x - \frac{1}{x}$ is equal to $x^5 - \frac{1}{x^5}$. This means that $x^4 + 1 + \frac{1}{x^4}$ can be expressed as $x-\frac{1}{x}$.
So, $P(z) = x^5 - \frac{1}{x^5} = (x - \frac{1}{x})(x^4 + 1 + \frac{1}{x^4}) = z(x^4 + 1 + \frac{1}{x^4})$
Now we substitute $z$ back to $x - \frac{1}{x}$:
$P(z) = z(x^4 + 1 + \frac{1}{x^4}) = z((x - \frac{1}{x})^4 + 1 + \frac{1}{(x - \frac{1}{x})^4})$
Therefore, $P(z) = \boxed{z((x - \frac{1}{x})^4 + 1 + \frac{1}{(x - \frac{1}{x})^4})}$.
To solve this problem, we can use a property of the binomial expansion.
Let's rewrite $x^5 - \frac{1}{x^5}$ as $\left(x - \frac{1}{x}\right)^5$ and expand using the binomial theorem:
$\left(x - \frac{1}{x}\right)^5 = \binom{5}{0} x^5 \left(-\frac{1}{x}\right)^0 + \binom{5}{1} x^4 \left(-\frac{1}{x}\right)^1 + \binom{5}{2} x^3 \left(-\frac{1}{x}\right)^2 + \binom{5}{3} x^2 \left(-\frac{1}{x}\right)^3 + \binom{5}{4} x \left(-\frac{1}{x}\right)^4 + \binom{5}{5} \left(-\frac{1}{x}\right)^5$
Simplifying each term, we have:
$x^5 - \frac{1}{x^5} = 1 \cdot x^5 - 5 \cdot x^3 + 10 \cdot x - 10 \cdot \frac{1}{x^3} + 5 \cdot \frac{1}{x} - \frac{1}{x^5}$
Now, let's express $x - \frac{1}{x}$ in terms of a new variable, $z$:
$x - \frac{1}{x} = z$
Substituting this expression into the expanded form, we have:
$x^5 - \frac{1}{x^5} = 1 \cdot x^5 - 5 \cdot x^3 + 10 \cdot x - 10 \cdot \frac{1}{x^3} + 5 \cdot \frac{1}{x} - \frac{1}{x^5} = P(z)$
So, $P(z) = x^5 - 5x^3 + 10x - 10\frac{1}{x^3} + 5\frac{1}{x} - \frac{1}{x^5}$ is the polynomial we are looking for.