If we inspect its derivative
$$f'=(2n+1)X^{2n} - (2n+1) = (2n+1)(X^{2n}-1)$$
we can deduce that
$$f'(w) = 0 \ \iff \ (w^{n})^{2} = 1 \ \iff \ w^n = 1 \ \ \text{or} \ \ w^n = -1$$
This means that $w \in G_n$ is an nth root of unity. Accordingly,
$$f(w) = w.(w^n)^2 - (2n+1)w + a = 0 \ \iff \ a = 2nw$$
Since $a \in \mathbb{R}$, we know that $a$ can only be $\pm 2n$,
Question. Suppose we drop the restriction $a \in \mathbb{R}$ and admit complex coefficients. What's up with the case when $w^n = -1$? If $n$ is odd, then $w$ can only be $-1$, right? But what happens when $n$ is even? Does it still hold that $w$ is an nth root of unity?