If $\alpha,\beta$ are roots of $x^2+px+q=0$ and also of $x^{2n}+p^nx^n+q^n=0$ and $\frac{\alpha}{\beta}$,$\frac{\beta}{\alpha}$ are the roots of $x^n+1+(x+1)^n=0$, then $n$ is
- Odd
- Even
- Any integer
$\alpha+\beta=-p$ and $\alpha\beta=q$
The third equation is $$x^n+1+\binom{n}{0}+\dbinom{n}{1}x+\cdots+\dbinom{n}{n}x^n=0$$
I cant comment about sum of roots of both second and third equation because in second equation, only 2 among $2n$ roots are given. Also, only 2 among $n$ roots are given for the third equation.
Assuming that you are working over the reals, note that $\alpha,\beta \neq 0$. Thus, $\alpha$ being a root of your second polynomial implies $\alpha^n +p^n +\beta ^n=0$, i.e. $-(\alpha^n +\beta^n)=p^n=(-(\alpha + \beta))^n$, where the second equation holds due to $\alpha + \beta = -p$. $\alpha/ \beta$ being a root of the third polynomial implies $-(\alpha^n + \beta ^n)=(\alpha + \beta)^n$. Putting both equations together gives $(-(\alpha + \beta))^n=(\alpha + \beta)^n$.
If $\alpha \neq -\beta$, $n$ must be even. In the other case, $-1$ must be a root of the third polynomial, i.e. $(-1)^n=-1$, and so $n$ must be odd.