If $\mathrm{\alpha }, \mathrm{\beta }$ the roots of ${\mathrm{x}}^2+\mathrm{px}+\mathrm{q}=0\text{ and }{\mathrm{x}}^{2\mathrm{n}}+{\mathrm{p}}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}+{\mathrm{q}}^{\mathrm{n}}=0$ and if $(\mathrm{\alpha }/\mathrm{\beta }),(\mathrm{\beta }/\mathrm{\alpha })$ are the roots of ${\mathrm{x}}^{\mathrm{n}}+1+(\mathrm{x}+1{)}^{\mathrm{n}}=0$ then $\mathrm{n}(\in \mathrm{N})$

We have 

$\mathrm{\alpha }+\mathrm{\beta }=-\mathrm{p}\text{ and }\mathrm{\alpha \beta }=\mathrm{q} \left(1\right)$

Also, since $\mathrm{\alpha }, \mathrm{\beta }$ are the roots of ${\mathrm{x}}^{2\mathrm{n}}+{\mathrm{p}}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}+{\mathrm{q}}^{\mathrm{n}}=0$,

We have ${\mathrm{\alpha }}^{2\mathrm{n}}+{\mathrm{p}}^{\mathrm{n}}{\mathrm{\alpha }}^{\mathrm{n}}+{\mathrm{q}}^{\mathrm{n}}=0\text{ and }{\mathrm{\beta }}^{2\mathrm{n}}+{\mathrm{p}}^{\mathrm{n}}{\mathrm{\beta }}^{\mathrm{n}}+{\mathrm{q}}^{\mathrm{n}}=0$

Subtracting the above relations, we get

$\left({\mathrm{\alpha }}^{2\mathrm{n}}-{\mathrm{\beta }}^{2\mathrm{n}}\right)+{\mathrm{p}}^{\mathrm{n}}\left({\mathrm{\alpha }}^{\mathrm{n}}-{\mathrm{\beta }}^{\mathrm{n}}\right)=0$

$\therefore {\mathrm{\alpha }}^{\mathrm{n}}+{\mathrm{\beta }}^{\mathrm{n}}=-{\mathrm{p}}^{\mathrm{n}} \left(2\right)$

Given, $\mathrm{\alpha }/\mathrm{\beta }\text{ or }\mathrm{\beta }/\mathrm{\alpha }\text{ is a root of }{\mathrm{x}}^{\mathrm{n}}+1+(\mathrm{x}+1{)}^{\mathrm{n}}=0$. So,

$\begin{array}{l} (\mathrm{\alpha }/\mathrm{\beta }{)}^{\mathrm{n}}+1+[(\mathrm{\alpha }/\mathrm{\beta })+1{]}^{\mathrm{n}}=0\\ \Rightarrow \left({\mathrm{\alpha }}^{\mathrm{n}}+{\mathrm{\beta }}^{\mathrm{n}}\right)+(\mathrm{\alpha }+\mathrm{\beta }{)}^{\mathrm{n}}=0\\ \Rightarrow -{\mathrm{p}}^{\mathrm{n}}+(-\mathrm{p}{)}^{\mathrm{n}}=0 [\mathrm{Using} \left(1\right) \mathrm{and} \left(2\right)]\end{array}$

It is possible only when n is even.