Question

If $\alpha$ and $\beta$ are the roots of the equations $x^2+px+q=0,$ $x^{2008}+p^{1004}{x}^{1004}+q^{1004}=0,$ then $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$ are the roots of $x^n+1+(x+1{)}^n=0.$ The value of $n$ must be

Answer 1

$\alpha ,\beta$ are roots of the equation $x^2+px+q=0$
$\Rightarrow \alpha +\beta =-p\dots \left(1\right)$
and $\alpha \beta =q\dots \left(2\right)$

$\alpha ,\beta$ are also roots of the equation $x^{2008}+p^{1004}{x}^{1004}+q^{1004}=0$
$\Rightarrow {\alpha }^{2008}+(\alpha +\beta {)}^{1004}\cdot {\alpha }^{1004}+(\alpha \cdot \beta {)}^{1004}=0$
$\Rightarrow {\alpha }^{2008}+{(\frac{\alpha }{\beta }+1)}^{1004}\cdot {\beta }^{1004}\cdot {\alpha }^{1004}+(\alpha \cdot \beta {)}^{1004}=0$

Dividing by $(\alpha \beta {)}^{1004},$ we get
${\left(\frac{\alpha }{\beta }\right)}^{1004}+{(\frac{\alpha }{\beta }+1)}^{1004}+1=0\dots \left(3\right)$
Given $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$ are the roots of $x^n+1+(x+1{)}^n=0$
$\Rightarrow {\left(\frac{\alpha }{\beta }\right)}^n+1+{(\frac{\alpha }{\beta }+1)}^n=0\dots \left(4\right)$

By comparing equation $\left(3\right)$ and $\left(4\right),$ we get
$n=1004$