October
6, 2006
Let
$\alpha$, $\beta$, $\gamma$, and $\delta$ be the roots of
$x^4+bx^3+cx^2+dx+e=0$. Compute
$(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$
in terms of $b$, $c$, $d$, and $e$.
Solution. We have
x^4+bx^3+cx^2+dx+e=(x-\alpha)(x-\beta)(x-\gamma)(x-\delta).
If we substitute $x=i$ above, we get
1-c+e-(b-d)i=(i-\alpha)(i-\beta)(i-\gamma)(i-\delta).
If we substitute instead $x=-i$, we get
$\begin{eqnarray}
1-c+e+(b-d)i&=&(-i-\alpha)(-i-\beta)(-i-\gamma)(-i-\delta) \\
&=& (-1)^4(i+\alpha)(i+\beta)(i+\gamma)(i+\delta) \\
&=& (i+\alpha)(i+\beta)(i+\gamma)(i+\delta)
\end{eqnarray}$
Noting that $(u+vi)(u-vi)=u^2+v^2$, multiplying the two equations
yields
(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)=(1-c+e)^2+(b-d)^2.