Question

If $a,b,c,...k$ are the roots of
$x^n+p_1{x}^{n-1}+p_2{x}^{n-2}+....+p_{n-1}x+p_n=0$,
show that
$(1+a^2)(1+b^2)....(1+k^2)=(1-p_2+p_4-....{)}^2+(p_1-p_3+p_5-....{)}^2$.

Answer 1

Given equation is $x^n+p_1{x}^{n-1}+p_2{x}^{n-2}+\dots +p_{n-1}x+p_n=0$

As $a,b,c,\dots ,n$ are the roots of the given equation,

$(x-a)(x-b)(x-c)+\dots (x-n)=x^n+p_1{x}^{n-1}+p_2{x}^{n-2}+\dots +p_{n-1}x+p_n\dots ..\left(1\right)$

Substitute roots of $x^2+1$ $=$ $0$ in the (1), we have

$(i-a)(i-b)(i-c)+\dots (i-n)=i^n+p_1{i}^{n-1}+p_2{i}^{n-2}+\dots +p_{n-1}i+p_n$

Or $(i-a)(i-b)(i-c)+\dots (i-n)=(1-p_2+p_4+\dots )-i(p_1-p_3+p_5\dots )\dots \dots \dots ..\left(2\right)$

$(-i-a)(-i-b)(-i-c)+\dots (-i-n)=(-i{)}^n+p_1(-i{)}^{n-1}+p_2(-i{)}^{n-2}+\dots +p_{n-1}(-i)+p_n$

Or $(i+a)(i+b)(i+c)+\dots (i+n)=(1-p_2+p_4+\dots )+i(p_1-p_3+p_5\dots )\dots \dots \dots ..\left(3\right)$

Multiplying (2) and (3), we get

$(1+a^2)(1+b^2)\dots (1+n^2)=(1-p_2+p_4+\dots {)}^2+(p_1-p_3+p_5\dots {)}^2$