1. ## Roots of equation?

If $\displaystyle x_1,x_2,x_3,\mbox{...},x_n$ are the roots of the equation $\displaystyle x^n + p_1x^{n - 1} + p_2x^{n - 2} + ... + p_{n - 1}x + p_n = 0$, where coefficient $\displaystyle p_1,p_2,p_3,\mbox{...},p_n$ are real, show that $\displaystyle (1 + x_1^2)(1 + x_2^2)(1 + x_3^2)\mbox{...}(1 + x_n^2) = (1 - p_1 + p_4 + \mbox{...})^2 + (p_1 - p_3 + p_5 + \mbox{...})^2$

2. Originally Posted by fardeen_gen
If $\displaystyle x_1,x_2,x_3,\mbox{...},x_n$ are the roots of the equation $\displaystyle x^n + p_1x^{n - 1} + p_2x^{n - 2} + ... + p_{n - 1}x + p_n = 0$, where coefficient $\displaystyle p_1,p_2,p_3,\mbox{...},p_n$ are real, show that $\displaystyle (1 + x_1^2)(1 + x_2^2)(1 + x_3^2)\mbox{...}(1 + x_n^2) = (1 - p_1 + p_4 + \mbox{...})^2 + (p_1 - p_3 + p_5 + \mbox{...})^2$
I think it should be $\displaystyle (1 + x_1^2)(1 + x_2^2)(1 + x_3^2)\mbox{...}(1 + x_n^2) = (1 - p_2 + p_4 + \mbox{...})^2 + (p_1 - p_3 + p_5 + \mbox{...})^2$

3. ## Induction

Base case $\displaystyle (n=1)$: $\displaystyle x+p_1=0~~\rightarrow~~1+x_1^2=1+(-p_1)^2=(1)^2+(p_1)^2$

Define $\displaystyle Q(x)=\sum_{i=0}^{n-1}q_ix^i=\prod_{i=1}^{n-1}(x-x_i)=0$ and $\displaystyle P(x)=\sum_{i=0}^np_ix^i=\prod_{i=1}^n(x-x_i)=$ $\displaystyle (x-x_n)Q(x)=xQ(x)-x_nQ(x)=0$ . So $\displaystyle p_i=q_i-x_nq_{i-1}$ for $\displaystyle i\in\mathbb{N}$ and $\displaystyle p_0=q_0=1$

Assume $\displaystyle (1+x_1^2)(1+x_2^2)...(1+x_{n-1}^2)=(1-q_2+q_4-q_6+...)^2+(q_1-q_3+q_5-q_7+...)^2=Q_{even}^2+Q_{odd}^2$

Show: $\displaystyle (1+x_1^2)(1+x_2^2)...(1+x_n^2)=(1-p_2+p_4-p_6+...)^2+(p_1-p_3+p_5-p_7+...)^2=P_{even}^2+P_{odd}^2$

Proof: $\displaystyle P_{even}^2+P_{odd}^2=(1-p_2+p_4-p_6+...)^2+(p_1-p_3+p_5-p_7+...)^2$

=$\displaystyle [1-(q_2-x_nq_1)+(q_4-x_nq_3)-(q_6-x_nq_5)+...]^2+$ $\displaystyle [(q_1-x_n)-(q_3-x_nq_2)+(q_5-x_nq_4)-(q_7-x_nq_6)+...]^2$

=$\displaystyle [(1-q_2+q_4-q_6+...)+x_n(q_1-q_3+q_5-q_7+...)]^2+$ $\displaystyle [(q_1-q_3+q_5-q_7+...)-x_n(1-q_2+q_4-q_6+...)]^2$

=$\displaystyle [Q_{even}+x_nQ_{odd}]^2+[Q_{odd}-x_nQ_{even}]^2$

=$\displaystyle [Q_{even}^2+Q_{odd}^2](1+x_n^2)$

=$\displaystyle (1+x_1^2)(1+x_2^2)...(1+x_{n-1}^2)(1+x_n^2)$

QED

4. $\displaystyle (x-x_{1})(x-x_{2}).....(x-x_{n})=x^n+p_{1}x^{n-1}+p_{2}x^{n-2}+...+p_{n}$
is an identity in x

Put $\displaystyle x=i$
$\displaystyle (i-x_{1})(i-x_{2}).....(i-x_{n})$ $\displaystyle =(p_{n}-p_{n-2}+p_{n-4}-p_{n-6}+...)+i(p_{n-1}-p_{n-3}+p_{n-5}-.....)$

Put $\displaystyle x=-i$
$\displaystyle (-i-x_{1})(-i-x_{2}).....(-i-x_{n})$ $\displaystyle =(p_{n}-p_{n-2}+p_{n-4}-p_{n-6}+...)-i(p_{n-1}-p_{n-3}+p_{n-5}-.....)$

Multiplying 1. and 2.
$\displaystyle (1+x_{1}^2)(1+x_{2}^2)...(1+x_{n}^2)=(p_{n}-p_{n-2}+p_{n-4}-p_{n-6}+...)^2+(p_{n-1}-p_{n-3}+p_{n-5}-.....)^2$

You will get the required result whatever n may be either even or odd.