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Consider the polynomial with real coefficients $P(x)=x^6+a x^5+b x^4+c x^3+b x^2+a x+1$, and let $x_1, x_2, \ldots, x_6$ be its zeros. Prove that $$ \prod_{k=1}^6\left(x_k^2+1\right)=(2 a-c)^2 $$

By vieta, we know that $$-a=\sum_{cyc} x_1$$ and $$-c=\sum_{cyc} x_1 x_2 x_3$$ Also, $$P(x)=\prod_{i=1}^6(x-x_i)$$ How do I proceed from here?

Ellie_Wong
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  • The given product is a symmetric polynomial in the roots of $P$. Any symmetric polynomial can be expressed as a polynomial in the elementary symmetric polynomials, which in this case are the coefficients of $P$. Wikipedia contains a constructive proof. – Servaes May 18 '23 at 20:08
  • @Servaes so we can write $P(x)$ as other elementary symmetric polynomials, but how do we use the fact to solve the problem though? And by express, does it include product of other elementary symmetric polynomials too? – Ellie_Wong May 18 '23 at 20:16
  • Yes, this includes products. If you read the Wikipedia page, you will find all information you need, and more. In particular, it describes in full detail a method for expressing any symmetric polynomial as a polynomial in the elementary ones. – Servaes May 18 '23 at 20:19

2 Answers2

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Divide the equation with $x^3$ and plug $t=x+{1\over x}$. We get an equation $$t^3+at^2+(b-3)t+c-2a=0$$

By Vietas formula we have $$t_1t_2t_3 = 2a-c$$

Let $x_1,x_2$ be solutions to $x+1/x = t_1$ and similary for other...

Then $$ \prod_{k=1}^6\left(x_k^2+1\right) = x_1x_2...x_6\cdot \prod_{k=1}^6\left(x_k+1/x_k\right)= 1\cdot t_1^2t_2^2t_3^2= (2 a-c)^2 $$

nonuser
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Observe that we can split apart the product term into two factors using difference of squares: $$\prod_{k = 1}^6 (x_k^2 + 1) = \left( \prod_{k = 1}^6 -(i - x_k) \right) \left( \prod_{k = 1}^6 -(-i - x_k) \right).$$ The negative inside both terms can be factored out and ignored because $(-1)^6$ is simply $1$. This leaves us with an expression that is simply $P(i) \cdot P(-i)$, which is assuredly real by the complex conjugate theorem. If we plug in both $i$ and $-i$ into the polynomial, the desired result falls out quite nicely to be \begin{align*} P(i) \cdot P(-i) &= i(2a - c) \cdot -i(2a - c) \\ &= (2a - c)^2. \end{align*}

chirpyboat73
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