Let $\alpha_1$, $\alpha_2$, $\alpha_3$, $\alpha_4$ be the roots of the following polynomial


Show that


I came across this problem on a facebook page as an challenge.The only way that Strike me is multiply the terms in LHS and put the corresponding values but this method would be very long as polynomial is of 4th degree.

If anybody know any other method then Please tell me. Thank you for your help!

  • 2
    $\begingroup$ I would first study en.wikipedia.org/wiki/Vieta%27s_formulas. Then, I would express $p, q,$ and $r$ in terms of $\alpha_1, \alpha_2, \alpha_3,$ and $\alpha_4.$ Then, I would compare the LHS and the RHS of the equation that you are being asked to prove. My instinct (which could easily be mistaken) is that the LHS and RHS will have to match up. $\endgroup$ Jul 28 '20 at 9:52
  • $\begingroup$ Addendum to previous comment: it is plausible to me that the match up might require you to additionally use that $\alpha_1, \alpha_2, \alpha_3,$ and $\alpha_4$ are each roots of the original polynomial; however, perhaps this additional usage will not be necessary. $\endgroup$ Jul 28 '20 at 9:55
  • $\begingroup$ Use twice the fact that if $R(x^2)=P(x)P(-x)$ then the zeros of $R(x)$ are the squares of the zeros $P(x)$. And also that $\beta$ is a zero of $Q(x)$ if and only if $1+\beta$ is a zero of $Q(x-1)$. And then the Vieta relation extracting the product of the zeros. $\endgroup$ Jul 28 '20 at 10:42

Rewrite $x^4+px^3+qx^2+rx+1=0$ as

$$x^2+\frac1{x^2}+q=-(px +\frac r x) $$

Square to get

$$x^4+\frac1{x^4} + (2+q^2-2pr)=(p^2-2q)x^2+ \frac{r^2-2q}{x^2}$$

Square again and rearrange to obtain the quartic equation in $x^4$ \begin{align} f(x^4)=&x^{16}+[2(2+q^2-2pr)-(p^2-2q)^2]x^{12}\\ &+[2+(2+q^2-2pr)^2-2(p^2-2q)(r^2-2q)]x^8\\ &+[2(2+q^2-2pr)-(r^2-2q)^2]x^4+1=0 \end{align}


$$(1+\alpha_1^4)(1+\alpha_2^4)(1+\alpha_3^4)(1+\alpha_4^4) =f(-1)=(p^2+r^2)^2+q^4-4pq^2r $$


Squaring both sides $$(x^4+qx^2+1)^2=x^2(px^2+r)^2$$

Let $x^2=y$


$$\iff y^4+y^2A+1=y^3B+yC$$

Let $z=1+y^2\implies y=\sqrt{z-1}$


$$ z^2+zE+F=\pm\sqrt{z-1}(zG+H)$$

Squaring both sides



$$\implies\prod_{r=1}(1+\alpha_r^4)=\prod_{r=1}z_r=\dfrac{F^2+H^2}1$$ using Vieta's formula


$\alpha^4+1=(\eta-\alpha)(\eta^2-\alpha)(\eta^5-\alpha)(\eta^7-\alpha)$ where $\eta=\exp(\pi i/4)=\frac1{\sqrt2}(1+i)$. Therefore $$\prod_{k=1}^4(1+\alpha_k^4)=P(\eta)P(\eta^3)P(\eta^5)P(\eta^7).$$ But $$P(\eta)P(\eta^5)=P(\eta)P(-\eta)$$ and $$P(x)P(-x)=(x^4+qx^2+1)^2-(px^3+rx)^2 =Q(x^2)$$ where $$Q(x)=x^4+(2q-p^2)x^3+(2+q^2-2pr)x^2+(2q-r^2)x+1.$$ So $$P(\eta)P(\eta^5)=Q(\eta^2)=Q(i)$$ and similarly $$P(\eta^3)P(\eta^7)=Q(-i).$$ Then $$\prod_{k=1}^4(1+\alpha_k^4)=Q(i)Q(-i)=(2pr-q^2)^2+(p^2-r^2)^2.$$ This should be re-arrangeable into your answer.


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