Four roots of the polynomial $x^4+px^3​+qx^2​+rx+1$ Let $a, b, c, d$ be four roots of the polynomial $x^4+px^3​+qx^2​+rx+1$ prove that $(a^4​+1)(b^4​+1)(c^4​+1)(d^4​+1)=(p^2​+r^2)​^2​+q^4-4pq^2​r$
Please provide hint. 
 A: Let


*

*$f(x) = x^4 + px^3 + qx^2 + rx + 1$.

*$\Lambda = \{ a, b, c, d \}$ be the set of roots of $f(x)$.

*$\Omega = \left\{ \frac{\pm 1 \pm i}{\sqrt{2}} \right\}$ be the set of roots of $x^4 + 1$. 


We have
$$\begin{align}\mathcal{P}\stackrel{def}{=}\prod_{\lambda \in \Lambda}(\lambda^4 + 1)
&= \prod_{\lambda \in \Lambda}\prod_{\omega \in \Omega}(\lambda - \omega)
= \prod_{\omega \in \Omega}\prod_{\lambda _in \Lambda}(\omega - \lambda)
= \prod_{\omega \in \Omega}f(\omega)\\
&= \prod_{\omega \in \Omega}(p \omega^3 + q \omega^2 + r\omega)
\end{align}
$$
Since $\prod_{\omega \in \Omega} \omega^2 = (-1)^2 = 1$ and $\omega^{-1} = \bar{\omega}$ for all $\omega \in \Omega$, we find
$$\begin{align}
\mathcal{P} = & \prod_{\omega \in \Omega}(q + p \omega + r \bar\omega)\\
= & \phantom{+}
\left(q + \frac1{\sqrt{2}}(p + r) + \frac{i}{\sqrt{2}}(p - r)\right)
\left(q + \frac1{\sqrt{2}}(p + r) - \frac{i}{\sqrt{2}}(p - r)\right)\\
& \times 
\left(q - \frac1{\sqrt{2}}(p + r) + \frac{i}{\sqrt{2}}(p - r)\right)
\left(q - \frac1{\sqrt{2}}(p + r) - \frac{i}{\sqrt{2}}(p - r)\right)\\
= & \left(q^2 + p^2 + r^2 + \sqrt{2}q(p+r)\right)
\left(q^2 + p^2 + r^2 - \sqrt{2}q(p+r)\right)\\
= & (q^2 + p^2 + r^2)^2 - 2q^2(p+r)^2\\
= & (p^2 + \color{red}{r}^2)^2 + q^4 - 4q^2 pr
\end{align}$$
Please note that there is a typo in the formula you want to prove.
A: We seek a quartic whose roots are $a^4, b^4, c^4 , d^4$.
Naively, this is $ x + px^{3/4} + qx^{1/2} + rx^{1/4} + 1 = 0 $, but the powers need to be integers.
Shifting terms, $x + qx^{1/2} + 1 = - px^{3/4} - rx^{1/4}$.
Squaring both sides, $x^2 + q^2x + 1 + 2x + 2qx^{3/2} + 2qx^{1/2} = p^2 x^{3/2} + 2pr x + r^2 x^{1/2}$.
Shifting terms, $x^2 + (q^2 - 2pr +2)x + 1 = x^{1/2} ((p^2-2q)x + (r^2 -2q))$.
Squaring again, $x^4 + (q^2 - 2pr +2)^2x^2 + 1 + 2(q^2 - 2pr +2)x^3 + 2(q^2 - 2pr +2)x + 2x^2 = x(  (p^2-2q)^2x^2 + (r^2 -2q)^2 + 2(p^2-2q)(r^2 -2q)x)$
Let 
$$ g(x) = x^4 + (q^2 - 2pr +2)^2x^2 + 1 + 2(q^2 - 2pr +2)x^3 + 2(q^2 - 2pr +2)x + 2x^2 - x(  (p^2-2q)^2x^2 + (r^2 -2q)^2 + 2(p^2-2q)(r^2 -2q)x).$$
Then $g(x) = ( x - a^4)(x-b^4)(x-c^4)(x-^4)$,
so $g(-1) = \prod ( -1 - a^4) = \prod (a^4 + 1) $.
On the other hand, by substituting in $ x = -1$,
$ g(-1) = 1 + (q^2 - 2pr +2)^2 + 1 - 2(q^2 - 2pr +2) - 2(q^2 - 2pr +2) + 2 + (  (p^2-2q)^2 + (r^2 -2q)^2 - 2(p^2-2q)(r^2 -2q))$
$ = p^4 + 2p^2r^2 + r^4 + q^4 - 4pq^2 r $ (E.g. expand by Wolfram)
which is the form that you want.
A: I use the following trick.

If $f(x)$ is a polynomial, then $f(x)f(-x)$ is a polynomial with only even degree terms, so there exists a polynomial $g(x)$ such that 
  $$g(x^2)=f(x)f(-x).$$ Then the zeros of $g(x)$ are exactly the squares of the zeros of $f(x)$.

With $f(x)=x^4+px^3+qx^2+rx+1$ we arrive at
$$
g(x)=1 + (2 q - r^2) x + (2 + q^2 - 2 p r) x^2 + (-p^2 + 2 q) x^3 + x^4.
$$
Repeating the dose we look for a polynomial $h(x)$ such that $h(x^2)=g(x)g(-x)$.
Expanding gives us
$$
\begin{aligned}
h(x)&=1 + (4 - 2 q^2 - 4 p r + 4 q r^2 - r^4) x\\
& + (6 + 4 p^2 q - 4 q^2 + 
    q^4 - 8 p r - 4 p q^2 r + 2 p^2 r^2 + 4 q r^2) x^2\\
& + (4 - p^4 + 
    4 p^2 q - 2 q^2 - 4 p r) x^3 + x^4.
\end{aligned}
$$
As explained in the comments, because
$$
h(x)=(x-a^4)(x-b^4)(x-c^4)(x-d^4)
$$ the answer is
$$
h(-1)=p^4+2 p^2 r^2-4 p q^2 r+q^4+r^4.
$$
A: Hint:
Use the fact that
$$x^4+1=-(px^2+qx+r)x,$$
expand the product and substitute Vieta's expressions.
A: $$\prod_{cyc}(a^4+1)=((a^2b^2-1)^2+(a^2+b^2)^2)((c^2d^2-1)^2+(c^2+d^2)^2)=$$
$$=((a^2b^2-1)(c^2+d^2)+(c^2d^2-1)(a^2+b^2))^2+$$
$$+((a^2b^2-1)(c^2d^2-1)-(a^2+b^2)(c^2+d^2))^2=$$
$$=\left(\sum_{cyc}a^2b^2c^2-\sum_{cyc}a^2\right)^2+\left(a^2b^2c^2d^2-\frac{1}{6}\sum_{sym}a^2b^2\right)^2.$$
Can you end it now?
