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. 

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.
