Relation of Roots of Bi-quadratic equation without using Vieta's formula

Question

If $ a, b, c, d $ are roots of the equation $$ x^4 + px^3 + qx^2 + rx + s=0 $$ Show that $$ (1+a^2)(1+b^2)(1+c^2)(1+d^2)=(1+s-q)^2 + (p-r)^2 $$

I was able to show this by using Vieta's formulas but the book from which I am doing this has not yet taught, till now it has just taught reminder, factor theorem and fundamental theorem of algebra so I want to ask that is there any other solution using only these or some more interesting way to do

Answer 1

So: $$ f(x)= (x-a)(x-b)(x-c)(x-d)$$

$$f(i) = (i-a)...\;\;\;\;{\rm and }\;\;\;\;f(-i)= (-i-a)...$$ thus $$f(i)f(-i) = (a^2+1)(b^2+1)(c^2+1)(d^2+1)$$

On the other side $$f(i)f(-i) = (1 -ip -q+ ri + s) (1 + ip + q - ri + s)$$ $$ = (1-q+s-i(p-r))(1-q+s+i(p-r))$$ $$= (1-q+s)^2+(p-r)^2$$

Answer 2

Also, we can use $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.$$ Indeed, $$(1+a^2)(1+b^2)(1+c^2)(1+d^2)=((a+b)^2+(1-ab)^2)((c+d)^2+(1-cd)^2)=$$ $$=((a+b)(c+d)-(1-ab)(1-cd))^2+((a+b)(1-cd)+(c+d)(1-ab))^2=$$ $$=(ab+ac+ad+bc+bd+cd-abcd-1)^2+(a+b+c+d-abc-abd-acd-bcd)^2=$$ $$=(q-s-1)^2+(-p+r)^2=(1+s-q)^2+(p-r)^2.$$