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

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