If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, prove the following. 
If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, show that $$(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2)=(1-q+s)^2+(p-r)^2.$$

I have tried putting
\begin{align}P(1)&=(\alpha-1)(\beta-1)(\gamma-1)(\delta-1)=1+p+q+r+s\\P(-1)&=(\alpha+1)(\beta+1)(\gamma+1)(\delta+1)=1-p+q-r+s.\end{align}
Then \begin{align}P(1)P(-1)&=(\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)\\&=(1+p+q+r+s)(1-p+q-r+s)=(1+q+s)^2-(p+r)^2\end{align}
Somehow it does not match the statement given.
 A: You correctly computed 
$$
(\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)=(1+q+s)^2-(p+r)^2
$$
only that this is not what was asked for.
But you were close: $\alpha^2+1 = (\alpha-i)(\alpha+i)$ etc, and therefore
$$
 (1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2) = P(i) P(-i)
$$
which expands to
$$
(1-ip-q+ir+s)(1+ip-q-ir+s) = (1-q+s)^2+(p-r)^2 \, .
$$
A: Denoting $\alpha=x_1$, $\beta=x_2$, $\gamma=x_3$ and $\delta=x_4$, we have $$\small\prod_{i=1}^4(1+x_i^2)-\prod_{i=1}^4(1-x_i^2)=\small2(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)+2(x_1+x_2+x_3+x_4)$$ so taking out a factor of $x_1x_2x_3x_4$ from the first term in brackets, $$\prod_{i=1}^4(1+x_i^2)=2\left(\prod_{i=1}^4x_i\right)\left(\sum_{i=1}^4\frac1{x_i}\right)+2\sum_{i=1}^4x_i+\prod_{i=1}^4(1-x_i^2)$$ which is evaluable by Vieta's formulas. For the sum of reciprocals, see this post.
A: Another way using Transformation of equation: 
We have $$x^4+qx^2+s=-x(px^2+r)$$
Squaring both sides
$$(x^2)^4+q^2(x^2)+s^2+2q(x^2)^3+2qs(x^2)+2s(x^2)^2=x^2(p^2(x^2)^2+r^2+2pr(x^2))$$
Replacing $x^2+1=y$
$$(y-1)^4+q^2(y-1)^2+s^2+2q(y-1)^3+2qs(y-1)+2s(y-1)^2=(y-1)(p^2(y-1)^2+r^2+2pr(y-1))$$
$$\iff y^4+\cdots+1+q^2+s^2-2q-2qs+2s+p^2+r^2-2pr=0$$
Using Vieta's formula $$\prod_{j=1}^4y_j=\dfrac{1+q^2+s^2-2q-2qs+2s+p^2+r^2-2pr}1=(1-q+s)^2+(p-r)^2$$
A: Your approach ends up proving $(α^2 - 1)(β^2 - 1)(γ^2 - 1)(δ^2 - 1) = (1 - q + s)^2 - (p - r)^2$. This means we will have to do something with that minus sign and to do that we can use the identity $(x + ιy)(x - ιy) = x^2 + y^2$. Putting $P(ι)$, and $P(-ι)$ you can easily solve the problem with the same approach.
