<s>[INST] 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.
 [/INST] 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.
</s>