Help me solve this olympiad challenge? Given:
$$p(x) = x^4 - 5773x^3 - 46464x^2 - 5773x + 46$$
What is the sum of all arctan of all the roots of $p(x)$?

Let $p_1,p_2,p_3,p_4$ be the roots of $p(x)=0$ 
So, we need $\arctan p_1+\arctan p_2+\arctan p_3+\arctan p_4$
Using Vieta's Formulae, 
$\sum p_i=p_1+p_2+p_3+p_4=\frac{5773}1$
$\sum p_ip_j=p_1p_2+p_1p_3+p_1p_4+p_2p_3+p_2p_4+p_3p_4=\frac{46464}1$
$\sum p_ip_jp_k=p_1p_2p_3+p_1p_2p_4+p_1p_3p_4+p_2p_3p_4=\frac{5773}1$
$p_1p_2p_3p_4=\frac{46}1$
We know, $$\tan(A+B+C+D)$$
$$=\frac{\tan A+\tan B+\tan C+\tan D -(\tan A\tan B\tan C+\tan A\tan B\tan D+\tan A\tan C\tan D+\tan B\tan C\tan D)}{1- (\tan A\tan B+\tan A\tan C+\tan A\tan D+\tan B\tan C+\tan B\tan D+\tan C\tan D)+ \tan A\tan B\tan C\tan D}$$
If we put $p_1=\tan A,p_2=\tan B$ etc.,
$$\tan(A+B+C+D)=\frac{p_1+p_2+p_3+p_4 -(p_1p_2p_3+p_1p_2p_4+p_1p_3p_4+p_2p_3p_4)}{1- (p_1p_2+p_1p_3+p_1p_4+p_2p_3+p_2p_4+p_3p_4)+ p_1p_2p_3p_4}$$
$$=\frac{5773-5773}{1-46464+46}=0$$
So, $\tan(A+B+C+D)=0\implies A+B+C+D=\arctan (0)$
$\implies \arctan p_1+\arctan p_2+\arctan p_3+\arctan p_4=\arctan (0)$
The general value of $\arctan (0)$ is $n\pi$ where $n$ is any integer, the special value being $0$(putting $n=0$)
