Interesting theoretic proof with polynomials Four polynomials with real coefficients $p_1, p_2, p_3, p_4$ are given. The sum of any two of them has no real roots. Prove that if the polynomial $p_1+p_2+p_3+p_4$ has a real root, then at least one of the polynomials $p_1, p_2, p_3, p_4$ doesn't have a real root. $$$$
Unfortunately I'm quite new to polynomials and don't have the best understanding of them, so I would greatly appreciate if someone could guide me through this with explanation.
 A: Let us use the fact that a polynomial with no real roots can be represented by one of the two forms:

*

*$\ \ \ P(x)^2+Q(x)^2$ [case (+)] ($P$ and $Q$ having no common roots).


*$−(P(x)^2+Q(x)^2)$ [case (-)]
(see there), corresponding to the cases where the dominant coefficient is positive (resp. negative).
The first hypothesis means that
$$p_1+p_2, \ \ \ \ p_1+p_3, \ \ \ \ p_1+p_4, \ \ \ \  p_2+p_3,  \ \ \ \ p_2+p_4,  \ \ \ \ p_3+p_4$$
are each one either in case (+) or in case (-).
Let us assume that
$$p_1+p_2+p_3+p_4 \ \ \ \text{has a real root.}\tag{1}$$
Therefore, we have as many (+) and (-) cases (see Remark 1 below), like this:
$$\begin{cases}(a)&p_1+p_2&=&A^2+B^2\\(b)&p_3+p_4&=&-(C^2+D^2)\end{cases}, 
\ \ \begin{cases}(c)&p_1+p_3&=&E^2+F^2\\(d)&p_2+p_4&=&-(G^2+H^2)\end{cases}, \ \ \begin{cases}(e)&p_1+p_4&=&I^2+J^2\\(f)&p_2+p_3&=&-(K^2+L^2)\end{cases}\tag{2}$$
for certain polynomials $A,B,C,\cdots K,L$.
Combining (a)+(c)-(f), one gets
$$2p_1=A^2+B^2+E^2+F^2+K^2>0, \ \ \text{therefore with no real roots} \tag{3}$$
as was desired.
Remark 1: In (2), if we have had for example $$\begin{cases}(a)&p_1+p_2&=&A^2+B^2\\(b)&p_3+p_4&=&(C^2+D^2)\end{cases},$$ the sum $p_1+p_2+p_3+p_4$ would have always been $>0$, contradicting (1).
Remark 2: One could object that the cases considered in (2) are one among many. In fact, we can WLOG attribute indices to the $p_k$s in such a way that equations (a),(b),(c),(d) hold. It remains the last two equations (e) and (f) for which we could have a (+) sign for (f) and (-) sign for (e). The reader will not have difficulty to see that in this case as well, one can conclude as in (3).
