1

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.

Harley
  • 71
  • 3
    Where did you come across this problem? Was there any context involved? It would help to know what sort of tools you might be expected to solve this problem with. – Steven Stadnicki Nov 02 '20 at 17:09
  • Nothing additional about the degrees of the $p_k$s ? – Jean Marie Nov 02 '20 at 17:23
  • 1
    @JeanMarie I mean, you can assume they all have even degree WLOG. – Rushabh Mehta Nov 02 '20 at 17:28
  • Does this post help with your problem?: https://math.stackexchange.com/questions/2578580/ – Andreas Nov 02 '20 at 17:31
  • Maybe use the fact that a polynomial with no real roots is either of the form $P(x)^2+Q(x)^2$ or of the form $-(P(x)^2+Q(x)^2)$ (https://math.stackexchange.com/q/823627) – Jean Marie Nov 02 '20 at 18:00
  • @DonThousand Couldn't one of them have odd degree? – Steven Stadnicki Nov 02 '20 at 18:36
  • No, because then it would have a real root, which would finish the question immediately. – Rushabh Mehta Nov 02 '20 at 18:50
  • @DonThousand Maybe I'm missing something obvious, but I don't see why. The sum of every pair could still have no real roots (since those can all be even degree e.g. if $p_1$ is cubic and $p_2, p_3, p_4$ quartic) and we're being asked to show that at least one of the $p$ has no real roots, so knowing that e.g. $p_1$ must have a real root doesn't tell us anything about the other $p_i$. – Steven Stadnicki Nov 02 '20 at 19:56
  • @StevenStadnicki, actually, I can't read. I read it as at least one of the polynomials has a real root. Oops – Rushabh Mehta Nov 02 '20 at 20:07
  • No worries! I'm glad I wasn't missing anything obvious. (In fact, I believe one could in principle even have two polynomials with odd degree, if the other two have higher even degrees and the two odd-degree polynomials have additively inverse leading coefficients — then all pairwise sums could have even degree and so potentially have no real roots) – Steven Stadnicki Nov 02 '20 at 20:12

1 Answers1

1

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).

Jean Marie
  • 81,803