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Do there exist four polynomials with real coefficients so that sum of every two doesn't have a real root while the sum of every three has a real root.

I have seen a similar problem with six polynomials which Ramsey's theorem could show that the answer is NO.

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1 Answer 1

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No, there are no such four polynomials.

We use two lemmas (the proof is written at the end of the answer) :

Lemma 1 : If $P(x)+Q(x)\gt 0, P(x)+R(x)\gt 0$ for all $x\in\mathbb R$, and $Q(x)+R(x)$ has no real roots, and there exists a real number $\alpha$ such that $P(\alpha)+Q(\alpha)+R(\alpha)=0$, then $Q(x)+R(x)\lt 0$ for all $x\in\mathbb R$.

Lemma 2 : If $P(x)+Q(x)\lt 0, P(x)+R(x)\lt 0$ for all $x\in\mathbb R$, and $Q(x)+R(x)$ has no real roots, and there exists a real number $\alpha$ such that $P(\alpha)+Q(\alpha)+R(\alpha)=0$, then $Q(x)+R(x)\gt 0$ for all $x\in\mathbb R$.


Suppose that four polynomials $P_1(x),P_2(x),P_3(x),P_4(x)$ satisfy our condition.

We may suppose that $$P_1(x)+P_2(x)\gt 0,\qquad P_1(x)+P_3(x)\gt 0$$ (At least two of $P_1(x)+P_2(x),P_1(x)+P_3(x),P_1(x)+P_4(x)$ have the same sign. We may suppose that $P_1(x)+P_2(x)$ and $P_1(x)+P_3(x)$ have the same sign. If both are negative, then we can name $-P_1(x),-P_2(x),-P_3(x),-P_4(x)$ as $P_1(x),P_2(x),P_3(x),P_4(x)$ respectively so that we have $P_1(x)+P_2(x)\gt 0$ and $P_1(x)+P_3(x)\gt 0$.)

From the lemma, we have $$P_2(x)+P_3(x)\lt 0$$

Suppose here that $P_1(x)+P_4(x)\gt 0$.

Then, since $P_1(x)+P_2(x)\gt 0$ and $P_1(x)+P_4(x)\gt 0$, it follows from the lemma that $P_2(x)+P_4(x)\lt 0$. Since $P_2(x)+P_3(x)\lt 0$ and $P_2(x)+P_4(x)\lt 0$, it follows from the lemma that $P_3(x)+P_4(x)\gt 0$. Since $P_1(x)+P_3(x)\gt 0$ and $P_3(x)+P_4(x)\gt 0$, it follows from the lemma that $P_1(x)+P_4(x)\lt 0$, which contradicts $P_1(x)+P_4(x)\gt 0$.

So, we have $P_1(x)+P_4(x)\lt 0$.

If $P_2(x)+P_4(x)\lt 0$, then since we have $P_2(x)+P_3(x)\lt 0$, it follows from the lemma that $P_3(x)+P_4(x)\gt 0$.

Now, we have two cases to consider :

Case 1 : $$P_1(x)+P_2(x)\gt 0,\quad P_1(x)+P_3(x)\gt 0,\quad P_2(x)+P_3(x)\lt 0$$ $$P_1(x)+P_4(x)\lt 0,\quad P_2(x)+P_4(x)\lt 0,\quad P_3(x)+P_4(x)\gt 0$$

In this case, we have $$P_1(x)\gt -P_2(x)\gt P_3(x)\gt -P_4(x)\gt P_1(x)$$ which is impossible.

Case 2 :

$$P_1(x)+P_2(x)\gt 0,\quad P_1(x)+P_3(x)\gt 0,\quad P_2(x)+P_3(x)\lt 0$$ $$P_1(x)+P_4(x)\lt 0,\quad P_2(x)+P_4(x)\gt 0$$ In this case, we have $$P_2(x)\lt -P_3(x)\lt P_1(x)\lt -P_4(x)\lt P_2(x)$$ which is impossible.


Finally, let us prove the lemmas.

Lemma 1 : If $P(x)+Q(x)\gt 0, P(x)+R(x)\gt 0$ for all $x\in\mathbb R$, and $Q(x)+R(x)$ has no real roots, and there exists a real number $\alpha$ such that $P(\alpha)+Q(\alpha)+R(\alpha)=0$, then $Q(x)+R(x)\lt 0$ for all $x\in\mathbb R$.

Proof for lemma 1 :

Since $$Q(\alpha)=-(P(\alpha)+R(\alpha))\lt 0,\qquad R(\alpha)=-(P(\alpha)+Q(\alpha))\lt 0$$ we have $Q(\alpha)+R(\alpha)\lt 0$.

Suppose here that there exists a real number $\beta$ such that $Q(\beta)+R(\beta)\gt 0$.

Then, by the intermediate value theorem, $Q(x)+R(x)$ has at least one real root, which contradicts that $Q(x)+R(x)$ has no real roots.

It follows from this that $Q(x)+R(x)\lt 0$ for all $x\in\mathbb R$.$\qquad\blacksquare$

Lemma 2 can be proven in the similar way as above.

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