If $P(x)>Q(x)^2+Q(x)+x^2-6$ $\forall$ $x\in \mathbb{R}$, show that there exists two roots which satisfy $| \alpha - \beta | <1$

Question

Someone suggested me that it's just factorisation followed by Pigeonhole Principle. I don't find any way to continue from there. Any help?

Answer 1

Let $a_1\le a_2\le\cdots\le a_6$ be the roots of $P(x)$.

Suppose that $$a_{i+1}-a_i\ge 1\qquad \text{for}\quad i=1,2,\cdots, 5$$ from which we have $$a_6-a_1\ge 5\tag1$$

Now, we have $$(P(a_i)=)\ \ 0\gt Q(a_i)^2+Q(a_i)+a_i^2-6$$ Considering the discriminant of the RHS, we have to have $$1^2-4\cdot 1\cdot (a_i^2-6)\gt 0,$$ i.e. $$-2.5\lt a_i\lt 2.5$$

It follows that $$a_6-a_1\lt 5$$ which contradicts $(1)$.