The polynomial $P$ has integer coefficients and $P(x)=5$ for five different integers $x$. Show that there is no integer $x$ such that $-6\le P(x)\le 4$ or $6\le P(x)\le 16$.
Problem
Source: baltic way 2008
Tags: algebra, polynomial, algebra unsolved
SimonM
14.11.2008 20:46
Let $ g(x) = P(x) - 5$
$ g(x) = (x - a)(x - b)(x - c)(x - d)(x - e)$
We need to show that for $ n \in \mathbb{Z}$, $ g(n) \in \{ 0 \} \cup \{ x : |x| > 11 \}$
However, since $ g(n) = a'b'c'd'e'$ where $ a' = x - n$ etc and all of $ a',b', \ldots, e'$ are different integers, the smallest magnitude (aside from 0) which can be obtained is $ |1 \times - 1 \times 2 \times - 2 \times \pm 3| = 12 > 11$
mathangel
15.11.2008 03:38
The degree of $ P$ might not be 5.
SimonM
15.11.2008 09:37
mathangel wrote: The degree of $ P$ might not be 5. Good point, multiply by some $ h(x)$ the product is still at least $ \pm 12$
Mathias_DK
24.11.2008 17:44
erdugan wrote: The polyminal P has integer coefficients and P(x)=5 for five different integers x.Show that there is no integer x such that -7<P(x)<5 or 5<P(x)<17 Let $ Q(x) = P(x)-5$. The $ Q(X) = r(x)(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)$ where $ r(x)$ has integer coefficients. So when $ Q(X) \neq 0$ we know that $ |Q(X)| = |r(x)(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)| \ge |r(x)2 \cdot 1 \cdot (-1) \cdot (-2) \cdot (-3)| \ge 12$...