Track / Stage 6 / 144 of 400 #1144 of 1964

Problem 1144

National olympiad, first round

Algebra Difficulty 6.2 Prove it

4. Given $n$ different integers $m_{1}, m_{2}, \cdots, m_{n}$ , prove that there exists an integer-coefficient polynomial $f(x)$ of degree $n$ that satisfies the following conditions:
(1) $f\left(m_{i}\right)=-1,1 \leqslant i \leqslant n$ ;
(2) $f(x)$ cannot be factored into the product of two integer-coefficient polynomials of degree at least 1.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

4. Let $f(x)=(x-m_{1})(x-m_{2}) \cdots(x-m_{n})-1$ , then $f(x)$ satisfies condition (1). Next, we prove that $f(x)$ satisfies condition (2). In fact, if $f(x)$ does not satisfy condition (2), i.e., there exist integer-coefficient polynomials $Q(x), R(x)$ with degree no less than 1, such that $f(x)=Q(x) R(x)$ .
Since for $i \in\{1,2, \cdots, n\}$ , we have $f\left(m_{i}\right)=-1$ .
Thus, $Q\left(m_{i}\right) R\left(m_{i}\right)=-1$ ,
Therefore, $\left\{\begin{array}{l}Q\left(m_{i}\right)=1 \\ R\left(m_{r}\right)=-1\end{array}\right.$ or $\left\{\begin{array}{l}Q\left(m_{i}\right)=-1 \\ R\left(m_{i}\right)=1 .\end{array}\right.$ .
Hence, $Q\left(m_{i}\right)+R\left(m_{i}\right)=0, i=1,2, \cdots, n$ .
Since the degree of $Q(x)+R(x)$ does not exceed $n-1$ , we have $Q(x)+R(x)=0$ ,
i.e., $Q(x) \equiv R(x)$ . Therefore, $f(x)=Q(x) R(x)=-[Q(x)]^{2}$ .
This contradicts the leading coefficient of $f(x)$ being 1, hence the conclusion holds.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.