Determine the lowest positive integer n such that following statement is true: If polynomial with integer coefficients gets value 2 for n different integers, then it can't take value 4 for any integer.
Problem
Source: Middle Europe Mathematical Olympiad TST Day 2 First problem
Tags: algebra, polynomial, absolute value, algebra unsolved
18.06.2009 16:27
The polynomial P(x)=4−x(x2+2x−1) has P(1)=P(−1)=P(−2)=2 and P(0)=4. Hence n>3. Next consider a polynomial P(x) with integer coefficients such that P(a)=P(b)=P(c)=P(d)=2 for four distinct integers a,b,c,d, and P(e)=4 for some integer e. Consider the polynomial Q(x)=4−P(x+e): Then Q(0)=0, and the four integers x1=a−e, x2=b−e, x3=c−e, and x4=d−e satisfy Q(xi)=2 for i=1,2,3,4. Since Q(0)=0, we have Q(x)=x∗R(x) for some integer polynomial R(x). Since x−y divides Q(x)−Q(y), we have xi−0 divides Q(xi)−Q(0)=2 for i=1,2,3,4. This implies that the xi are a permutation of 2,1,−1,−2. Then R(2)=1 and R(−2)=−1, and 4=2−(−2) must divide R(2)−R(−2)=2. Contradiction. Hence n≤4.
27.08.2009 16:25
Under the notations above, it follows P(x)=(x−a)(x−b)(x−c)(x−d)T(x)+2. But then 4=P(e)=(e−a)(e−b)(e−c)(e−d)T(e)+2, so |(e−a)(e−b)(e−c)(e−d)|≤2. However, the absolute value of the product of four distinct not-null integers is at least 4.
21.05.2017 00:17
Suppose that, for a given n, there exists such a polynomial. We define Q(x) as P(x)-2 for simplicity. We know Q(x) has n integer zeros, so we write: Q(x)=(x-x1)(x-x2)....(x-xn)*R(x) We want to find an integer x such that Q(x)=2, so since each of {x-x1, x-x2, ..., x-xn} is an integer, so is R(x). We also know that each of {x-x1, x-x2, ..., x-xn, R(x)} has to divide 2, and that each of {x-x1, x-x2, ..., x-xn} is distinct, but 2 can be written as a product of at most 3 different integers (e.g. (-2)*1*(-1)) so n<=3. Now we need a construction for every n<=3, and that's not hard. For n=3: Q(x)=(x-1)(x+1)(x-2) For n=2: Q(x)=-2*(x-1)*(x+1) For n=1: Q(x)=x. Therefore, n=4.
07.05.2022 10:51
We claim the answer is n=4. First, consider the polynomial (x−1)(x−3)(x−4)+2. It equals 2 for the 3 integers 1,3,4 and it takes on the value of 4 at x=2. Thus, n>3. Now, suppose n≥4 and the property is false. Let P(x) be a polynomial for which the property fails. Then, P(x)=α(x−r1)(x−r2)⋯(x−rn)+2 where r1,…,rn are distinct integers. Since the property fails, there is some integer m for which P(m)=4. Then, 2=α(m−r1)(m−r2)⋯(m−rn). It follows |m−ri|∈{1,2} for all i. Indeed, at most 1 of them can be 2, so at least three are 1. Hence, there exist i≠j for which m−ri=m−rj, contradicting ri≠rj. Therefore, the property never fails for n=4, as desired. ◼