Determine the smallest positive integer M with the following property: For every choice of integers a,b,c, there exists a polynomial P(x) with integer coefficients so that P(1)=aM and P(2)=bM and P(4)=cM. Proposed by Gerhard Woeginger, Austria
Problem
Source: 2021 Mediterranean Mathematical Olympiad P1 MMC
Tags: algebra, polynomial, Integer Polynomial
11.09.2021 21:07
The answer is 6. Indeed, assuming 3∤ and 2\nmid c-b we must have 4-1|(c-a)M\implies 3|M and 4-2|(c-b)M\implies 2|M and so 6|M. Now, by Lagrange interpolation, we take the parabola that satisfies the three equations. In the denominators of the formula there will be (4-1), (4-2), (2-1) which all get cancelled by M, since it is a multiple of 6, and so the polynomial has integer coefficients. Therefore M=6 is the minimal such number
11.09.2021 21:56
Note that we must have 2\mid M(c-b) and 3\mid M(c-a), thus 6\mid M, taking c=3,b=2,a=1. Now, we claim that M=6 works. Indeed, consider polynomial P_{abc}=(2a+c-3b)x^2+(15b-12a-3c)x+(16a+2c-12b).We are done.
11.08.2024 03:53
GYATTERRIFIC PROBLEM!!! Using polynomial and divisibility, it is clear that 3 | P(4)-P(1) = M(c-a) and 2 | P(4)-P(2) = M(c-b). It is clearly possible to make 3 not a factor of c-a, and 2 a factor of c-b, by making c and a different values mod 3 and c and b different values mod 2. Then, it is necessary that 2|M and 3|M so 6|M. Now M=6 works: Either the solution is a linear function which has clearly integer coefficients, or it is quadratic. Consider dx^2+ex+f a quadratic equation, so that d+e+f=6a, 4d+2e+f=6b, 9d+3e+f=6c Solving this system of simultaneous equations gives the coefficients d, e, f as integers. Then M=6 is possible too.