HMMT 二月 2000 · ORAL 赛 · 第 8 题

HMMT February 2000 — ORAL Round — Problem 8

[55] f is a polynomial of degree n with integer coefficients and f ( x ) = x + 1 for x = 1 , 2 , ..., n . What are the possible values for f (0) ? − → − → − → − → − →

解析

Let f be one such polynomial, for example, the obvious one: f ( x ) = x + 1. Then all such polynomials must be of the form f ( x ) + g ( x ) where g ( x ) = 0 at x = 1 , 2 , ..., n . g ( x ) must of the form C ( x − 1)( x − 2) ... ( x − n ) and since f + g has to have integer coefficients, n C must be an integer. At 0 this gives us ( f + g )(0) = f (0) + g (0) = 1 + C ( − 1) n !. Since C can be any positive or negative integer, the answer is that f(0) can be any integer congruent to 1(mod n !), i.e. of the form 1 + Cn !, C an integer.