HMMT 二月 2001 · 团队赛 · 第 10 题

HMMT February 2001 — Team Round — Problem 10

Define a monic irreducible polynomial with integral coefficients to be a polynomial with leading coefficient 1 that cannot be factored, and the prime factorization of a polynomial with leading coefficient 1 as the factorization into monic irreducible polynomials. How many not necessarily distinct monic irreducible polynomials are there in the prime factorization of 8 4 8 2 ( x + x + 1)( x + x + 1) (for instance, ( x + 1) has two prime factors)?

解析

Define a monic irreducible polynomial with integral coefficients to be a polynomial with leading coefficient 1 that cannot be factored, and the prime factorization of a polynomial with leading coefficient 1 as the factorization into monic irreducible polynomials. How many not necessarily distinct monic irreducible polynomials are there in the prime factorization of 8 4 8 2 ( x + x + 1)( x + x + 1) (for instance, ( x + 1) has two prime factors)? 8 4 8 4 4 4 2 2 2 4 2 4 2 Solution: x + x +1 = ( x +2 x +1) − x = ( x +1) − ( x ) = ( x − x +1)( x + x +1) = 4 2 2 2 8 2 6 5 3 2 ( x − x +1)( x + x +1)( x − x +1), and x + x +1 = ( x + x +1)( x − x + x − x +1). If an integer n polynomial f ( x ) = a x + · · · + a (mod p ), where p does not divide a , has no zeros, then f n 0 n 6 5 3 2 has no rational roots. Taking p = 2, we find x − x + x − x + 1 is irreducible. The prime 4 2 2 2 2 6 5 3 2 factorization of our polynomial is thus ( x − x +1)( x − x +1)( x + x +1) ( x − x + x − x +1), so the answer is 5 .