HMMT 二月 2001 · 团队赛 · 第 10 题
HMMT February 2001 — Team Round — Problem 10
题目详情
英文原题
- Define a monic irreducible polynomial with integral coefficients to be a polynomialwith leading coefficient 1 that cannot be factored, and the prime factorization of a polynomialwith leading coefficient 1 as the factorization into monic irreducible polynomials. How manynot 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
英文解析
- Define a monic irreducible polynomial with integral coefficients to be a polynomialwith leading coefficient 1 that cannot be factored, and the prime factorization of a polynomialwith leading coefficient 1 as the factorization into monic irreducible polynomials. How manynot 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 integernpolynomial f ( x ) = a x + · · · + a (mod p ), where p does not divide a , has no zeros, then fn 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 .