HMMT 二月 2015 · 代数 · 第 2 题

英文解析

2. The fraction has a unique “(restricted) partial fraction decomposition” of the form
   2015
   1 a b c = + + ,
   2015 5 13 31
   where a, b, c are integers with 0 ≤ a < 5 and 0 ≤ b < 13. Find a + b .
   Answer: 14 This is equivalent to 1 = 13 · 31 a + 5 · 31 b + 5 · 13 c . Taking modulo 5 gives 1 ≡ 3 · 1 a 1
   (mod 5), so a ≡ 2 (mod 5). Taking modulo 13 gives 1 ≡ 5 · 5 b = 25 b ≡ − b (mod 13), so b ≡ 12
   (mod 13). The size constraints on a, b give a = 2, b = 12, so a + b = 14.
   Remark. This problem illustrates the analogy between polynomials and integers, with prime powers
   1 1 1 1 2 3
   (here 5 , 13 , 31 ) taking the role of powers of irreducible polynomials (such as ( x − 1) or ( x + 1) ,
   when working with polynomials over the real numbers).
   Remark. The “partial fraction decomposition” needs to be restricted since it’s only unique “modulo
   1”. Abstractly, the abelian group (or Z -module) Q / Z has a “prime power direct sum decomposition”
   (more or less equivalent to Bezout’s identity, or the Chinese remainder theorem), but Q itself (as anabelian group under addition) does not.
   You may wonder whether there’s a similar “prime power decomposition” of Q that accounts not justfor addition, but also for multiplication (i.e. the full ring structure of the rationals). In some sense,
   the “adeles/ideles” serve this purpose, but it’s not as clean as the partial fraction decomposition (foradditive structure alone)—in fact, the subtlety of adeles/ideles reflects much of the difficulty in numbertheory!