PUMaC 2013 · 代数（B 组） · 第 1 题

PUMaC 2013 — Algebra (Division B) — Problem 1

[ 3 ] Suppose a, b, c > 0 are integers such that abc − bc − ac − ab + a + b + c = 2013 . Find the number of possibilities for the ordered triple ( a, b, c ).

解析

Suppose there are 0 real roots. Then f = ( t − 2 x t + 1)( t − 2 x t + 1) for some 1 2 x , x ∈ ( − 1 , +1) such that 4 x x and 2 x + 2 x are both integers, and any such 1 2 1 2 1 2 x , x will do. Therefore, the two-way map 1 2 2 2 ( t − 2 x t + 1)( t − 2 x t + 1) ↔ ( u − 2 x )( u − 2 x ) (B.20) 1 2 1 2 2 is a bijection between possibilities for f and polynomials g ( u ) = u + bu + c such that b, c are integers and the roots of g belong to ( − 2 , +2) . By the quadratic formula, √ 2 2 the condition on the roots works out to b − 4 < ± b − 4 c < b + 4 , meaning b ≥ 4 c and − 2 b − 4 < c < +2 b − 4 . But we also know b, c ∈ ( − 4 , +4) from expressing b, c in terms of x , x . Altogether, the only possibilities for ( b, c ) are 1 2 (0 , 0) , (1 , − 3) , (B.21) (2 , − 3) , (2 , − 2) , (2 , − 1) , (3 , − 3) , (3 , − 2) , (3 , − 1) , (3 , 0) , (3 , 1) , a total of 10 possibilities for f . 5