PUMaC 2011 · 代数（B 组） · 第 3 题

PUMaC 2011 — Algebra (Division B) — Problem 3

[ 4 ] Let f ( x ) = x − 7 x + 16 x − 10. As x ranges over all integers, find the sum of distinct prime values taken on by f ( x ).

解析

By inspection, we see that 1 is a root of this polynomial. Factoring out ( x − 1), we have 2 2 2 f ( x ) = ( x − 1)( x − 6 x + 10). Since x − 6 x + 10 = ( x − 3) + 1, then for any x < 0 or 2 x > 3, both | x − 1 | , | x − 6 x + 10 | ≥ 2, so their product cannot be prime. Trying directly, 2 f (0) = − 10 , f (1) = 0 are not prime. If x ≥ 2, we need at least one of x − 1 and ( x − 3) + 1 to be equal to 1, so we only need to consider the cases x = 2 and x = 3. At both of these, f (2) = f (3) = 2, so the sum of all distinct primes values taken on by f ( x ) is 2 .