HMMT 二月 2015 · 冲刺赛 · 第 13 题

HMMT February 2015 — Guts Round — Problem 13

[ 8 ] Let P ( x ) = x + ax + bx + 2015 be a polynomial all of whose roots are integers. Given that P ( x ) ≥ 0 for all x ≥ 0, find the sum of all possible values of P ( − 1).

解析

[ 8 ] Let P ( x ) = x + ax + bx + 2015 be a polynomial all of whose roots are integers. Given that P ( x ) ≥ 0 for all x ≥ 0, find the sum of all possible values of P ( − 1). Answer: 9496 Since all the roots of P ( x ) are integers, we can factor it as P ( x ) = ( x − r )( x − s )( x − t ) for integers r, s, t . By Viete’s formula, the product of the roots is rst = − 2015, so we need three integers to multiply to − 2015. P ( x ) cannot have two distinct positive roots u, v since otherwise, P ( x ) would be negative at least in some infinitesimal region x < u or x > v , or P ( x ) < 0 for u < x < v . Thus, in order to have two Guts positive roots, we must have a double root. Since 2015 = 5 × 13 × 31, the only positive double root is 2 a perfect square factor of 2015, which is at x = 1, giving us a possibility of P ( x ) = ( x − 1) ( x + 2015). Now we can consider when P ( x ) only has negative roots. The possible unordered triplets are ( − 1 , − 1 , − 2015) , ( − 1 , − 5 , − 4 ( − 1 , − 31 , − 65) , ( − 5 , − 13 , − 31), which yield the polynomials 2 ( x + 1) ( x + 2015) , ( x + 1)( x + 5)( x + 403), ( x + 1)( x + 13)( x + 155) , ( x + 1)( x + 31)( x + 65) , ( x + 5)( x + 13)( x + 31), respectively. Noticing that P ( − 1) = 0 for four of these polynomials, we see that the nonzero values are P ( − 1) = 2 ( − 1 − 1) (2014) , (5 − 1)(13 − 1)(31 − 1), which sum to 8056 + 1440 = 9496.