HMMT 十一月 2016 · 团队赛 · 第 10 题

HMMT November 2016 — Team Round — Problem 10

[ 7 ] Determine the largest integer n such that there exist monic quadratic polynomials p ( x ), p ( x ), 1 2 p ( x ) with integer coefficients so that for all integers i ∈ [1 , n ] there exists some j ∈ [1 , 3] and m ∈ Z 3 such that p ( m ) = i . j

解析

[ 7 ] Determine the largest integer n such that there exist monic quadratic polynomials p ( x ), p ( x ), 1 2 p ( x ) with integer coefficients so that for all integers i ∈ [1 , n ] there exists some j ∈ [1 , 3] and m ∈ Z 3 such that p ( m ) = i . j Proposed by: Eshaan Nichani Answer: 9 2 2 2 The construction for n = 9 can be achieved with the polynomials x + x + 1, x + x + 2, and x + 5. 2 First we consider what kinds of polynomials we can have. Let p ( x ) = ( x + h ) + k . h is either an integer or half an integer. Let k = 0. If h is an integer then p ( x ) hits the perfect squares 0, 1, 4, 9, etc. If h is half an integer, then let k = 1 / 4. Then p ( x ) hits the product of two consecutive integers, i.e. 0, 2, 6, 12, etc. Assume there is a construction for n = 10. In both of the cases above, the most a polynomial can hit out of 10 is 4, in the 0 , 1 , 4 , 9 case. Thus p must hit 1 , 2 , 5 , 10, and p and p hit 3 integers each, out 1 2 3 of 3 , 4 , 6 , 7 , 8 , 9. The only ways we can hit 3 out of 7 consecutive integers is with the sequences 0 , 2 , 6 or 0 , 1 , 4. The only way a 0 , 2 , 6 works is if it hits 3, 5, and 9, which doesn’t work since 5 was hit by p . 2 Otherwise, p is 0 , 1 , 4, which doesn’t work as p hits 3, 4, and 7, and p must hit 6, 8, and 9, which 2 2 3 is impossible. Thus no construction for n = 10 exists.