HMMT 二月 2007 · TEAM1 赛 · 第 1 题

英文解析

1. = P ( α − 3 α + 3 α ). Because P is nonconstant, it has at least one zero. Because P has finite degree,
   3 2
   there exist minimal and maximal roots of P . Writing α − 3 α + 3 α ≥ α ⇐⇒ α ( α − 1)( α − 2) ≥ 0,
   we see that the largest zero of P cannot exceed 2. Likewise, the smallest zero cannot be negative, so
   ′ 3 2
   all of the zeroes of P lie in [0 , 2]. Moreover, if α / ∈ { 0 , 1 , 2 } is a zero of P , then α = α − 3 α + 3 α isanother zero of P that lies strictly between α and 1. Because P has only finitely many zeroes, all ofp q rits zeroes must lie in { 0 , 1 , 2 } . Now write P ( x ) = kx ( x − 1) ( x − 2) for nonnegative integers p, q andr having a positive sum. The given equation becomes
   2 p q r 2 p 2 q 2 r 2
   k ( x + 1) x ( x − 1) ( x − x + 1) ( x − x ) ( x − x − 1) = P ( x + 1) P ( x − x + 1)
   3 3 p 3 q 3 r = P ( x + 1) = k ( x + 1) x ( x − 1) . ( ∗ )
   For the leading coefficients to agree, we require k = k . Because the leading coefficient is nonzero, P2
   must be monic. In ( ∗ ), r must be zero lest P ( x + 1) = 0 have complex roots. Then q must be zero as 3
   2 3
   well. For, if q is positive, then P ( x − x + 1) = 0 has 1 as a root while P ( x + 1) does not. Finally, thepremaining possibilities are P ( x ) = x for p an arbitrary positive integer. It is easily seen that thesepolynomials are satisfactory.