HMMT 二月 2007 · TEAM1 赛 · 第 1 题

HMMT February 2007 — TEAM1 Round — Problem 1

[ 15 ] Evaluate the functions φ ( n ) , σ ( n ) , and τ ( n ) for n = 12 , n = 2007 , and n = 2 . 2

解析

= 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 is another zero of P that lies strictly between α and 1. Because P has only finitely many zeroes, all of p q r its zeroes must lie in { 0 , 1 , 2 } . Now write P ( x ) = kx ( x − 1) ( x − 2) for nonnegative integers p, q and r 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) . ( ∗ ) 2 For the leading coefficients to agree, we require k = k . Because the leading coefficient is nonzero, P 3 must be monic. In ( ∗ ), r must be zero lest P ( x + 1) = 0 have complex roots. Then q must be zero as 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, the p remaining possibilities are P ( x ) = x for p an arbitrary positive integer. It is easily seen that these polynomials are satisfactory.