HMMT 二月 2015 · 冲刺赛 · 第 18 题

英文解析

18. [ 11 ] Let f : Z → Z be a function such that for any integers x, y , we have
    2 2 2 2
    f ( x − 3 y ) + f ( x + y ) = 2( x + y ) f ( x − y ) .
    Suppose that f ( n ) > 0 for all n > 0 and that f (2015) · f (2016) is a perfect square. Find the minimumpossible value of f (1) + f (2).
    Answer: 246 Plugging in − y in place of y in the equation and comparing the result with theoriginal equation gives
    ( x − y ) f ( x + y ) = ( x + y ) f ( x − y )
    This shows that whenever a, b ∈ Z − { 0 } with a ≡ b (mod 2), we havef ( a ) f ( b )
    a b=
    which implies that there are constants α = f (1) ∈ Z , β = f (2) ∈ Z for which f satisfies the > 0 > 0
    equation ( ∗ ):
    {
    n · α when 2 ∤ nf ( n ) =
    · β when 2 | nn
    4 2 22
    Therefore, f (2015) f (2016) = 2015 α · 1008 β = 2 · 3 · 5 · 7 · 13 · 31 αβ , so αβ = 5 · 7 · 13 · 31 · t forsome t ∈ Z . We claim that ( α, β, t ) = (5 · 31 , 7 · 13 , 1) is a triple which gives the minimum α + β . In > 0
    particular, we claim α + β ≥ 246.
    √
    √
    Consider the case t ≥ 2 first. We have, by AM-GM, α + β ≥ 2 · αβ ≥ 4 · 14105 > 246. Suppose
    2 2
    t = 1. We have α · β = 5 · 7 · 13 · 31. Because ( α + β ) − ( α − β ) = 4 αβ is fixed, we want to have αas close as β as possible. This happens when one of α, β is 5 · 31 and the other is 7 · 13. In this case,
    α + β = 91 + 155 = 246.
    Guts
    Finally, we note that the equality f (1) + f (2) = 246 can be attained. Consider f : Z → Z such thatf ( n ) = 91 n for every odd n ∈ Z and f ( n ) = n for every even n ∈ Z . It can be verified that f 155
    satisfies the condition in the problem and f (1) + f (2) = 246 as claimed.2
    n 2