PUMaC 2023 · 代数（B 组） · 第 5 题

PUMaC 2023 — Algebra (Division B) — Problem 5

Let P ( x ) be a polynomial with integer coefficients satisfying 2 2 ( x + 1) P ( x − 1) = ( x − 10 x + 26) P ( x ) for all real numbers x . Find the sum of all possible values of P (0) between 1 and 5000, inclusive. 4 3 2 2 2 2

解析

Let P ( x ) be a polynomial with integer coefficients satisfying 2 2 ( x + 1) P ( x − 1) = ( x − 10 x + 26) P ( x ) for all real numbers x . Find the sum of all possible values of P (0) between 1 and 5000, inclusive. Proposed by Sunay Joshi Answer: 5100 It is clear that the only constant solution is P ≡ 0, for which P (0) is not in the desired range. Therefore we assume P is nonconstant in what follows. Note that since the functional equation 2 holds for all reals, it holds for all complex numbers. Next, note that the roots of x + 1 are 2 ± i , while the roots of x − 10 x + 26 are ± i + 5. Plugging in x = i , we find P ( i ) = 0. Plugging in x = i + 1, we find P ( i + 1) = 0. Plugging in x = i + 2, we find P ( i + 3) = 0. Lastly, plugging in x = i + 3, we find P ( i + 4) = 0. Since P has real coefficients, its roots also include the conjugates − i, − i + 1 , − i + 2 , − i + 3 , − i + 4. Therefore P ( x ) can be written as 2 2 2 2 2 P ( x ) = Q ( x )( x + 1)( x − 2 x + 2)( x − 4 x + 5)( x − 6 x + 10)( x − 8 x + 17). We now claim that Q ( x ) is a nonzero constant. Plugging our expression for P into our functional equation, we find Q ( x − 1) = Q ( x ) for all x , hence Q ( x ) ≡ c ̸ = 0 is a constant. To finish, set x = 0 to find P ( x ) = 1700 c . The only integer multiples of 1700 between 1 and 5000 are 1700 and 3400, hence our answer is 1700 + 3400 = 5100. 4 3 2 2 2 2