PUMaC 2009 · 代数（B 组） · 第 2 题

PUMaC 2009 — Algebra (Division B) — Problem 2

Let p ( x ) be the polynomial with leading coefficent 1 and rational coefficents, such that √ √ √ p ( 3 + 3 + 3 + . . . ) = 0 , and with the least degree among all such polynomials. Find p (5).

解析

Let p ( x ) be the polynomial with least degree, leading coefficent 1, rational coefficents, and √ √ √ p ( 3 + 3 + 3 + . . . ) = 0. Find p (5). √ √ √ Solution. 17. Let x = 3 + 3 + 3 + . . . . Using methods of calculus, one can show that x is a well defined real number. Here, however we do not worry about such questions but rather formally manipulate the expression with the square roots, trying to get some relation between √ √ 2 the powers of x . We have x = 3 + 3 + 3 + . . . = 3 + x hence x satisfies the equation 2 x − x − 3 = 0. From this it follows that x is not rational (since the roots of the quadratic equation above aren’t), and so it does not satisfy any equation of the type x − n = 0, with n 2 rational. Hence p ( x ) is at least of degree 2, so p ( x ) = x − x − 3. Then p (5) = 25 − 5 − 3 = 17.