Question

Let $P(x)$ be a nonzero polynomial such that $(x - 1)P(x + 1) = (x + 2)P(x)$ for every real $x,$ and $\left(P(2)\right)^2 = P(3).$ Then $P\left(\tfrac{7}{2}\right) = \tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

Solution:

Setting $x = 1$ in the identity gives $0 = 3P(1),$ so $P(1) = 0.$ Setting $x = 0$ gives $-P(1) = 2P(0),$ so $P(0) = 0,$ and setting $x = -2$ gives $-3P(-1) = 0,$ so $P(-1) = 0.$ Hence $P(x) = x(x - 1)(x + 1)L(x)$ for some polynomial $L.$

Substituting back, $(x - 1)\,(x + 1)x(x + 2)L(x + 1) = (x + 2)\,x(x - 1)(x + 1)L(x),$ so $L(x + 1) = L(x)$ for all real $x,$ which forces $L$ to be a constant $c.$ The normalization $\left(P(2)\right)^2 = P(3)$ reads $(6c)^2 = 24c,$ so $c = \frac{2}{3}.$

Then $P\left(\tfrac{7}{2}\right) = \frac{2}{3} \cdot \frac{7}{2} \cdot \frac{5}{2} \cdot \frac{9}{2} = \frac{105}{4},$ and $m + n = 105 + 4 = 109.$

2016 AIME I Problem 11

Solution:

Problem 11 in Other Years