The polynomials pn(x) are defined as follows: p0(x) = 1; pn+1'(x) = (n + 1) pn(x + 1), pn+1(0) = 0 for n ≥ 0. Factorize p100(1) into distinct primes.
Solution
Answer: (101)99.
We show that pn(x) = x (x + n)n-1 for n > 0. In fact, this is an easy induction. We have pn+1'(x) = (n + 1) (x + n + 1)n - n(n + 1)(x + n + 1)n-1. Integrating: pn+1(x) = (x + n + 1)n+1 - (n + 1)(x + n + 1)n = x(x + n + 1)n.
© John Scholes
jscholes@kalva.demon.co.uk
7 Jan 2001