Problem:
Let P(x) be a nonzero polynomial such that (x−1)P(x+1)=(x+2)P(x) for every real x, and (P(2))2=P(3). Then P(27​)=nm​, where m and n are relatively prime positive integers. Find m+n.
Solution:
The given condition (x−1)P(x+1)=(x+2)P(x) implies that P(x) is divisible by x−1. From this it follows that x divides P(x+1), and the given condition then implies that x divides P(x). Substituting x−1 in place of x in the given condition yields (x−2)P(x)=(x+1)P(x−1), which implies that P(x) is divisible by x+1. Thus there is a polynomial L(x) such that P(x)=x(x−1)(x+1)L(x). Substituting this for P(x) in the given condition yields (x−1)(x+1)x(x+2)L(x+1)=(x+2)x(x−1)(x+1)L(x) or L(x+1)=L(x), implying that there is a constant c such that L(x)=c for all x. It follows that P(x)=c(x3−x). Now the condition (P(2))2=P(3) implies that c=32​, so P(27​)=4105​. The requested sum is 105+4=109​.
The problems on this page are the property of the MAA's American Mathematics Competitions