Skip to main content

2013/14 AHSMC Part 1 Problem 13

Problem 13 of 16

Let f(x)=x2+x+1f(x) = x^2 + x + 1. Let nn be the positive integer such that f(n)=f(20)f(21)f(n) = f(20)f(21). Then the number of distinct prime divisors of nn is

Answer choices

Show solution

Solution

Note that f(m−1)f(m)=(m2−m+1)(m2+m+1)=m4+m2+1=f(m2)f(m-1)f(m) = (m^2 - m + 1)(m^2 + m + 1) = m^4 + m^2 + 1 = f(m^2). Substituting m=21m = 21 yields f(20)f(21)=f(441)f(20)f(21) = f(441). Therefore, n=441=3272n = 441 = 3^2 7^2. The answer is (B).

University of Alberta