MathDB

4

Part of 2013 Chile National Olympiad

Problems(1)

nf(2n + 1) = (2n + 1)(f(n) + n) , f(2n) = 2f(n)

Source: Chile Finals 2013 L2 p4

10/5/2022
Consider a function f defined on the positive integers that meets the following conditions: f(1)=1 ,  f(2n)=2f(n) ,  nf(2n+1)=(2n+1)(f(n)+n)f(1) = 1 \, , \,\, f(2n) = 2f(n) \, , \,\, nf(2n + 1) = (2n + 1)(f(n) + n) for all n≥1n \ge 1. a) Prove that f(n)f(n) is an integer for all nn. b) Find all positive integers mm less than 20132013 that satisfy the equation f(m)=2mf(m) = 2m.
functionalfunctional equationalgebranumber theory