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for n, exists p so that p|n and f(n) = f(n/p)-f(p)

Source: JBMO 2008 Shortlist N6

October 14, 2017
JBMOnumber theory

Problem Statement

Let f:N→Rf : N \to R be a function, satisfying the following condition: for every integer n>1n > 1, there exists a prime divisor pp of nn such that f(n)=f(np)−f(p)f(n) = f \big(\frac{n}{p}\big)-f(p). If f(22007)+f(32008)+f(52009)=2006f(2^{2007}) + f(3^{2008}) + f(5^{2009}) = 2006, determine the value of f(20072)+f(20083)+f(20095)f(2007^2) + f(2008^3) + f(2009^5)
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