f(n) is a function defined on the positive integers with positive integer values such that f(ab) = f(a)f(b) when a, b are relatively prime and f(p+q) = f(p)+f(q) for all primes p, q. Show that f(2) = 2, f(3) = 3 and f(1999) = 1999.
f(6) = f(3+3) = f(3) + f(3) and = f(23) = f(2)f(3). Hence f(2) = 2. Hence f(4) = f(2+2) = f(2) + f(2) = 4.
Now f(12) = f(4)f(3) = 4f(3). Also = f(5) + f(7) = f(5) + (f(5) + 2) = 2(f(3) + 2) + 2. Hence f(3) = 3.
We have 2002 = 271113, and f(1999) = f(2002) - 3. We have f(5) = 3 + 2 = 5, f(7) = 5 + 2 = 7. Also f(15) = 53 = 15, and f(15) = f(13) + 2, so f(13) = 13. Similarly f(11) = f(13) - 2 = 11. Hence f(2002) = 2002 and f(1999) = 1999.