A function $f$ maps from the positive integers to the positive integers, with the following properties: $$f(ab)=f(a)f(b)$$ where $a$ and $b$ are coprime, and $$f(p+q)=f(p)+f(q)$$ for all prime numbers $p$ and $q$. Prove that $f(2)=2, f(3)=3$, and $f(1999)=1999$.
It is simple enough to prove that $f(2)=2$ and $f(3)=3$, but I'm struggling with $f(1999)=1999$. I tried proving the general solution of $f(n)=n$ for all $n$ with a proof by contradiction: suppose $x$ is the smallest $x$ such that $f(x)<x$. I'm struggling find a way to show that no such $x$ exists if $x=p^m$ for $p$ a prime.
Can anyone help me finish off the $p^m$ case, or else show me another way of finding the answer? Computers and calculators are not allowed.