Fixed points of the arithmetic derivative 
Let $f: \mathbb{N} \mapsto \mathbb{N}$ be a function such that: 
i) $f(1)=0$. 
ii) $f(p)=1,  \forall p\in\mathbb{P}$.
iii) $f(ab)=af(b)+bf(a)$.  
Solve for $n$, such that $f(n)=n$.

Solution: Note that the first condition follows immediately and was not required to be given in the question. It can be proved by putting $a=b=1$.
Now, put $a=b=p$, where $p$ is a prime. We get, $f(p^2)=2pf(p)$.
Lemma: $f(p^n)=np^{n-1}$, where $p$ is a prime.
Proof: We proceed by induction. $$f(p^{n+1})=pf(p^n)+p^nf(p) \implies  f(p^{n+1})=pnp^{n-1}+p^n=(n+1)p^n.$$ 
Hence proved. $\square$
Claim: All numbers of the form $p^p$ where $p$ is a prime are solutions to the given equation.
Proof: Putting $n=p$ and using the above lemma, we get that $f(p^p)=p^p$. Hence proved. $\square$
Now, observe that $f(abc)=af(bc)+bcf(a)=ab+ac+bc$, where $a,b,c$ are all primes. Using induction, we can easily show that for prime $p_1,p_2,\ldots,p_n$, $f\left(p_1p_2\ldots p_n\right)$ is the cyclic sum of products $p_2\ldots p_n$. Note that this function behaves exactly like the derivative.
Claim: No other number except those of the form of $p^p$ for prime $p$ satisfy the equation.
Proof: Assume the contrary. Let $A$ be a solution not of the form $p^p$, where $p$ is a prime. Write the prime factorization of $A$. Let it be $p_1^{a_1}p_2^{a_2}\ldots p_n^{a_n}$.
We have assumed that $f(A)=A$. So, $$p_1^{a_1}f\left(p_2^{a_2}\ldots\right)+a_1p^{a_1-1}\left(p_2^{a_2}\ldots\right)=p_1^{a_1}p_2^{a_1}\ldots$$ This implies that $p_1f\left(p_2^{a_2}\ldots\right)+a_1\left(p_2^{a_2}\ldots\right)=p_1p_2^{a_2}\ldots$. RHS is divisible by $p_1$. LHS is divisible by $p_1$ iff $p$ divides $a_1$, as all other $p_i$'s are primes.
Thus, either $a_1=0$ or $a_1>p_1$
Now, $f(A)$ can also be written as $$a_1p_1^{a_1-1}\left(p_2^{a_2}\ldots\right)+a_2p_1^{a_2-1}\left(p_1^{a_1}\ldots\right)+\ldots=p_1^{a_1}p_2^{a_2}\ldots$$ Divide both sides by $A$ to get :
$$\boxed{\frac{a_1}{p_1}+\frac{a_2}{p_2}+\ldots+\frac{a_n}{p_n}=1.}$$ As all the terms in the RHS are positive, we get that $\boxed{a_i<p_i}$, $\forall i \in {1,\ldots,n}$. The only possibility left is that $a_1$ is $0$. It will proceed similarly. Thus, we get a contradiction and hence proved.
Thus, the solution set for this problem is $$\left\{a:a\in \mathbb{N};a=p^p, p\in\mathbb{P}\right\}.$$ $\blacksquare$
Is my solution correct?
 A: Almost correct. Having just $a_i<p_i$ doesn’t immediately solve the problem, since perhaps you could have something like this $$\frac{1}{6}+\frac{1}{3}+\frac{1}{2}=1.$$ However, by using that the $p_i$ are prime, you can easily conclude. Here’s the final part of the argument, somewhat rewritten.
Dividing your second-to-last equation by $p_1^{a_1-1}p_2^{a_2-1}\ldots p_n^{a_n-1}$, you get $$\sum_{\text{cyc}}a_1p_2p_3\ldots p_n=p_1p_2p_3\ldots p_n,$$ which implies $p_i\mid a_i$. Now, either $a_i=0$ or $a_i=p$, since your last equation precisely implies $a_i\leq p_i$. From here, we find that all $a_i$ must be zero, except for one that must be $p_i$. This is the same as saying that our original number was of the form $p^p$, and and we’re done.
A: 
Thus, either $a_1=0$ or $a_1>p_1$

This is slightly incorrect, you should replace that with : $a_1=0$ or $a_1\ge p_1$
The rest seems correct to me.
A: Near the end you divided by $A$. Doing that right away takes you to "logarithmic derivatives", so to speak:
Let $g(n)=\frac{f(n)}n$. Then 


*

*$g(1)=0$

*$g(p)=\frac 1p$

*$g(ab)=g(a)+g(b)$
Solve for $n$ with $g(n)=1$.
This gives you much faster that
$$g\left (\prod_p p^{a_p}\right) =\sum_p\frac{a_p}p.$$
This sum can only be $=1$ if is it in fact just one summand $\frac{a_p}{p}$ and $a_p=p$ for it. Indeed, in all other cases, multiplying the sum with the product of all but one of the finitely many primes occurring will turn all summands but one into integers, contradiction.
