Arithmetic Derivative of a number being equal to the number itself 
The arithmetic derivative $D(n)$ of a positive integer $n$ is defined via the following rules:

(a) $D(1)=0$,
(b) $D(p)=1$ for all primes $p$,
(c) $D(a b)=b D(a)+a D(b)$ for all positive integers $a$ and $b$.
Find the sum of all positive integers $n$ below $2019$ satisfying $D(n)=n$.
Through experimentation, I have found $4$ is such a number. For some prime $P$, $$\begin{aligned}
D\left(P^k\right)=D\left(D \cdot P^{k-1}\right) & =P \cdot D\left(P^{k-1}\right)+P^{k-1} \cdot D(P) \\
& =P \cdot D\left(P^{k-1}\right)+P^{k-1}
\end{aligned}$$
So we get a sort of recursive equation which I don't think looks useful at all.
If $n=p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_k^{\alpha_k}$, then $$D\left(p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_k^{\alpha_k}\right)=p_2^{\alpha_2} \ldots p_k^{\alpha_k} \cdot D\left(p_1^{\alpha_1}\right)+p_1^{\alpha_1} D\left(p_2^{\alpha_2} \ldots p_k^{\alpha_k}\right)$$
I can't figure out how to proceed.
 A: Claim: The solutions are exactly $p^p$ for prime $p$.
I won't give a rigorous proof but here is a general outline:
Lemma 1: For prime $p$, $D(p^k) = \frac{k}{p}*p^k$.
This can be shown with induction, starting with the base case of $D(p^1) = 1$.
Lemma 2: If $m = p_1^{k_1}p_2^{k_2}...p_n^{k_n}$, for distinct primes $p_i$, then $D(m) = \frac{k_1}{p_1}*m + \frac{k_2}{p_2}*m +\ldots+ \frac{k_n}{p_n}*m$.
This can also be shown by inducting on the number of distinct primes in the unique prime factorization, starting with the base case of $D(p_1^{k_1}) = \frac{k_1}{p_1}*p_1^{k_1}$ per Lemma 1.
Finally, let's go back to our condition that $D(m) = m = p_1^{k_1}p_2^{k_2}...p_n^{k_n}$. By Lemma 2, this is equivalent to
$$\frac{k_1}{p_1}*m + \frac{k_2}{p_2}*m +\ldots+ \frac{k_n}{p_n}*m = m$$
which simplifies to
$$\frac{k_1}{p_1} + \frac{k_2}{p_2} +\ldots+ \frac{k_n}{p_n} = 1$$
Thus $0 < k_i \leq p_i$. Clearing the denominators yields
$$k_1*p_2...p_n + k_2*p_1p_3...p_n +\ldots+ k_n*p_1...p_{n-1} = p_1p_2...p_n$$
Taking this expression $\mod{p_i}$ for any $i$, we see that the only term remaining is
$$k_i * p_1...p_{i-1}p_{i+1}...p_n = 0$$
implying $p_i \mid k_i$, which, with our bounds on $k_i$ means that $k_i = p_i$ and consequently $i=1$.
So $m = p^{p}$.
