On the arithmetic differential equation $n''=n'$ 
If $n'$ denotes the arithmetic derivative of non-negative integer $n$, and $n''=(n')'$, then solve the following equation $$n''=n'.$$

What I have found, you can read in one minute! I have tried to explain it very detailed so anyone, even with a little knowledge of elementary number theory (like me), can follow the steps. 
$n=0$ and $n=1$ are solutions. It is known that for a natural number $n>1$ with prime factorization of $\prod_{i=1}^{k} p_i^{a_i}$ arithmetic derivative is $$n'=n \sum_{i=1}^{k} \frac{a_i}{p_i}. \tag{1}$$
Let $m=n'$, then equation becomes $m'=m$. Let prime factorization of $m$ be $\prod_{j=1}^{l} q_j^{b_j}$. Then from equation $(1)$ we get $$\frac{b_1}{q_1}+ \frac{b_2}{q_2}+... + \frac{b_l}{q_l}=1. \tag{2}$$
This equation implies that $q_j \ge b_j$. Multiply both sides of the equation $(2)$ by $q_1 q_2 ... q_{l-1}$. It follows that $q_1 q_2 ... q_{l-1}\frac{b_l}{q_l}$ is an integer. Thus $q_l | b_l$. Hence $b_l \ge q_l$ and $b_l=q_l$. Subsequently $b_1=b_2=...=b_{l-1}=0$ and $m=q^q$ for some prime number $q$. 
Thus we have $n'=m=q^q$ and $n\sum_{i=1}^{k} \frac{a_i}{p_i}=q^q$ or $$\prod_{i=1}^{k} p_i^{a_i-1}\sum_{i=1}^{k} \left( p_1 p_2 ... p_k \frac{a_i}{p_i} \right)=q^q. \tag{3}$$ Notice that if $p_i \neq q$ is a prime divisor of $n$, then $a_i=1$. We claim that if $q$ is a prime divisor of $n$, then its the only one. If $q \mid n$ then $n$ is in the form $$n=p_1p_2...p_kq^a,$$ 
Where $\gcd(q, p_i)=1$. Now its easy to see from equation $(3)$ that $a \le q$ and dividing both sides of it by $q^{a-1}$ gives $$q\sum_{i=1}^{k} \left( \frac{p_1 p_2 ... p_k}{p_i} \right)+p_1 p_2 ... p_k a=q^{q-a+1}.$$Therefore, $q|a$, which leads to $a \ge q$ and $q=a$. Thus $$\sum_{i=1}^{k} \left( \frac{p_1 p_2 ... p_k}{p_i} \right)+p_1 p_2 ... p_k=1,$$Which is a contradiction and $n=q^q$. Thus $n=q^q$ is a solution to the original equation, where $q$ is a prime number.
If $q \nmid n$, then equation $(3)$ gives 
$$\sum_{i=1}^{k} \left( \frac{p_1 p_2 ... p_k}{p_i} \right)=q^q,$$Where I am stuck with.
Edit: According to @user49640 comment, there are some solutions of the form $n=2p$, where $p=q^q-2$ is a prime. For example for $q=7$ and $q=19$. See also @Thomas Andrews answer for an another solution not in the form $n=p^p$.
Look at this solution I found: 
$$(2\times17431\times147288828839626635378984008187404125879)'=29^{29}$$

In my proof below I will disregard the only even solution $2^2$, as it is an exception. With primes $a, b$ we have:
'$(ab) = a + b$
If $a, b$ are both odd or even, the sum is even; no solutions. Therefore exactly one of $a, b$ must be odd. Given that and $'(ab\cdot n) =\ '(ab)\cdot n + ab\cdot '(n)$. If $n$ is even both $ab$ and $n$ are even, making the sum even. That can't be a solution therefore $n$ must be odd.
So every solution with $2$ or more primes must be of the form $2\cdot n$ where $n$ is odd.
A number of the form $2n$ has arithmetic derivative $(2n)' = 2'n + 2n' = n + 2n'$. We're looking for an odd number $n$ such that
$$p^p = n + 2\ 'n = n  + 2n\sum_{i=1}^{k} \frac{a_i}{p_i}$$
If $n = q$ for prime $q$, we get:
$$p^p = q + 2\longrightarrow p^p-2 = q$$
If $n = qr$ for primes $q, r$ we get:
$$p^p = 3(q + r)\longrightarrow \text{no solutions due to p = 3}$$
If $n = qrs$ for primes $q, r, s$ we get:
$$p^p = qrs + 2(qr + qs + rs)$$
At this point I don't even know whether solutions exist.
