Find all functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $\varphi(f(x+y))=\varphi(f(x))+\varphi(f(y))\quad\forall x,y\in\mathbb{N}$ I was studying about Cauchy's Functional Equation, and suddenly I had the following question in my mind:
Find all functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $$\varphi(f(x+y))=\varphi(f(x))+\varphi(f(y))\quad\forall x,y\in\mathbb{N}$$
Here, $\varphi(n)$ is Euler's Totient Function.
I first tried to find some trivial cases, and intuitively I felt that if $f(x)=c$, where $c$ is a constant, then $c$ must be a perfect power of a prime. And, in this case, the only trivial solution I found was $f(x)=2^k$ where $k$ is a positive integer. 
I really have no idea whether other solutions do exist or not.
 A: If $0$ is in $\mathbb{N}$ with the convention $\varphi(0)=0$, then there is a unique solution $f(n)=0$ for every $n\in\mathbb{N}$.  From now on, I assume that $\mathbb{N}$ is defined to be $\{1,2,3,\ldots\}$.  I claim that there does not exist such a function $f$.
Suppose on the contrary that $f$ exists.  Let $g:=\varphi\circ f$.  Then, $g$ satisfies Cauchy's functional equation.  That is, there exists $c\in\mathbb{N}$ such that $g(n)=cn$ for all $n\in\mathbb{N}$.  
Let $d_1,d_2,\ldots,d_k$ be all natural numbers that divide $c$.  Pick pairwise distinct primes $p_1,p_2,\ldots,p_{2k}$ that do not divide $c$.  Consider the system of congruences
$$d_ix\equiv-1\pmod{p_{2i-1}p_{2i}}$$
for $i=1,2,\ldots,k$.  This congruence has a unique solution
$$x\equiv x_0\pmod{P}\,,$$
where $P:=p_1p_2\cdots p_{2k}$ and $x_0\in\mathbb{Z}_{>0}$.  By Dirichtlet's Theorem, there exists a prime natural number $p>c$ such that $p\equiv x_0\pmod{P}$, noting that $\gcd(x_0,P)=1$.  It follows that the equation
$$\varphi(N)=cp$$
has no solution $N\in\mathbb{N}$.  Thus, $\varphi\big(f(p)\big)=g(p)=cp$ is impossible.
