Representation of Euler phi function Reading through Strayer's 'Elementary Number Theory', after the following theorems are proved:


*

*If $f$ is a multiplicative arithmetic function, then so is $F(n)=\sum_{d|n, d>0}f(d)$.

*Euler's phi function $\varphi(n)$ is multiplicative.
there is a remark that the proof that $\varphi(n)$ is multiplicative is longer than some other proofs of multiplicativity because $\varphi$ cannot be represented by $\sum_{d|n, d>0}f(d)$ for some multiplicative $f$.
Is this actually a theorem? Or just some offhand remark that $\varphi$ isn't easily  seen to have such a representation?
 A: 
Theorem: $\varphi$ cannot be represented by $\sum_{d|n, d>0}f(d)$ for some funtion $f$.

Proof: Assume that there exists a function $f$, such that $\varphi(n)=\sum_{d|n, d>0}f(d)$. Consider some special cases.
1) For $n=1$,  we have $\{d\in \mathbb{Z} : d\mid 1, d>0\}=\{1\}$ and $\varphi(1)=f(1)=1$
2) For a prime number $p$, we have $\{d\in \mathbb{Z} : d\mid p, d>0\}=\{1,p\}$. Thus $$\varphi(p)=f(1)+f(p)=1+f(p)=p-1.$$ So $f(p)=p-2$.
3) For $n=pq$, where $p,q$ are prime numbers, we have $\{d\in \mathbb{Z} : d\mid pq, d>0\}=\{1,p,q,pq\}$,
$$\varphi(pq)=\varphi(p) \varphi(q)=(p-1)(q-1)=pq-p-q+1,$$
and 
$$\sum_{d|n, d>0}f(d)=f(1)+f(p)+f(q)+f(pq)=f(pq)+p-2+q-2+1=f(pq)+p+q-3.$$
If $\varphi(pq)=\sum_{d|n, d>0}f(d)$ then
$$f(pq)+p+q-3=pq-p-q+1$$
and thus
$$f(pq)=pq-2p-2q+4=f(p)f(q).$$
4)For $n=p^k$ where $k\ge 1$, we have  $\{d\in \mathbb{Z} : d\mid p^k, d>0\}=\{p^i\mid i=0,\cdots,k\}$,
$$\varphi(p^k)=p^{k-1}(p-1)$$
and
$$\sum_{d|n, d>0}f(d)=\sum_{i=0}^{k}f(p^i).$$
If $\varphi(p^k)=\sum_{d|n, d>0}f(d)$ then
$$p^{k-1}(p-1)=\sum_{i=0}^{k}f(p^i)$$
and thus
$$p^{k}(p-1)=\sum_{i=0}^{k+1}f(p^i).$$
Hence $$f(p^{k+1})=p^k(p-1)-p^{k-1}(p-1)=p^{k-1}(p-1)^2.$$
To sum up, for any $n=p^k\ge 1$, we have 
$$f(p^k)=\begin{cases}
1, &k=0, \\
p-2, &k=1, \\
p^{k-2}(p-1)^2, &k\ge 2.
\end{cases}$$
Thans for @Francis Adams 's comment. The rest is to discuss the case $n=p^sq^t$ where $p\ne q$ are two primes. To be continued...
