# Inverse of Euler's phi (totient) function

How do you calculate the inverse of Euler's totient function?

For example, what is $\phi^{-1}(12)$?

I'm confused as my lecture notes give no explanation, they just state that e.g. $\phi^{-1}(2)=-1$ and $\phi^{-1}(3)=-2$.

When I search for Inverse of Euler's totient function I get answers for how to solve $\phi(n)=k$, which is not what I'm looking for, so maybe I'm asking the wrong question?

I'm more confused by the fact the answer that I'm given is $\phi^{-1}(12)=2$ because if $\phi^{-1}(2)=-1$ and $\phi^{-1}(3)=-2$, assuming $\phi^{-1}$ is multiplicative, $\phi^{-1}(12)=(\phi^{-1}(2))^2\phi^{-1}(3)=(-1)^2(-2)=-2.$

• The solutions to $\phi(n)=12$ are $13,21,26,28,36,42$. But if this is not what you mean by inverse, what do you mean? – Henry May 31 '17 at 21:00
• Well, yes. They are equivalent problems to solve. If you know that $f(x)=y$ then we know that $f^{-1}(y)=x$. In your case, solving $\phi^{-1}(12)=x$ is equivalent to solving $\phi(x) =12$ – Brevan Ellefsen May 31 '17 at 21:03
• Did you get your notes from a lecturer as part of a course? It would be best to ask that person for clarification. – Matthew Conroy May 31 '17 at 21:13

They meant the Dirichlet inverse of $\phi(n)$, which is a multiplicative function : $$gcd(n,m) = 1 \quad\implies\quad \phi(nm) = \phi(n)\phi(m)$$ From $\phi(p^k)= p^{k-1}(p-1)$ we obtain $$\sum_{d | n} \phi(d) = n, \qquad \phi(n) = \sum_{d | n} \mu(d) \frac{n}{d}$$ where $\mu(n)$ is the Möbius function. thus $$\phi^{-1}(n) = \sum_{d | n} d \mu(d)$$ Which is multiplicative too. Therefore
$$\phi^{-1}(3) = 1-3 = -2, \qquad \phi^{-1}(2) = 1-2 = -1$$
$$\phi^{-1}(12) = \phi^{-1}(3)\phi^{-1}(4)= (1-3)(1-2) = 2$$
• Excellent, thank you. That clears things up. Please could you clarify why $\phi^{-1}(12)= 2$ and not $-2$? I would have thought $\phi^{-1}(6)= 2$ but may be mistaken. – math_apprentice Jun 1 '17 at 9:54
• @math_apprentice I wrote how I get $\phi^{-1}(12)=2$. It is not true that $\phi^{-1}(12) = \phi^{-1}(3) (\phi^{-1}(2))^2$ : $\phi^{-1}(n)$ is not completely multiplicative. – reuns Jun 1 '17 at 10:33