On which natural numbers do we have injectivity of Euler's totient function?

I was wondering about the Euler's function and I would like to know which are all the natural numbers $$m$$ such that:

$$\phi (m) \neq \phi (n) \forall n \in \mathbb N\setminus \{m\}$$

After putting some thought into it:

$$m$$ cannot be odd, because $$\phi(m) = \phi(2m)$$.

$$m$$ cannot be twice an odd number for the previous reason. So no prime numbers.

$$m$$ must be a multiple of $$4$$, but which ones?

We know that

$$m = p_1^{\alpha_1} \cdot ... \cdot p_n^{\alpha_n} \implies \phi(m) = p_1^{\alpha_1-1} \cdot ... \cdot p_n^{\alpha_n-1}(p_1-1)\cdot...\cdot(p_n-1)$$

EDIT: I think I got something, $$m$$ must be a composite number, let's say it is $$(4p) \cdot q$$ with $$gcd(4p,q)=1$$, so we have that

$$\phi (m) = \phi(4p) \cdot \phi (q)$$

but we know for a fact that there is some $$q_2 \in \mathbb N$$ such that $$\phi (q_2) = \phi (q)$$ if we could know for sure that $$\gcd(q_2,4p)=1$$, then we would have that $$\phi$$ is never injective for any natural number, as $$\phi (4p \cdot q)$$ would be the same as $$\phi (4p \cdot q_2)$$.

• I expect you mean $m \neq \phi(n), \forall n \in \Bbb{N}$ in your first display, because $\phi(m)$ is definitely $\phi(n)$ when $n = m$. Mar 4 at 2:42
• @EricTowers injectivity, I assumed implicitly that $m \neq n$. Gonna change that Mar 4 at 2:43
• Mar 4 at 2:45

We don't expect there to be any such $$m$$, and this is a famous unsolved problem. See here.