2
$\begingroup$

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 $\phi (4p \cdot q_2)$.

$\endgroup$
3
  • $\begingroup$ 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 $n = m$. $\endgroup$ Mar 4 at 2:42
  • $\begingroup$ @EricTowers injectivity, I assumed implicitly that $m \neq n$. Gonna change that $\endgroup$ Mar 4 at 2:43
  • 1
    $\begingroup$ Dupe of math.stackexchange.com/questions/4186699/… . $\endgroup$ Mar 4 at 2:45

1 Answer 1

3
$\begingroup$

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

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .