I know that there is a similar post, but I 'm trying a different proof. Also I will define $P$ be the set of all positive prime numbers.

Question: If $\phi$ is Euler's Phi Function, we want to find all $n \in \mathbb{Z^+}$ such that $\phi(n)=4$.

Answer: Let $n=p_1^{n_1}\cdot...\cdot p_k^{n_k}\in \mathbb{Z}^+$ be the factorisation of $n$ in to primes. Then $$\phi(n)=p_1^{n_1-1} \dotsb p_k^{n_k-1}\cdot(p_1-1) \dotsb (p_k-1)=4.$$

So, $\forall i \in \{1,2,\cdots,k\} \implies p_i-1|4 $. Hence, $$p_i-1\in\{1,2,4\} \iff p_i\in \{2,3,5\} \subset P$$ Now, we can see the primes that $n$ contains: $n=2^{n_1}3^{n_2}5^{n_3}$, where $n_1,n_2,n_3 \in \mathbb{Z}^+$. So,

\begin{align*} \phi(2^{n_1}3^{n_2}5^{n_3})=4 \iff \phi(2^{n_1})\phi(3^{n_2})\phi(5^{n_3})=4 \tag{*} \end{align*}

The possible cases for $n_i$ are:

  • $n_1=1,2,3\implies \phi(2)=1,\phi(2^2)=2, \phi(2^3)=4$ respectively
  • $n_2=1 \implies \phi(3)=2$
  • $n_3=1 \implies \phi(5)=4$

All the posible combinations for the relation (*) are $\phi(5),\ \phi(5)\phi(2),\ \phi(3)\phi(2^2),\ \phi(2^3)$. So, $n \in \{5,10,12,8\}.$

Is this completely right?

Thank you.

  • 2
    $\begingroup$ Look good to me. $\endgroup$ – Igor Rivin Jan 2 '17 at 20:26
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    $\begingroup$ Yes, its right! ( only a typo: n_2=n_3=1) $\endgroup$ – Martín Vacas Vignolo Jan 2 '17 at 20:29
  • $\begingroup$ Thank you for your answers. I fixed the typo. $\endgroup$ – Chris Jan 2 '17 at 20:35

This seems to be completely correct to me.

  • $\begingroup$ Thank you for your answer. Can we work similarly to find all $n\in \mathbb{Z} ^+: \phi(n)=x$ for some $x>4$? $\endgroup$ – Chris Jan 2 '17 at 20:46
  • 2
    $\begingroup$ Yes, this technique is general, though the casework can be a difficult computation. $\endgroup$ – Stella Biderman Jan 2 '17 at 20:53
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    $\begingroup$ @Chris, for the general case, see math.stackexchange.com/questions/23947/…. $\endgroup$ – lhf Jan 2 '17 at 21:08

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