# Find all $n \in \mathbb{Z^+}$ such that $\phi(n)=4$

I know that there is a similar post, but I 'm trying a different proof. I will write $$P$$ for 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 $$n=p_1^{n_1}\dotsb p_k^{n_k}\in \mathbb{Z}^+$$ be the factorisation of $$n$$ in to primes. Then $$\phi(n)=p_1^{n_1-1} \dotsb p_k^{n_k-1}(p_1-1) \dotsb (p_k-1)=4.$$

So, for any $$i \in \{1,2,\cdots,k\}$$ we have $$p_i-1|4$$. 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.

• Look good to me. Jan 2, 2017 at 20:26
• Yes, its right! ( only a typo: n_2=n_3=1) Jan 2, 2017 at 20:29
• Thank you for your answers. I fixed the typo. Jan 2, 2017 at 20:35

• Thank you for your answer. Can we work similarly to find all $n\in \mathbb{Z} ^+: \phi(n)=x$ for some $x>4$? Jan 2, 2017 at 20:46