6
$\begingroup$

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 Fuction, we want to find all $n \in \mathbb{Z^+} : \phi(n)=4$.

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

So, $\forall i \in \{1,2,...,k \} \implies p_i-1|4 $ . And from this, we have that

$$p_i-1\in\{1,2,4 \} \implies p_i\in \{2,3,5\} \in P$$ Now, we can see the primes that $n$ containts: $n=2^{n_1}3^{n_2}5^{n_3}, \ n_1,n_2,n_3 \in \mathbb{Z}^+$. So, $$\phi(2^{n_1}3^{n_2}5^{n_3})=4 \iff \phi(2^{n_1})\phi(3^{n_2})\phi(5^{n_3})=4 \ (*)$$

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
    Look good to me. – 2017-01-02
  • 2
    Yes, its right! ( only a typo: n_2=n_3=1) – 2017-01-02
  • 0
    Thank you for your answers. I fixed the typo. – 2017-01-02

1 Answers 1

3

This seems to be completely correct to me.

  • 0
    Thank you for your answer. Can we work similarly to find all $n\in \mathbb{Z} ^+: \phi(n)=x$ for some $x>4$? – 2017-01-02
  • 2
    Yes, this technique is general, though the casework can be a difficult computation. – 2017-01-02
  • 3
    @Chris, for the general case, see http://math.stackexchange.com/questions/23947/how-to-solve-the-equation-phin-k. – 2017-01-02