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Determin all integers for which $(\mathbb{Z}/n\mathbb{Z})^*$ is cyclic of order 4.

What I did:

I know that $|(\mathbb{Z}/n\mathbb{Z})^*| = \varphi(n)$ the Euler function. but how to solve $\varphi(n)=4$?

Many thanks.

Conjecture
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    If $n$ has a prime factor $\ge7$ then $\phi(n)\ge 7-1=6$. – Angina Seng Apr 30 '18 at 18:26
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    There are not that many natural numbers $n$ with $\varphi(n)=4$. All such numbers are less than $20$. Furthermore, if $(\mathbb{Z}/n\mathbb{Z})^\times$ is cyclic, $n$ is equal to $1$, $2$, $4$, $p^k$, or $2p^k$ for some odd prime $p$ and for some integer $k>0$. – Batominovski Apr 30 '18 at 18:26
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    All $n$ with $\phi(n)=4$ have been determined here, for example. – Dietrich Burde Apr 30 '18 at 18:35

1 Answers1

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You already know that $\#(\mathbb{Z}/n\mathbb{Z})^\times=\phi(n)$, where $\phi$ is the Euler-totient function. In general, I am unaware if there is any known direct 'inverse' for the Euler-totient function (whatever that means since the function is not injective). But this is an area of some research, see this paper.

However, $\lim_{n \to \infty} \phi(n)=\infty$ and $\phi(n)$ is 'eventually' increasing (in that the function takes the same value finitely many times). So one would only need to test a few integers $n$. See the table here to see that $\phi(n)=4$ only for $n \in \{5,8,10,12\}$.

mathematics2x2life
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