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For the non-prime numbers $n$ up to $20$ I listed the number of units of $\mathbb{Z}/n\mathbb{Z}$, the number of square roots of unity of $\mathbb{Z}/n\mathbb{Z}$ and the number of divisors of $n$.

\begin{array}{c|c|c} \# & \text{units} & \text{square roots of unity} & \text{divisors} \\ \hline 4 & 2 & 2&3\\ \hline 6 & 2 & 2&4\\ \hline 8 & 4 &4 &4\\ \hline 9 & 6 & 2&3\\ \hline 10 & 4 & 2&4\\ \hline 12 & 4 & 4&6\\ \hline 14 & 6 & 2&4\\ \hline 15 & 8 & 4&4\\ \hline 16 & 8 & 4&5\\ \hline 18 & 6 & 2&6\\ \hline 20 & 8 & 4&6\\ \hline \end{array}

The number of units is just $\varphi(n)$ (Euler's totient function), i.e. the number of numbers $m < n$ that are relatively prime to $n$.

But I don't see the correlation pattern between the numbers: How do the numbers of units, square roots of unity and divisors relate for general $n$?

Or more generally: Are there expressions for the number of square roots of unity and for the number of divisors (like for the number of units = $\varphi(n)$)?

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  • $\begingroup$ What do you mean by a "root of unity"? Every unit is a root of unity. $\endgroup$ – Slade Dec 14 '18 at 12:18
  • $\begingroup$ I think your counts of roots of unity are not consistent. $\endgroup$ – Slade Dec 14 '18 at 12:22
  • $\begingroup$ I must have understood something wrong: in my understanding $3$ is a unit of $\mathbb{Z}/20\mathbb{Z}$ because $3\cdot 7 \equiv 1 \pmod {20}$, but it's not a root of unity because $3\cdot 3 \not \equiv 1 \pmod {20}$. The units of $\mathbb{Z}/20\mathbb{Z}$ are $1,3,7,9,11,13,17,19$ while the roots of unity are $1,9,11,19$. So every root of unity is a unit but not vice versa. Who of us is wrong? $\endgroup$ – Hans-Peter Stricker Dec 14 '18 at 12:34
  • $\begingroup$ Ah, I see: I forgot to mention that I meant square roots of unity. I edited the question. $\endgroup$ – Hans-Peter Stricker Dec 14 '18 at 13:02
  • $\begingroup$ See my updated answer. I believe it answers all of your questions. $\endgroup$ – Slade Dec 14 '18 at 14:05
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The number of divisors of $n$ is usually considered to be a primitive function in number theory, so there is most likely not a formula in terms of something simpler. However, if $n$ has prime factorization $\prod_i p_i^{a_i}$, we can write $d(n) = \prod_i (a_i + 1)$.

The total number of roots of unity in $\mathbb{Z}/n\mathbb{Z}$ is just $\varphi(n)$, since every unit is a root of unity of some order.

If $n$ is $2$, $4$, a power of an odd prime, or twice a power of an odd prime, then the units of $\mathbb{Z}/n\mathbb{Z}$ are a cyclic group of order $\varphi(n)$, so the number of generators is $\varphi(\varphi(n))$, which counts the number of primitive roots of unity. For all other $n$, the group of units is not cyclic and there are no primitive roots of unity.

EDIT: The question has changed to ask about square roots of unity.

In general, the structure of the group of units is known, as a decomposition into cyclic groups. See, for example, Wikipedia. From this, you can see how many square roots of unity there are.

I'll summarize the result. Write $n=2^a m$, where $m$ is an odd integer. Then, the number of square roots of unity in $\mathbb{Z}/n\mathbb{Z}$ is $2^{k+l}$, where $k$ is the number of prime factors of $m$, and $l$ is $0$ if $a\leq 1$, $1$ if $a=2$, and $2$ if $a\geq 3$.

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In ${\Bbb Z}_n$, $n\geq 2$, the number of units plus the number of zero divisors is $n-1$, where $0$ is not considered as a zero divisor. The reason is that for each $0\ne a\in{\Bbb Z}_n$, if $\gcd(a,n)=1$, then $a$ is a unit; otherwise, $a$ is a zero divisor.

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These functions are quite complicated due to being combinatorial in nature. Part of the Wikipedia page

https://en.m.wikipedia.org/wiki/Root_of_unity_modulo_n

on roots of unity modulo $n$ which is a good place to start is reproduced below.

enter image description here

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