\documentclass[12pt]{article}

\usepackage[table]{xcolor}
\usepackage{amsmath, amsthm}
\usepackage{amsfonts,amssymb}
\usepackage{hyperref}
\usepackage{authblk}
\usepackage[all]{xy}
\usepackage{enumitem}
\usepackage{stmaryrd}
\usepackage[makeroom]{cancel}
\usepackage{bbm}
\usepackage{fancyvrb}
\usepackage{xstring} % For the replace command
\usepackage{tikz}
%\usetikzlibrary{intersections}

\usepackage{xsim}

\xsimsetup{solution/print=true} %%%%%%

\usepackage{xeCJK}

\newtheorem*{rk}{Remark}


\usepackage{xlop}

\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\f}{\mathfrak{f}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\g}{\mathfrak{g}}
\newcommand{\K}{\mathbb{K}}
\renewcommand{\l}{\mathfrak{l}}
\newcommand{\p}{\mathfrak{p}}
\renewcommand{\P}{\mathfrak{P}}
\newcommand{\PP}{\mathbb{P}}
\newcommand{\lcm}{\operatorname{lcm}}
\newcommand{\MO}{\operatorname{MO}}

\topmargin -2.5cm
\textheight 25cm
\textwidth 15cm


\begin{document}

\begin{center}

{\Huge Introduction to number theory \\ Exercise sheet 2}

\vspace{0.5cm}

\url{https://www.maths.tcd.ie/~mascotn/teaching/2021/MAU22301/index.html}

\vspace{0.5cm}

Version: \today

\vspace{1cm}

Email your answers to \href{mailto:makindeo@tcd.ie}{makindeo@tcd.ie} by Friday October 22nd, 2PM.

The use of electronic calculators and computer algebra software is allowed.

\end{center}

\vspace{1cm}

\begin{exercise}[subtitle=No primitive roots (100 pts)]

In this exercise, whenever $x \in \Z$ and $n \in \N$, we write $x \bmod n$ for $x$ seen as an element of~$\Z/n\Z$. If furthermore $x$ and $n$ are coprime, so that $x \bmod n$ actually lies in $(\Z/n\Z)^\times$, we denote by $\MO(x \bmod n)$ the multiplicative order of $x \bmod n$.

We fix two distinct odd primes $p \neq q \in \N$, and we set $n=pq$.

\begin{enumerate}

\item (30 pts)  Let $x \in \N$ be coprime to $n$. Prove that
\[ \MO(x \bmod n) = \lcm\big( \MO(x \bmod p), \MO(x \bmod q)\big). \]

\emph{Hint:中國餘數定理.}

\medskip

Suppose from now on that $g \in \Z$ is a primitive root mod $n$ (in particular, $g$ is coprime to $n$).

\item (20 pts) Express $\MO(g \bmod n)$ in terms of $p$ and $q$.
\item (20 pts) Prove that $g$ is a primitive root both mod $p$ and mod $q$.

\emph{Hint: 相同的提示.}

\item (20 pts) Prove that $\MO(g \bmod p)$ and $\MO(g \bmod q)$ are both even.
\item (10 pts) Deduce that such a primitive root $g$ mod $n$ cannot exist.
\end{enumerate}
\end{exercise}

\begin{solution}
\begin{enumerate}
\item Let $k \in \N$. By Chinese remainders (valid since $p$ and $q$, being distinct primes, are coprime), we have that $x^k \equiv 1 \bmod n=pq$ iff. $x^k \equiv 1 \bmod p$ and $x^k \equiv 1 \bmod q$, which is equivalent to $k$ being simultaneously a multiple of $\MO(x \bmod p)$ and of $\MO(x \bmod q)$. The multiplicative order of $x \bmod n$, which is the smallest such $k$, is therefore $\lcm\big( \MO(x \bmod p), \MO(x \bmod q)\big)$.

\item Since $g$ is a primitive root mod $n$, we have $\MO(g \bmod n) = \phi(n) = \phi(pq) = \phi(p) \phi(q) = (p-1)(q-1)$ by multiplicativity of $\phi$ and the fact that $p$ and $q$ are prime.
\item Let $u \in (\Z/p\Z)^\times$ and $v \in (\Z/q\Z)^\times$. By Chinese remainders, there exists $w \in (\Z/n\Z)^\times$ such that $w \bmod p = u$ and $w \bmod q = v$. As $g$ is a primitive root mod $n$, this $w$ must be a power of $g$, say $w = g^k$ for some $k \in \Z$. But then $u=g^k \bmod p$ and $v=g^k \bmod q$. As $u$ (resp. $v$) was arbitrary, this shows that every element of $(\Z/p\Z)^\times$ (resp. $(\Z/q\Z)^\times$) is a power of $g$, which means that $g$ is a primitive root mod $p$ (resp. mod $q$).
\item Since $g$ is a primitive root mod $p$, we have $\MO(g \bmod p) = \phi(p) = p-1$ as $p$ is prime, and $p-1$ is even since $p$ is odd. Similarly, $\MO(g \bmod q) = \phi(q) = q-1$ is even.
\item By the previous questions,
\[ (p-1)(q-1) = \MO(x \bmod n) = \lcm\big( \MO(x \bmod p), \MO(x \bmod q)\big)\] \[ = \lcm(p-1,q-1) = \frac{(p-1)(q-1)}{\gcd(p-1,q-1)} \leq \frac{(p-1)(q-1)}2 < (p-1)(q-1) \]
since $\gcd(p-1,q-1) \geq 2$ as $p-1$ and $q-1$ are both even. This is absurd.
\end{enumerate}
\end{solution}

\bigskip

\textbf{This was the only mandatory exercise, that you must submit before the deadline. The following exercises are not mandatory; they are not worth any points, and you do not have to submit them. However, I highly recommend that you try to solve them for practice, and you are welcome to \href{mailto:mascotn@tcd.ie}{email me} if you have questions about them. The solutions will be made available with the solution to the mandatory exercise.}

%\bigskip
%
%\begin{center}
%\noindent\rule{12cm}{1.5pt}
%\end{center}

\bigskip

\begin{exercise}[subtitle=An inverse]
\begin{enumerate}
\item Use Euclid's algorithm to determine whether $47$ is invertible mod $111$, and to find its inverse if it is.
\item Solve the equation $47x \equiv 5 \bmod{111}$ in $\Z / 111 \Z$.
\end{enumerate}
\end{exercise}
\begin{solution}
\begin{enumerate}
\item We know that 47 is invertible mod 111 if and only if 47 and 111 are coprime. If they are, we need to look for $u$ and $v \in \Z$ such that $47u+111v=1$; indeed, $u$ will then be an inverse of $47 \bmod{111}$. To find $u$ and $v$, we either spot them directly\footnote{It can happen sometimes, but here there are no obvious candidates}, or we use the Euclidean algorithm. This algorithm will also tell us if the gcd of 47 and 111 is not 1, so let's apply it:
\begin{align*}
111 &= 2 \times 47 +17 \\
47 &= 2 \times 17 +13 \\
17 &= 13 +4 \\
13&=3 \times 4+1 \\
\end{align*}
So the gcd is 1, so 47 is invertible mod 111. To find it, we write
\begin{align*}
1 &= 13 - 3\times 4 \\
 &= 13 - 3(17-13) = 4 \times 13 - 3 \times 17 \\
&= 4(47 - 2 \times 17)-3 \times 17 = 4 \times 47 - 11 \times 17 \\
&= 4 \times 47 - 11 \times (111 - 2 \times 47) = 26 \times 47 - 11 \times 111,\\
\end{align*}
whence $26 \bmod 111$ is the inverse of $47 \bmod 111$.

\item Since $47$ is invertible mod $111$, the only solution is \[ x=5 \times 47^{-1} = 5 \times 26 = 130 \equiv 19 \bmod{111}. \]
\end{enumerate}
\end{solution}

\bigskip

\begin{exercise}[subtitle=More inverses]
\begin{enumerate}
\item Fix $n \in \N$, let $x \in (\Z/n\Z)^\times$ be invertible, and let $y \in \Z/n\Z$ be the inverse of~$x$. Prove that $x^2$, $-x$, and $y$ are also invertible, and find their inverses.
\item Find all the elements of $(\Z/15\Z)^\times$, and give the inverse of each of them. What is $\phi(15)$ ?

\emph{Hint: Use the previous question to save your effort!}
\end{enumerate}
\end{exercise}
\begin{solution}
\begin{enumerate}
\item We are tempted to write
\[ \frac1{x^2} = \left( \frac1x \right)^2 = y^2, \quad \frac1{-x} = - \frac1x = -y, \quad \frac1y = \frac1{1/x} = x, \]
and to conclude that $x^2$, $-x$, and $y$ are invertible, of respective inverses $y^2$, $-y$, and $x$. Let us check:
\[ x^2 y^2 = (xy)^2 = 1^2 = 1, \quad (-x)(-y) = xy = 1, \quad yx = xy =1 \]
since $xy = 1$ as $y$ is the inverse of $x$.
This proves that our intuition is correct.

\item The elements of $\Z/15\Z$ can be represented by $-7,-6, \cdots, 5,6,7$. Among these, the invertible ones are the ones that are coprime with 15, namely $-7,-4,-2,-1,1,2,4,7$. That's 8 of them, so $\phi(15)=8$.

 NB we could have determined $\phi(15)$ directly from the factorisation of $15$: since $15=3 \times 5$, we have $\phi(15) = 15(1-1/3)(1-1/5)=8$. But since the question was asking to list the elements of $(\Z/15\Z)^\times$, it was simpler to just count them!

Let us now match each element with its inverse, Obviously, 1 and $-1$ are their own inverses. Also, the inverse of $2$ is $8=-7$ since $2 \cdot 8 = 16 =1$ in $\Z/15\Z$. By the previous question, we immediately get that the inverses of $4$, $-2$ and $-7$ are respectively $(-7)^2=49=4$, $--7=7$, and $2$. It is now easy to complete the following table:

\[ \begin{array}{l|cccccccc}
x & -7 & -4 & -2 & -1 & 1 & 2 & 4 & 7  \\
\hline
x^{-1} & 2 & -4 & 7 & -1 & 1 & -7 & 4 & -2 \\
\end{array} \]

\end{enumerate}
\end{solution}

\bigskip

\begin{exercise}[subtitle=A system of congruences]
Find an integer $x \in \Z$ such that $x \equiv 12 \bmod 7$ and $x \equiv 7 \bmod{12}$.
\end{exercise}
\begin{solution}
 This is Chinese remainders: as 12 and 7 are coprime, we have a 1:1 correspondence
\[ \Z/84 \Z \longleftrightarrow \Z / 7 \Z \times \Z/ 12 \Z,  \tag{中}\label{CRT} \]
and we are looking for a pre-image of $(12 \bmod 7, 7 \bmod 12)$ under this correspondence.

Let us start by finding $u$ and $v$ such that $7u+12v=1$. Either we spot them right away, or we use the Euclidean algorithm:
\begin{align*}
12 &= 7 +5 \\
7 &= 5 +2 \\
5 &= 2 \times 2 + 1 \\
\end{align*}
whence
\[ 1 = 5 - 2 \times 2 = 5 - 2 \times (7-5) = 3 \times 5 - 2 \times 7 = 3 (12-7) - 2 \times 7 = 3 \times 12 - 5 \times 7. \]
So, under the correspondence \eqref{CRT}, $3 \times 12 = 36$ is a preimage of $(1 \bmod 7, 0 \bmod 12)$, and $-5 \times 7 = -35$ is a preimage of $(0 \bmod 7, 1 \bmod 12)$. As a result, since
\begin{align*} (12 \bmod 7, 7 \bmod 12) &= (5 \bmod 7, 7 \bmod 12) \\
&= 5 \times (1 \bmod 7, 0 \bmod 12) + 7 \times (0 \bmod 7, 1 \bmod 12),
\end{align*}
a preimage for $(12 \bmod 7, 7 \bmod 12)$ is $x = 5 \times 36 + 7 \times -35 = -65$.

\emph{Remark: Of course, any integer congruent to $-65 \bmod{84}$ works (for instance, 19). In fact, the Chinese remainder theorem tells us that the solutions are exactly the numbers that are congruent to $-65 \bmod{84}$; no more, no less.}

\end{solution}

\bigskip

\begin{exercise}[subtitle=Primes mod 6]
\begin{enumerate}
\item Let $p$ be a prime number which is neither 2 nor 3. Prove that either $p \equiv 1 \bmod 6$ or $p \equiv -1 \bmod 6$.
\item Prove that there are infinitely many primes $p$ such that $p \equiv -1 \bmod 6$.

\emph{Hint: Suppose on the contrary that there are finitely many, say $p_1, \cdots, p_k$. Let $N=6p_1 \cdots p_k-1$, and consider a prime divisor of $N$.}
\item Why does the same proof fail to show that there are infinitely may primes $p$ such that $p \equiv 1 \bmod 6$?
\item \emph{Dirichlet's theorem on primes in arithmetic progressions}, which is way beyond the scope of this course, states that for all coprime positive integers $a$ and $b$, there are infinitely many primes $p$ such that $p \equiv a \bmod b$; in particular, there are in fact infinitely many primes $p$ such that $p \equiv 1 \bmod 6$. Why, in the statement of this theorem, is it necessary to assume that $a$ and $b$ are coprime?
\end{enumerate}
\end{exercise}
\begin{solution}
\begin{enumerate}
\item If $p$ is neither $2$ nor $3$, then $p \nmid 6$, so $p$ and $6$ are coprime. But the only invertibles in $\Z/6\Z$ are $\pm 1$ (either see it by inspection of all 6 elements, or use the fact that $\phi(6)=2$), so we must have $p \equiv \pm 1 \bmod 6$.

\item Suppose that $p_1, \cdots, p_k$ are the only such primes, and let $N=6p_1 \cdots p_k-1$. Clearly, neither 2 nor 3 divide $N$ (since $N \equiv -1 \bmod 6$), so the primes dividing $N$ are all $\equiv \pm1 \bmod 6$ by the previous question. If they were all $\equiv +1 \bmod 6$, then $N$, their product, would also be $\equiv +1 \bmod 6$, which is not the case. So at least one of them, say $p$, is $\equiv -1 \bmod 6$. But this $p$ cannot be one of $p_1,\cdots,p_k$, else we would have $p \mid (6 p_1 \cdots p_k - N)=1$. We therefore have reached a contradiction.

\item We could suppose by contradiction that $p_1, \cdots, p_k$ are the only primes $\equiv 1 \bmod 6$, and consider a prime divisor of $N=6 p_1 \cdots p_k - 1$ (or $N=6 p_1 \cdots p_k + 1$). Such a prime could not be any of the $p_i$ for the same reason as above, but there is no reason why it would have to be $\equiv 1 \bmod 6$; indeed, nothing prevents the divisors of $N$ from being all $\equiv -1 \bmod 6$. So we are stuck.


\item If $p \equiv a \bmod b$, then $p = bx+a$ for some $x \in \Z$, so $\gcd(a,b) \mid p$; and obviously, if $\gcd(a,b) >1$, this can only happen for at most one prime $p$ (exactly one if $\gcd(a,b)=p$ is itself prime, and none else).
\end{enumerate}
\end{solution}

\bigskip

\begin{exercise}[subtitle=Euler]
Compute $\phi(2020)$ and $\phi(2021)$.

\noindent \emph{Hint: $45^2=2025$.}

\end{exercise}
\begin{solution}
In order to use the formulas for $\phi$, we must factor $2020$. We see easily that $2020=2^2 \times 5 \times 101$; and since neither 2 nor 3 nor 5 nor 7 divides 101, we conclude that 101 is prime since it is not divisible by any prime $\leq \sqrt{101} < 11$. So we can conclude that
\[ \phi(2020) = \phi(2^2 \times 5 \times 101) = \phi(2^2) \times \phi(5) \times \phi(101) = 2 \times 4 \times 100 = 800 \]
as $\phi(p^v)=p^{v-1}(p-1)$ for $p$ prime; in particular $\phi(p)=p-1$ (this reflects that for $p$ prime, only $0$ is not invertible in $\Z/p\Z$).

Similarly, in order to evaluate $\phi(2021)$, we must fact $2021$; but this time, the factorisation is not so clear. The hint shows the way out: $2021 = 2025-4=45^2-2^2 = (45-2)(45+2) = 43 \times 47$, which is the complete factorisation since one easily checks that 43 and 47 are both prime (it is enough to rule out divisibility by the primes up to $7$). It follows by multiplicativity that $\phi(2021) = \phi(43) \phi(47) = 42 \times 46 = 1932$.
\end{solution}

\bigskip

\begin{exercise}[subtitle=Factoring with Euler]
Let $n=93011$. Given than $n=pq$ is a product of two distinct primes, and that $\phi(n)=92400$, find $p$ and $q$.

\emph{Hint: Consider the polynomial $(x-p)(x-q)$.}
\end{exercise}
\begin{solution}
We observe that $p$ and $q$ are the roots of the polynomial $f(x)=(x-p)(x-q) = x^2-(p+q)x+n$. We already know $n$; besides, $\phi(n) = (p-1)(q-1) = pq-p-q+1$, so $p+q = n+1 - \phi(n) = 612$. Therefore $p$ and $q$ are the solutions of
\[ x^2-612x+93011. \]
We compute $\Delta = 2500 = 50^2$, so
\[ p,q = \frac{612 \pm 50}{2} = 281, 331. \]

\emph{Remark: This shows that in general, computing $\phi(n)$ is as hard as factoring $n$; in other words, if we want to evaluate $\phi(n)$, there is no way around factoring $n$.}
\end{solution}

\bigskip

\begin{exercise}[subtitle=A really large number]

What is the remainder of $22^{7^{2020}}$ when divided by $17$?

\noindent \emph{Just to be clear: $a^{b^{c}}$ means $a^{(b^{c})}$, as opposed to $(a^b)^c = a^{bc}$.}

\end{exercise}
\begin{solution}
We want to reduce $22^{7^{2020}}$ mod 17.

First, we have $22 \equiv 5 \bmod 17$, so $22^{7^{2020}} \equiv 5^{7^{2020}} \bmod 17$. 

Next, by Fermat we know that since $\gcd(5,17)=1$, $5^{\phi(17)} \equiv 1 \bmod 17$, so the value of $5^m \bmod 17$ only depends on $m \bmod \phi(17)$. As $17$ is prime, we have $\phi(17)=17-1=16$, so we want to determine $7^{2020} \bmod 16$.

Again by Fermat, since $\gcd(7,16)=1$, we have $7^{\phi(16)} \equiv 1 \bmod 16$, so $7^m \bmod 16$ only depends on $m \bmod \phi(16)$. And as $16=2^4$, we have $\phi(16)=2^{4-1}(2-1)=8$.

So we now want to determine $2020 \bmod 8$. That one is easy, we find that $2020 \equiv 4 \bmod 8$.

So $7^{2020} \equiv 7^4 \bmod 16$. In order to compute that, we begin by $7^2=49 \equiv 1 \bmod 16$, whence $7^4 = (7^2)^2 \equiv 1^2 = 1 \bmod 16$.

Thus $5^{7^{2020}} \equiv 5^1 = 5 \bmod 17$. Conclusion:
\[ 22^{7^{2020}} \equiv 5 \bmod 17, \]
so the remainder is 5.
\end{solution}

\bigskip

\begin{exercise}[subtitle=Inverse Euler]
The goal of this exercise is to find all integers $n \in \N$ such that $\phi(n)=4$.
\begin{enumerate}
\item Prove that if $p \in \N$ is a prime and $v \in \N$ is such that $p^v \mid n$, then $(p-1)p^{v-1} \mid \phi(n)$.

\emph{Hint: When $p$ is prime, what is $\phi(p^v)$?}

\item Using the previous question, prove that if $\phi(n)=4$, then $n$ cannot be divisible by a prime $p \ge 7$. Also prove that $3^2, \ 5^2 \nmid n$. 
\item Find all $n$ such that $\phi(n)=4$.

\emph{Hint: Think in terms of the factorisation of $n$. You should find that there are four such $n$ --- but you are required to prove this as part of this question!}
\end{enumerate}
\end{exercise}
\begin{solution}
\begin{enumerate}
\item Let $p^{v'}$ be the exact power of $p$ that divides $n$, so that $v' = v_p(n) \ge v$ and we may write $n=p^{v'} m$ with $\gcd(p^{v'},m)=1$. Since $\phi$ is multiplicative, $\phi(n) = \phi(p^{v'}) \phi(m)$ is divisible by $\phi(p^{v'}) = (p-1)p^{v'-1}$, whence also by $(p-1)p^{v-1}$ since $v'-1 \ge v-1$.

\item Let $p^v \mid n$, then by the previous question $(p-1) p^{v-1} \mid \phi(n) = 4$, so in particular $p-1 \le 4$, i.e. $p \le 5$. Besides, if $v =2$, then we get $(p-1)p \mid 4$, so $p \mid 4$, so $p=2$. In other words, we cannot have $p^2 \mid n$ unless $p=2$.

\item According to the previous question, the only possible prime factors of $n$ are 2, 3, and 5; and 3 and 5 cannot be repeated factors. So we must have $n=2^a 3^b 5^c$ for $a \ge 0$ and $b,c = 0$ or $1$. Since $\phi$ is multiplicative and since powers of distinct primes are coprime, we must then have $4 = \phi(n) = \phi(2^a) \phi(3^b) \phi(5^c)$. Now observe that
\[ \phi(2^a) = \left\{ \begin{array}{ll} 1, & \text{ if } a = 0, \\ 2^{a-1}, & \text{ if } a \ge 1, \end{array} \right. \]
\[ \phi(3^b) = \left\{ \begin{array}{ll} 1, & \text{ if } b = 0, \\ 2, & \text{ if } b=1, \end{array} \right. \]
and
\[ \phi(5^c) = \left\{ \begin{array}{ll} 1, & \text{ if } c = 0, \\ 4, & \text{ if } c= 1. \end{array} \right. \]
These observations clearly imply that the only combinations of $(a,b,c)$ for which $\phi(n)=4$ correspond to $n=2^3$, $2^2 3$, $5$, or $2 \cdot 5$.

In conclusion, we have exactly four solutions: $n=5$ or $8$ or $10$ or $12$. 
\end{enumerate}
\end{solution}

\bigskip

\begin{exercise}[subtitle=More inverse Eulers]

\emph{This exercise is a bit more difficult, but still doable. The questions are independent from each other.}

\begin{enumerate}
\item Using the fact that $2018 = 2 \times 1009$ and that $1009$ is prime, prove that there is no $n \in \N$ such that $\phi(n)=2018$.

\emph{Hint: Suppose $1009$ is a factor of $\phi(n)$. Where can this factor come from?}
\item Prove that for all $m \in \N$, there are at most finitely many\footnote{This means either finitely many or none.} $n \in \N$ such that $\phi(n) = m$.

\emph{Hint: Try to bound the prime factors of $n$ in terms of $m$.}

\item Prove that $\phi(n)$ is even for all $n \ge 3$.

\emph{Hint: Start with the case when $n$ is a prime power.}
\end{enumerate}
\end{exercise}
\begin{solution}
\begin{enumerate}
\item Suppose $n$ is such that $\phi(n) = 2018 = 2 \cdot 1009$. Factoring $n = \prod p_i^{v_i}$ with the $p_i$ distinct primes and the $v_i \ge 1$, we get that
\[ 1009 \mid 2018 = \phi(n) = \prod \phi(p_i^{v_i}) \]
since $\phi$ is multiplicative. Since $1009$ is prime, Euclid's lemma tells us that 1009 must divide one of the factors $\phi(p_i^{v_i}) = (p_i-1)p_i^{v_i-1}$, and then another use of Euclid gives us $1009 \mid (p_i-1)$ or $1009 \mid p_i^{v_i-1}$. We are now going to prove that neither of these can happen.

Indeed, if $1009 \mid (p_i-1)$, then $p_i \equiv 1 \bmod 1009$, so $p_i=1+1009x$ for some integer $x \ge 0$. Actually, $1+1009x$ is not prime for $x=0$ nor for $x=1$, and not for $x=2$ either since $1+1009 \cdot 2 = 2019$ si divisible by 3 (sum of digits). So we must have\footnote{Actually, a computer search shows that the smallest $x$ such that $1+1009x$ is prime is $x=10$.} $x \ge 3$. But then $p_i > 3 \cdot 2018$, which is absurd since the fact that $(p_i -1) \mid \phi(p_i^{v_i}) \mid \phi(n) = 2018$ implies that $p_i-1 \le 2018$.

And if $1009 \mid p_i^{v_i-1}$, then by uniqueness of factorisation we must have $p_i=1009$ and $v_i=2$; but this is absurd since in this case $\phi(p_i^{v_i}) = (p_i-1)p_i = 1008 \cdot 1009$ is clearly way too large to divide $\phi(n)=2018$.

So we have reached a contradiction, which means that no such $n$ exists.

\item Fix $m$, and let $n$ be such that $\phi(n)=m$. Let $p$ be a prime factor of $n$, and let $v=v_p(n)$ be the corresponding exponent, so that $n = p^v q$ with $q \in \N$ coprime to $p$. Then $m=\phi(n) = \phi(p^v) \phi(q)$ is divisible by $\phi(p^v)=(p-1)p^{v-1}$, so $(p-1)p^{v-1} \le m$. This forces $p -1 \le m$, i.e. $p \le m+1$. and also shows that $v$ cannot be arbitrarily large.

So $n$ has finitely many possible prime factors (the primes $\le m+1$), and for each such prime $p$, $v_p(n)$ is bounded. So $n$ has finitely many possible factorisations, i.e. there are finitely many candidates for such $n$.

\emph{Remark: This is enough to prove that there are at most finitely many $n$ such that $\phi(n)=m$, but of course, we have been rather sloppy, and as result we have probably grossly overestimated the number of such $n$. Estimating this number precisely would be much more complicated, cf. the previous exercise for instance!}

\item Since $n \ge 3$, $n \neq 1$, so $n$ has at least one prime factor $p$. Write again $n=p^v q$, where $v=v_p(n)$ so that $\gcd(p,q)=1$. Then $(p-1)p^{v-1} = \phi(p^v) \mid \phi(n)$.

Now clearly $p-1$ is even, except if $p=2$. So we are done if we can take $p \neq 2$. In other words, the only case for which we have not yet proved that $\phi(n0$ is even is the case where $n=2^v$ is a power of 2. But since $n \ge 3$, we have $v \ge 2$, so $\phi(n) = \phi(2^v) = 2^{v-1}$ is again even as $v-1 > 0$.

\emph{Remark: If you know group theory, there is a shorter way to prove this: the element $-1$ has (multiplicative) order $2$ in the multiplicative group $(\Z/n\Z)^\times$, so this group must have even order by Lagrange. Note that this logic requires $n \geq 3$, in that it assumes that $-1 \neq 1$!}

\end{enumerate}
\end{solution}

\bigskip

\begin{exercise}[subtitle=Divisibility criteria]
Let $n \in \N$.

\begin{enumerate}
\item Prove that $n$ is congruent mod $9$ to the sum of its digits. In other words, if $n_0, n_1, n_2, \cdots$ are the digits of $n$ from right to left, so that
\[ n = n_0 + 10 n_1 + 100 n_2 + \cdots = \sum_i n_i 10^i, \]
then $n \equiv n_0 + n_1 + n_2 + \cdots \bmod 9$.
\item Prove that $9 \mid n$ iff. $9$ divides the sum of digits of $n$.
\item Find a similar criterion to test whether $11 \mid n$.
\end{enumerate}
\end{exercise}

\begin{solution}

The key is the following observation: if $n_0, n_1, n_2, \cdots$ are the digits of $n$ from right to left, so that
\[ n = n_0 + 10 n_1 + 100 n_2 + \cdots = \sum n_i 10^i, \]
and since $10 \equiv 1 \bmod 9$, we have
\[ n = \sum n_i 10^i \equiv \sum n_i 1^i = \sum n_i \bmod 9;\]
in other words, $n$ is congruent to the sum of its digits (mod 9). In particular, this congruence also holds mod 3 since $3 \mid 9$.

So we have
\[ 9 \mid n \Longleftrightarrow n \equiv 0 \bmod 9 \Longleftrightarrow \sum n_i \equiv 0 \bmod 9 \Longleftrightarrow 9 \mid \sum n_i  \]
and
\[ 3 \mid n \Longleftrightarrow n \equiv 0 \bmod 3 \Longleftrightarrow \sum n_i \equiv 0 \bmod 3 \Longleftrightarrow 3 \mid \sum n_i.  \]

For divisibility by 11, we notice that $10 \equiv -1 \bmod{11}$, so that
\[ n =  \sum n_i 10^i \equiv \sum n_i (-1)^i = n_0 - n_1 + n_2 - n_3 + \cdots \bmod{11}. \]
As a result, $11 | n$ if and only if the expression
\[ n_0 - n_1 + n_2 - n_3 + \cdots, \]
which we may call the \emph{alternate} sum of digits of $n$, is divisible by 11.

\end{solution}

\bigskip

\begin{exercise}[subtitle=A huge number!]
\emph{In this exercise, you may use the first question of the previous exercise: every integer is congruent mod $9$ to the sum of its digits.}

Let $A=4444^{4444}$, let $B$ be the sum of the digits of $A$, let $C$ be the sum of the digits of $B$, and finally let $D$ be the sum of the digits of $C$.
\begin{enumerate}
\item Compute $D \bmod 9$.
\item Prove that $D \le14$.

\emph{Hint: Start with the upper bound $A < 10000^{5000}=10^{20000}$.}
\item What is the exact value of $D$ (as opposed to just mod $9$)?
\end{enumerate}
\end{exercise}

\begin{solution}
\begin{enumerate}
\item Since every integer is congruent mod $9$ to the sum of its digits, we have $D \equiv C \equiv B \equiv A \bmod 9$, so we can just as well compute $A \bmod 9$.

Now $4444 \equiv 16 \equiv -2 \bmod 9$, so $A \equiv (-2)^{4444} \bmod 9$. Now $-2$ and $9$ are coprime, so by Fermat's little theorem we have $(-2)^{\phi(9)} \equiv 1 \bmod 9$. We have $\phi(9)=6$, so we can replace the exponent 4444 by anything congruent to it mod 6. Since $4444 \equiv 4 \bmod 6$, we deduce that $A \equiv (-2)^4 = 16 \equiv 7 \bmod{9}$. 

\item We are going to estimate roughly the size of $D$. First of all, we have \[A < 10000^{5000} = 10^{20000}, \] so $A$ has at most $20000$ digits, so \[ B \leqslant 9 \times 20000 = 180000. \] So either $B$ has 6 digits and the first one is a 1, or it has 5 digits or less; either way \[ C \leqslant 1+6 \times 9 = 55. \]
Therefore $C$ has at most 2 digits and the first one is at most 5, so \[ D \leqslant 5+9=14. \]

\item Since we know that $D \equiv 7 \bmod 9$ and that $D \le 14$, we conclude that in fact $D=7$.
\end{enumerate}
\end{solution}

\bigskip

\begin{exercise}[subtitle=Primitive roots mod 43]
\begin{enumerate}
\item Suppose you choose an element of $(\Z/43\Z)^\times$ at random. What is the probability that this element is a primitive root? In other words, what is the proportion of elements of $(\Z/43\Z)^\times$ that are primitive roots?
\item Find a primitive root $g \in (\Z/43\Z)^\times$.
\item What is the multiplicative order of $g^{2020}$, where $g$ is the primitive root found in the previous question?
\item Same question for $g^43$.
\item Prove that every primitive root in $(\Z/43\Z)^\times$ is a power of $g$.
\item For which $m \in \Z$ is $g^m$ a primitive root?
\end{enumerate}
\end{exercise}

\begin{solution}
\begin{enumerate}
\item Since 43 is prime, primitive roots exist. More precisely, there are exactly $\phi(\phi(43))=\phi(42)=12$ of them. Compared to the $\phi(43)=42$ elements of $(\Z/43\Z)^\times$, that's a proportion $12/42 = 2/7$ that are primitive roots.

\item We are just going to try values of $g$ until we find one. Since $42 = 2 \cdot 3 \cdot 7$, we know that $g$ is a primitive root if and only if $g^{2 \cdot 3} = g^6$, $g^{2 \cdot 7} = g^{14}$, and $g^{3 \cdot 7} = g^{21}$ are all $\neq 1$.

Obviously $g=1$ is not a primitive root (quite the opposite!), so let us try $g=2$. We compute in $\Z/43\Z$ that
\[ 2^6 = 64 = 21 \neq 1, \]
but
\[ 2^{14} = (2^7)^2 = (2 \cdot 21)^2 = 42^2 = -1^2 = 1 \]
so 2 is not a primitive root.

Let us try $g=3$:
\[ 3^6 = 3^4 3^2 = 81 \cdot 9 = -5\cdot 9 = -45 = -2 \neq 1, \]
\[ 3^{14} = 3^2 3^6 3^6 = 9 \cdot -2 \cdot -2 = 36 = -7 \neq 1, \]
\[ 3^{21} = 3 \cdot 3^6 3^{14} = 3 \cdot -2 \cdot -7 = 42 = -1 \neq 1 \]
so $g=3$ is one of the 12 primitive roots.

\item Since $g$ is a primitive root, it has order $\phi(43)=42$. Thus
\[ \MO(g^{2020}) = \frac{\MO(g)}{\gcd(MO(g),2020)} =  \frac{42}{\gcd(2 \times 3 \times 7, 2^2 \times 5 \times 101)} = \frac{42}{2} = 21. \] 
In particular, $g^{2020}$ is not a primitive root since $21 < 42$.

\item Do not fall into the trap of thinking that $g^43=1$! Indeed, $\MO(g)$ is $42$ not $43$; therefore $g^43 = g^1 = g$, so $\MO(g^43) = \MO(g) = 42$.

\item Since $g$ is a primitive root, every element of $(\Z/43\Z)^\times$ is a power of $g$; in particular this includes the primitive roots. And this has nothing to do with $43$!

\item We know that $MO(g^m) = \frac{42}{\gcd(42,m)}$, so $g^m$ is a primitive root iff. $MO(g^m)=42$ iff. $m$ is coprime to $42$.
\end{enumerate}
\end{solution}

\bigskip

\begin{exercise}[subtitle=More primitive roots]
\begin{enumerate}
\item Find a primitive root for $\Z/7\Z$. Justify your answer in detail.
\item Same question for $\Z/11\Z$.
\item Same question for $\Z/23\Z$.
\end{enumerate}
\end{exercise}
\begin{solution}
\begin{enumerate}
\item Fermat's little theorem tells us that every $x \in (\Z/7\Z)^\times$ has order dividing $7-1=6=2 \times 3$. Therefore, $x$ is a primitive root iff. it satisfies $x^2 \neq 1$ and $x^3 \neq 1$.

Let us try $x=2$. We have $2^2 =4 \neq 1$, but $2^3 = 8 =1$ so $2$ is not a primitive root (in fact, since $2 \neq 1$ it does not have order 1, and since 3 is prime, the identity $2^3=1$ tells us that the multiplicative order of 2 is 3).

Let us try again, with $x=3$. We find $3^2=9 \neq 1$ and $3^3=27=-1 \neq 1$, so $3$ is a primitive root.

\emph{Remark: we know that there are in fact $\phi(6)=2$ primitive roots; the other one is $3^{-1} = 5$.}

\item We have $11-1=10=2 \times 5$, so we are looking for an $x \neq 0$ such that $x^2 \neq 1$ and $x^5 \neq 1$.

Let us try $x=2$. This time we are luckier: we have $2^2 = 4 \neq 1$ and $2^5=32=-1 \neq 1$, so $2$ is a primitive root.

\emph{Remark: we know that there are in fact $\phi(10)=4$ primitive roots; they are the $2^m$ where $m \in (\Z/10\Z)^*$, in other words, $2$, $8$, $7$, and $6$.}

\item We have $23-1=22=2 \times 11$, so we are looking for an $x \neq 0$ such that $x^2 \neq 1$ and $x^{11}\neq 1$.

Let us try $x=2$. Bad luck: we have $2^2 = 4 \neq 1$, but $2^{11}=1$, so $2$ is a not primitive root.

Let us try again with $x=3$: we have $3^2=9 \neq 1$, but again $3^{11}=1$, so 3 is not a primitive root either.

The next value is $x=4$, however we can see directly that $4^{11}=(2^2)^{11}=2^{22}=(2^{11})^2=1$, so 4 is not going to work either.

But let us not give up! For $x=5$ we have $5^2=25=2 \neq 1$, and $5^{11}=-1 \neq 1$, so $5$ is a primitive root. 

\emph{Remark: we know that there are in fact $\phi(22)=10$ primitive roots; they are the $5^m$ where $m \in (\Z/22\Z)^*$. Also, to compute $x^{11}$, it is a good idea to write something like $x^11=x \times (x^5)^2$, and to reduce mod 23 at every step.}
\end{enumerate}

\emph{Final remarks: in $\Z/p\Z$, we can only have $x^2=1$ when $x=\pm1$. So as long as we did not consider $x=-1$ ($x=1$ would be really too silly), we didn't have to care about $x^2$ being $\neq 1$. Also, when we see Legendre symbols in chapter 3, we'll see that we have $x^{\frac{p-1}2}=\left( \frac{x}{p} \right) = \pm 1$; this explains why $x^{\frac{p-1}2}=-1$ whenever $x$ is a primitive root.}
\end{solution}

\bigskip

%\begin{exercise}[subtitle=Even more primitive roots]
%Let $p \in \N$ be prime, and let $g \in (\Z/p\Z)^\times$ be a primitive root.
%\begin{enumerate}
%\item Let $a \in \Z$. Give a necessary and sufficient condition on $a$ for $g^a$ to be a primitive root in $\Z/p\Z$.
%\item Prove that if $a$ is prime, then $g^a$ is a primitive root in $\Z/p\Z$ if and only if $p \not \equiv 1 \bmod a$.
%\item Show that the previous assertion is no longer valid when $a$ is not assumed to be prime, by finding a counterexample.
%\item Is every primitive root of $\Z/p\Z$ of the form $g^a$ for some $a \in \Z$? Justify your answer.
%\end{enumerate}
%\end{exercise}
%\begin{solution}
%\begin{enumerate}
%\item By definition of $g$, the multiplicative order of $g$ is $p-1$. As a consequence, for every $a$ the multiplicative order of $g^a$ is $\frac{p-1}{\gcd(p-1,a)}$. Therefore, $g^a$ is a primitive root iff. $a$ and $p-1$ are coprime.
%\item By the previous question, $g^a$ is a primitive root iff. $p-1$ and $a$ are coprime. Since $a$ is prime, this is equivalent to $a$ not dividing $p-1$, which is equivalent to $p$ not being congruent to $1$ mod $a$.
%\item In view of the previous questions, we will get a counterexample if we can find $a$ and $p$ such that $a$ does not divide $p-1$ and yet $\gcd(a,p-1)\neq 1$, i.e. such that $1 < \gcd(p-1,a) < a$.
%
%So for instance we can take $a=4$, $p=7$. Indeed, we then have that $p \not equiv 1 \bmod a$, and yet the multiplicative order of $g^a$ is $\frac{6}{\gcd(4,6)}=3 < 6$. (To be even more concrete, we can take $g=3$ as in the previous exercise, and then $g^a=3^4=81=11=4$ is not a primitive element since $4^3=64=1+63=1$.)
%
%\item Yes, simply because by definition of primitive roots, every nonzero element of $\Z/p\Z$, primitive root or not, is of the form $g^a$ for some $a \in \N$.
%\end{enumerate}
%\end{solution}
%
%\bigskip

\begin{exercise}[subtitle=A multiplicative sequence]
The goal of this exercise is to understand the behavior of the sequence $t_n=2^n$ in $\Z/40\Z$.
\begin{enumerate}
\item Why cannot we say that $t_n$ is periodic mod 40 ``as usual''?
\item Find a formula for the values of $t_n$ mod 5 in terms of $n$. You answer should have the form ``if $n$ is like this, then $t_n=$ this; if $n$ is like that, then $t_n=$ that; if  \dots''.
\item Find a formula for the values of $t_n$ mod 8 in terms of $n$.

\emph{Hint: Compute $t_n$ for $n \le 4$ ``by hand''.}

\item Deduce a formula for $t_n \bmod 40$. What is the period? What is the length of the ``tail''?

\emph{Hint: Chinese.}
\end{enumerate}
\end{exercise}

\begin{solution}
\begin{enumerate}
\item We know that the sequence $1,x,x^2,\cdots$ is periodic for all $x \in (\Z/N\Z)^\times$, but 2 is not invertible mod 40 (their gcd is 2) so this argument does not apply.

\item Now 2 is invertible mod 5, so we know that $2^n \bmod 5$ is periodic, of period dividing $\phi(5)=4$ by Fermat's little theorem. Let us see: in $\Z/5\Z$, we have $t_0=1$, $t_1=2$, $t_2=4=-1$, $t_3=-2$, and $t_4=-4=1=t_0$. So the period is exactly 4 (in other words 2 is a primitive root in $\Z/5\Z$) and
\[ t_n \bmod 5 = \left\{ \begin{array}{rr} 1 & \text{ if } n \equiv 0 \bmod 4, \\ 2 & \text{ if } n \equiv 1 \bmod 4, \\ -1 & \text{ if } n \equiv 2 \bmod 4, \\ -2 & \text{ if } n \equiv 3 \bmod 4. \\ \end{array} \right. \]

\item Since 2 is not invertible mod 8, the theory seen in class does not apply. Let us compute a few terms anyway. Mod 8, we have $t_0=1$, $t_1=2$, $t_2=4$, and $t_3=8=0$. So $t_n\equiv 0 \bmod 8$ for all $n \geqslant 3$. Thus 
\[ t_n \bmod 8 = \left\{ \begin{array}{rr} 1 & \text{ if } n =0 \\ 2 & \text{ if } n=1, \\ 4 & \text{ if } n =2, \\ 0 & \text{ if } n \geqslant 3 . \\ \end{array} \right. \]
\item The hint is to use CRT. Indeed, for $n \geqslant 3$ we have $t_n \equiv 0 \bmod 8$ whereas $t_n \bmod 5$ is given by the formula found in question 2. By putting this information together, we can deduce $t_n \bmod 40$ for $n \geqslant 3$; and for $n < 3$ we can just compute $t_n$ by hand. We find  
\[ t_n \bmod 40 = \left\{ \begin{array}{rr} 1 & \text{ if } n =0 \\ 2 & \text{ if } n=1, \\ 4 & \text{ if } n =2, \\ 8 & \text{ if } n \geqslant 3  \text{ and } n \equiv 3 \bmod 4, \\ 16 & \text{ if } n \geqslant 3  \text{ and } n \equiv 0 \bmod 4, \\ 32 & \text{ if }n \geqslant 3  \text{ and }  n \equiv 1 \bmod 4, \\ 24 & \text{ if }n \geqslant 3  \text{ and } n \equiv 2 \bmod 4. \\ \end{array} \right. \]
We see that we have a tail of length 3, after which we enter a cycle of length~4.
\[ \xymatrix{ 16 \ar[r] & 32 \ar[d] \\
8 \ar[u] & 24 \ar[l] \\
4 \ar[u] & \\
2 \ar[u] & 1 \ar[l] } \]
\end{enumerate}
\end{solution}

\bigskip

\begin{exercise}[subtitle=A divisibility relation]

Prove that $2^{3n+5}+3^{n+1}$ is divisible by $5$ for all $n \in \N$.

\noindent \emph{Hint: Multiplicative orders.}
\end{exercise}
\begin{solution}
Since $2 \in (\Z/5\Z)^\times$, its multiplicative order mod 5 is a divisor of 4 (in fact, it can be checked that its order is exactly 4, i.e. 2 is a primitive root mod 5), so $2^m \bmod 5$ only depends on $m \bmod 4$. And since $3n+5 \bmod 4$ only depends on $n \bmod 4$, we have that $2^{3n+5} \bmod 5$ only depends on $n \bmod 4$.

Similarly, the multiplicative order of $3 \bmod 5$ divides 4 (its in is fact again exactly 4), so $3^m \bmod 5$ only depends on $m \bmod 4$, and so $3^{n+1} \bmod 5$ only depends on $n \bmod 4$. As a result, the expression $2^{3n+5}+3^{n+1} \bmod 5$ only depends on $n \bmod 4$. Thus all we have to do is check that $2^{3n+5}+3^{n+1} \equiv 0 \bmod 5$ for 4 values of $n$ \textbf{representing all 4 elements of $\Z/4\Z$}, such as $0,1,2,3$, or even cleverer, $-1,0,1,2$.

Other solution: instead of checking for 4 values of $n$, which is easy but still a bit tedious, we can directly compute that $3n +5 \equiv -n+1 \bmod 4$, so that
\[ 2^{3n+5} \equiv 2^{-n+1} \equiv 2 \times 3^n \bmod 5 \]
since $3$ is the inverse of $2 \bmod 5$; as a result, we have
\[ 2^{3n+5}+3^{n+1} \equiv 2 \times 3^n + 3 \times 3^n = 5 \times 3^n \equiv 0 \bmod 5 .\]
\end{solution}

\bigskip

%\begin{exercise}[subtitle=Largest possible orders]
%
%\begin{enumerate}
%\item Prove that for every $n \in \N$, and for every $x \in \Z/n\Z$, the additive order of $x$ is at most $n$.
%
%\item Prove that for every $n \in \N$, there exists an $x \in \Z/n\Z$ whose additive order is exactly $n$.
%
%\item Prove that for every $n \in \N$, and for every $x \in (\Z/n\Z)^\times$, the multiplicative order of $x$ is at most $\phi(n)$.
%
%\item Find an $n \in \N$ such that every $x \in (\Z/n\Z)^\times$ has multiplicative order $< \phi(n)$.
%
%\end{enumerate}
%
%\end{exercise}
%\begin{solution}
%\begin{enumerate}
%\item We have $x+x+\cdots+x$ ($n$ times) $=nx = 0x = 0$, so the additive order of $x$ divides $n$.
%
%\item Remember that the additive order of $x$ is the smallest $m \in \N$ such that $mx=0 \in \Z/n\Z$. So if we take $x=1$, then we have $mx=0$ iff. $m=0$ iff. $n \mid m$, so the additive order of $1$ is $n$. (In fact, we can show similarly that the additive order of any \emph{invertible} element of $\Z/n\Z$ is exactly $n$, and that conversely the only elements of additive order $n$ are the invertibles.)
%
%\item The multiplicative order of $x$ is the smallest $m \in \N$ such that $x^m=1 \in (\Z/n\Z)^\times$. We have $x^\phi(n)=1$ by Fermat's little theorem, so the multiplicative order of $x$ divides $\phi(n)$.
%
%\item We can take $n=8$: then every $x \in (\Z/8\Z)^\times$ satisfies $x^2=1$, so is of order $\leqslant 2$, whereas $\phi(8)=4 > 2$.
%\end{enumerate}
%\end{solution}

\begin{exercise}[subtitle=Possible orders]

\begin{enumerate}
\item Let $n \in \N$. Explain why the additive order of any $x \in \Z/n\Z$ is a divisor of $n$, and prove that for any $d \mid n$, there exists an $x \in \Z/n\Z$ of order $d$.

\item Let $p \in \N$ be a prime. Explain why the multiplicative order of any $x \in (\Z/p\Z)^\times$ is a divisor of $p-1$, and prove that for any $d \mid (p-1)$, there exists an $x \in (\Z/p\Z)^\times$ of multiplicative order $d$.

\item Let $n \in \N$. Is it true that for any $d \mid \phi(n)$, there exists an $x \in (\Z/n\Z)^\times$ of multiplicative order $d$?

\item Suppose that $n \geqslant 2$, and that there exists an $x \in (\Z/n\Z)^\times$ of multiplicative  order $n-1$. Prove that $n$ must be prime.

\end{enumerate}

\end{exercise}
\begin{solution}
\begin{enumerate}
\item For all $x$, we have $nx=0x=0$ so the additive order of $x$ divides $n$. If $d \mid n$, then we can consider $x=\frac{n}d \in \Z/n\Z$, and it is clear that $mx=0 \in \Z/n\Z$ precisely when $d \mid m$, so this $x$ is of additive order exactly $d$.

\item By Fermat's little theorem, the multiplicative order of $x$ divides $\phi(p)$, and $\phi(p)=p-1$ since $p$ is prime. Let now $g \in (\Z/p\Z)^\times$ be a primitive root (there exists at least one since $p$ is prime), then by definition $g^m=1$ iff. $(p-1) \mid m$. So if $d \mid (p-1)$, then $x=g^{\frac{p-1}d}$ satisfies
\[ x^m = 1 \Longleftrightarrow g^{\frac{p-1}d m} = 1 \Longleftrightarrow (p-1) \mid\frac{p-1}d m \Longleftrightarrow d \mid m, \]
which shows that the multiplicative order of $x$ is exactly $d$.

\item No. In fact, this is false when $d=\phi(n)$: in this case, such $x$ are precisely primitive roots, and those do not exist for most $n$.

\item Since $x \in (\Z/n\Z)^\times$ is invertible, all its powers are also invertible (of inverse the same power of the inverse of $x$). But $x$ has multiplicative order $n-1$, so the sequence of its power is periodic of period exactly $n-1$, so $x$ has $n-1$ distinct powers. So we have at least $n-1$ invertibles in $\Z/n\Z$. But in $\Z/n\Z$ there are $n$ elements, and clearly $0$ cannot be invertible, so we see that all nonzero elements of $\Z/n\Z$ are invertible. This means that $\Z/n\Z$ is a field, so $n$ must be prime.
\end{enumerate}
\end{solution}

\end{document}
