Congruences 2

Theorem 1 (Fermat’s Little Theorem). Let $p$ be a prime. Then for any integer $a$, $$a^p\equiv a\mod p$$ and for any integer $a$ not divisible by $p$,
$$a^{p-1}\equiv 1\mod p.$$

Proof. We prove the second statement first. Suppose that $p\not| a$. If
$i\equiv j\mod p$ then clearly $ai\equiv aj\mod p$. Conversely, if
$ai\equiv aj\mod p$ then $$p|ai-aj=a(i-j).$$ Since $(a,p)=1$, $p|i-j$, i.e., $i\equiv j\mod p$. So, $0a,1a,2a,\cdots,(p-1)a$ are a complete set of residues $\mod p$, that is, they are rearrangement of $0,1,2,\cdots,p-1$ when considered $\mod p$. Hence, $$(p-1)!a^{p-1}\equiv(p-1)!\mod p,$$
i.e., $p|(p-1)!(a^{p-1}-1)$. Since $p\not|(p-1)!$ then $p|a^{p-1}-1$, i.e., $$a^{p-1}\equiv 1\mod p.$$ If $a$ is divisible by $p$, then clearly
$$a^p\equiv a\mod p$$ since either side is congruent to $0\mod p$.

Corollary 2. If $p\not| a$ and if $n\equiv m\mod p-1$ then $$a^n\equiv a^m\mod p.$$

Proof. Say $n>m$. Since $n\equiv m\mod p-1$, $$n=m+(p-1)c$$ for some $c\in\mathbb{Z}$. By Fermat’s Little Theorem (Theorem 1),
$$a^{p-1}\equiv 1\mod p$$ and so
$$a^{c(p-1)}\equiv 1\mod p\Longrightarrow a^n=a^{m+(p-1)c}\equiv a^m\mod p.$$

Example. Find the last base-$7$ digit in $2^{1000000}$.

Solution: Let $p=7$. Since $1000000$ leaves a remainder of $4$ when divided by $p-1=6$, $$1000000\equiv 4\mod 6=7-1$$ and $7\not|2$. By Corollary 2, $$2^{1000000}\equiv 2^4=16\equiv 2\mod 7.$$
Thus, $2$ is the answer.

Lemma 3. If $a\equiv b\mod m$, $a\equiv b\mod n$, and
$(m,n)=1$, then $a\equiv b\mod mn$.

Proof. Since $a\equiv b\mod m$, $b-a=md$ for some $d\in\mathbb{Z}$.
\begin{align*} a\equiv b\mod n&\Longrightarrow n|md\\ &\Longrightarrow n|d\ \mbox{since}\ (m,n)=1\\ &\Longrightarrow d=ns\ \mbox{for some}\ s\in\mathbb{Z}\\ &\Longrightarrow b-a=mns\\ &\Longrightarrow a\equiv b\mod mn. \end{align*}

Definition (Euler phi function).
For any positive integer $n$, $\phi(n)$ is defined to be the number
of integers $a$, $1\leq a\leq n$, such that $(a,n)=1$. The function
$\phi$ is called the Euler phi function.

Example. $\phi(1)=1$ by definition.

To calculate $\phi(6)$, we write all positive integers that are
less than or equal to $6$ and then cross out the ones that are not
relatively prime to $6$. Counting the remaining numbers will give
$\phi(6)$: $$1,\not 2,\not 3,\not 4,5,\not 6.$$ Hence, $\phi(6)=2$.

Since $7$ is a prime number, any positive integer that are less than
$7$ are relatively prime to $7$, so $\phi(7)=6$. In general, if $p$
is a prime number then $$\label{eq:phiprime} \phi(p)=p-1.$$

In order to calculate $\phi(12)$, we write $$1,\not 2,\not 3,\not 4,5,\not 6,7,\not 8,\not 9,\not 10,11,\not 12$$ Thus, $\phi(12)=4$.

As one can easily imagine, if a number $n$ gets bigger, computing
$\phi(n)$ will become really a hardship.

The following theorem provides a short cut to computing $\phi(n)$.

Theorem 4. If $k$ and $a$ are positive integers such that all primes dividing
$k$ also divide $a$, then $$\label{eq:phi}\phi(ka)=k\phi(a).$$

Proof. We first write all positive integers that are less than or equal to $ka$: $$\begin{array}{cccc} 1 & 2 & \cdots & a\\a+1 & a+2 & \cdots & 2a\\& & \vdots &\\(k-1)a+1 & (k-1)a+2 & \cdots & ka\end{array}$$ and then cross out the entries that are not relatively prime to $ka$. It is easy to see that this amounts to crossing out everything not relatively prime to $a$. Clearly, any prime that divides $a$ divides $ka$. Conversely, if a prime divides $ka$, then it divides $k$ or $a$. If the prime divides $k$ then by the assumption, it divides $a$ as well. By Theorem 5 here, the pattern of crossing out is the same in each row. Since there are $\phi(a)$ entries left in the first row, there must be $k\phi(a)$ entries left in all. So, $$\phi(ka)=k\phi(a).$$

It would be also interesting to study the relationship between $\phi(pa)$ and $\phi(a)$ when $p\not|a$. First note that if $(a,n)>1$ then clearly $(pa,n)>1$ but the converse need not be true. If a prime number $q>1$ divides $pa$ then $q|p$ or $q|a$. If $q|p$ then $q=p$. That is, if $(pa,n)>1$ then $p|n$ or $(a,n)>1$. After crossing out all $n$ such that $(a,n)>1$, $p\phi(a)$ integers left and yet more to be eliminated, namely, the multiples of $p$, $1p,2p,\cdots,ap$. Of these, those $kp$ such that $(k,a)>1$ have
already been eliminated, and there are just $\phi(a)$ more to cross out. Thus, \begin{align*} \phi(pa)&=p\phi(a)-\phi(a)\\ &=(p-1)\phi(a). \end{align*} Hence, we proved the following theorem.

Theorem 5. Let $a$ be a positive integer and $p$ a prime such that $p\not|a$. Then $$\label{eq:phi2}\phi(pa)=(p-1)\phi(a).$$

Example. \begin{align*} \phi(10^6)&=\phi(10^5\cdot 10)\\ &=10^5\phi(10)\ \mbox{by \eqref{eq:phi}}\\ &=10^5\phi(2\cdot 5)\\ &=10^5(2-1)\phi(5)\ \mbox{by \eqref{eq:phi2}}\\ &=10^5\cdot 4 \ \mbox{by \eqref{eq:phiprime}}\\&=400,000.\\\phi(60)&=\phi(2^2\cdot 3\cdot 5)\\ &=2\phi(2\cdot 3\cdot 5) \ \mbox{by \eqref{eq:phi}}\\ &=2\cdot(2-1)\phi(3\cdot 5) \ \mbox{by \eqref{eq:phi2}}\\ &=2(3-1)\phi(5)\ \mbox{by \eqref{eq:phi2}}\\ &=16 \ \mbox{by \eqref{eq:phiprime}}.\\ \phi(62)&=\phi(2\cdot 31)\\ &=(2-1)\phi(31)\ \mbox{by \eqref{eq:phi2}}\\ &=30\ \mbox{by \eqref{eq:phiprime}}. \end{align*}

Theorem 6. Let $p$ be a prime and $\alpha$ a positive integer. Then
$$\label{eq:phiprime2}\phi(p^\alpha)=p^\alpha\left(1-\frac{1}{p}\right).$$

Proof. It follows immediately from \eqref{eq:phi} and \eqref{eq:phiprime}.\begin{align*}\phi(p^\alpha)&=p\phi(p^{\alpha-1})\\ &=p^2\phi(p^{\alpha-2})\\ &=\cdots\\ &=p^{\alpha-1}\phi(p)\\ &=p^{\alpha-1}(p-1)\\ &=p^{\alpha}\left(1-\frac{1}{p}\right). \end{align*}

Theorem 7 (Chinese Remainder Theorem) Suppose that we want to solve a system of congruences to different moduli: \begin{aligned}x&\equiv a_1\mod m_1,\\x&\equiv a_2\mod m_2,\\&\cdots\ \ \ \ \ \cdots\\x&\equiv a_r\mod m_r.\end{aligned}\label{eq:congsys}
Suppose that each pair of moduli is relatively prime: $(m_i,m_j)=1$ for $i\ne j$. Then there exists a simultaneous solution $x$ to all of the congruences, and any two solutions are congruent to one another modulo $M=m_1m_2\cdots m_r$.

Proof. First we prove uniqueness modulo $M$. Suppose that $x’$ and $x^{\prime\prime}$
are two solutions to the system of congruences \eqref{eq:congsys}. Let $x=x’-x^{\prime\prime}$. Then for all $i=1,\cdots,r$,\begin{align*}x’&\equiv a_i\mod m_i,\\ x^{\prime\prime}&\equiv a_i\mod m_i.\end{align*}This implies that \begin{align*} x=x’-x^{\prime\prime}\equiv 0\mod m_i&\Longrightarrow x\equiv 0\mod M\ \mbox{by Lemma 3}\\&\Longrightarrow x’\equiv x^{\prime\prime}\mod M. \end{align*} We now show how to construct a solution $x$. For each $i=1,\cdots,r$, define $M_i:=M/m_i$. Then $(m_i,M_i)=1$, so by Bézout’s Lemma there exist $y,z\in\mathbb{Z}$ such that $$M_iy+m_iz=1.$$ This implies that $$M_iy\equiv 1\mod m_i.$$ Write $y:=N_i$, i.e., $$M_iN_i\equiv 1\mod m_i.$$ Set $$x:=\sum_{i=1}^ra_iM_iN_i.$$ Since $m_i|M_j$ whenever $i\ne j$, $$x\equiv a_iM_iN_i\equiv a_i\mod m_i.$$

Corollary 8. The Euler phi function is multiplicative, i.e., $$\phi(mn)=\phi(m)\phi(n)$$ whenever $(m,n)=1$.

Proof. We must count the number of integers between $0$ and $mn-1$ which have no common factor with $mn$. For each $0\leq j\leq mn-1$, let $j_1$ be its least nonnegative residue$\mod m$ (i.e., $0\leq j_1<m$ and $j\equiv j_1\mod m$) and $j_2$ be its least nonnegative residue$\mod n$ (i.e., $0\leq j_2<n$ and $j\equiv j_2\mod n$). It follows from the Chinese Remainder Theorem (Theorem 7) that for each pair $j_1,j_2$ there is one and only one $j$ between $0$ and $mn-1$ for which $$j\equiv j_1\mod m,\ j\equiv j_2\mod n.$$ Notice that $j$ has no common factor with $mn$ if and only if it has no common factor with $m$ (which is equivalent to $j_1$ having no common factor with $m$) and $j$ has no common factor with $n$ (which is equivalent to $j_2$ having no common factor with $n$). Thus, the $j$’s which we must count are in $1:1$ correspondence with the pairs $j_1,j_2$ for which $$0\leq j_1<m,\ (j_1,m)=1;\ 0\leq j_2<n,(j_2,n)=1.$$ The number of possible $j_1$’s is $\phi(m)$ and the number of possible $j_2$’s is $\phi(n)$. So, the number of pairs is $\phi(m)\phi(n)$.

Corollary 9. For any positive integer $n$,$$\phi(n)=n\prod_{p|n}\left(1-\frac{1}{p}\right).$$

Proof. Let $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_r^{\alpha_r}$ be the prime factorization of $n$. Then by Corollary 8, \begin{align*}\phi(n)&=\phi(p_1^{\alpha_1})\phi(p_2^{\alpha_2})\cdots\phi(p_r^{\alpha_r})\\ &=p_1^{\alpha_1}\left(1-\frac{1}{p_1}\right)p_2^{\alpha_2}\left(1-\frac{1}{p_2}\right)\cdots p_r^{\alpha_r}\left(1-\frac{1}{p_r}\right)\ \mbox{by \eqref{eq:phiprime2}}\\&=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_r^{\alpha_r}\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\cdots\left(1-\frac{1}{p_r}\right)\\&=n\prod_{p|n}\left(1-\frac{1}{p}\right).\end{align*}

Example. \begin{align*}\phi(60)&=\phi(2^2\cdot 3\cdot 5)\\&=60\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\\&=16.\\\phi(62)&=\phi(2\cdot 31)\\ &=62\left(1-\frac{1}{2}\right)\left(1-\frac{1}{31}\right)\\&=30.\\\phi(360)&=\phi(2^3\cdot 3^2\cdot 5)\\&=360\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\\&=96.\end{align*}

The following theorem is a generalization of Fermat’s Little Theorem
(Theorem 1) due to Euler.

Theorem 10 (Euler-Fermat Theorem). If $(a,m)=1$ then $$\label{eq:euler} a^{\phi(m)}\equiv 1\mod m.$$

Proof. First consider the case when $m$ is a prime power, say $m=p^\alpha$.
We use induction on $\alpha$. The case $\alpha=1$ is precisely Fermat’s Little Theorem. Suppose that $\alpha\geq 2$ and the formula
\eqref{eq:euler} holds for the $(\alpha-1)$-st power of $p$. Then \begin{align*} \phi(p^{\alpha-1})&=p^{\alpha-1}\left(1-\frac{1}{p}\right)\\&=p^{\alpha-1}-p^{\alpha-2}\end{align*} and \begin{align*}a^{\phi(p^{\alpha-1})}&=a^{p^{\alpha-1}-p^{\alpha-2}}\\&\equiv 1\mod p^{\alpha-1}.\end{align*} That is,
$$a^{p^{\alpha-1}-p^{\alpha-2}}=1+p^{\alpha-1}b\ \mbox{for some}\ b\in\mathbb{Z}.$$ Thus, \begin{align*}a^{\phi(p^\alpha)}&=(a^{p^{\alpha-1}-p^{\alpha-2}})^p\\&=(1+p^{\alpha-1}b)^p.\end{align*} Note that the binomial coefficients of $(1+x)^p$ are each divisible by $p$ except in the $1$ and $x^p$ in the ends. This means the binomial coefficients of $(1+p^{\alpha-1}b)^p$ are each divisible by $p^\alpha$ except in the $1$ and $p^{p(\alpha-1)}b^p$ in the ends. Now, $$p^{p(\alpha-1)}b^p=p^\alpha p^{p(\alpha-1)-\alpha}b^p.$$ Since $p$ is a prime, $p\geq 2$. So, \begin{align*}p(\alpha-1)-\alpha&\geq 2(\alpha-1)-\alpha\\&=\alpha-2\geq 0\ \mbox{by assumption}.\end{align*}This means $p^{p(\alpha-1)}b^p$ is also divisible by $p^\alpha$. Hence, $$a^{\phi(p^\alpha)}\equiv 1\mod p^\alpha.$$ Let $m=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_r^{\alpha_r}$ be the prime factorization of $m$. Then by the multiplicity of $\phi$ (Corollary 8), \begin{align*}\phi(m)&=\phi(p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_r^{\alpha_r})\\&=\phi(p_1^{\alpha_1})\cdots \phi(p_r^{\alpha_r}).\end{align*} For each $i=1,\cdots,r$, $$a^{\phi(p_i^{\alpha_i})}\equiv 1\mod p_i^{\alpha_i}.$$
Let $M_i:=\phi(m)/\phi(p_i^{\alpha_i})$ for $i=1,\cdots,r$. Then \begin{align*} a^{\phi(m)}&=a^{\phi(p_1^{\alpha_1})\cdots \phi(p_r^{\alpha_r})}\\&=(a^{\phi(p_i^{\alpha_i})})^{M_i}\\&\equiv 1\mod p_i^{\alpha_i}.\end{align*}
Since $(p_i^{\alpha_i},p_j^{\alpha_j})=1$ if $i\ne j$, $$a^{\phi(m)}\equiv 1\mod m=p_1^{\alpha_1}\cdots p_r^{\alpha_r}$$ by Lemma 3.

Remark. Suppose that $a$ is any positive integer such that $(a,105)=1$.
\begin{align*}\phi(105)&=\phi(3\cdot 5\cdot 7)\\&=105\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\\&=48.\end{align*}So, by Euler’s Theorem (Theorem 10) $$a^{48}\equiv 1\mod 105.$$ As we have seen in the proof of Theorem 10, there is a smaller power of $a$ which gives $1\mod m=105=3\cdot 5\cdot 7$. \begin{align*}a^{\phi(3)}=a^2\equiv 1\mod 3,\\ a^{\phi(5)}=a^4\equiv 1\mod 5,\\ a^{\phi(7)}=a^6\equiv 1\mod 7.\end{align*}
The least common multiple of $\phi(3)=2,\phi(5)=4,\phi(7)=6$ is 12 and so
\begin{align*}a^{12}&\equiv 1\mod 3,\\ a^{12}&\equiv 1\mod 5,\\ a^{12}&\equiv 1 \mod 7.\end{align*} By Lemma 3, $$a^{12}\equiv 1\mod3\cdot 5\cdot 7=105.$$

Example. Compute $2^{1000000}\mod 77$.

Solution: $77=7\cdot 11$ and $\phi(7)=6,\ \phi(11)=10$. The least common multiple of $6$ and $10$ is $30$. So, $$2^{30}\equiv 1\mod 77.$$ Since $1000000=30\cdot 33333+10$, $$2^{1000000}\equiv 2^{10}\mod 77\equiv 23\mod 77.$$

There is an alternative way to calculate $2^{1000000}\mod 77$. First compute $2^{1000000}\mod 7$. $$2^{\phi(7)}=2^6\equiv 1\mod 7.$$ Since $1000000=6\cdot 166666+4$, $$2^{1000000}\equiv 2^4\equiv 2\mod 7.$$ On the other hand,
$$2^{10}\equiv 1\mod 11.$$ So, $$2^{1000000}\equiv 1\mod 11.$$ We now find $0\leq x\leq 76$ which satisfies \begin{align*}x&\equiv 2\mod 7\\ x&\equiv 1\mod 11\end{align*} by the Chinese Remainder Theorem (Theorem 7). Let $M=7\cdot 11=77$, $M_1=11$, and $M_2=7$. Since $(7,11)=1$, there exist $y,z\in\mathbb{Z}$ such that $7y+11z=1$. By Euclidean algorithm, one can find solution $y=-3$, $z=2$. Thus, \begin{align*}a_1&=2,\ m_1=7,\ M_1=11,\ N_1=2,\\ a_2&=1,\ m_2=11,\ M_2=7,\ N_2=-3.\end{align*} Hence, \begin{align}x&=\sum_{i=1}^2a_iM_iN_i\ &=23\mod 77.\end{align}

Theorem. $$\sum_{d|n}\phi(d)=n.$$

Proof. Let $$f(n):=\sum_{d|n}\phi(d).$$ First we show that $f(n)$ is multiplicative. Suppose that $(m,n)=1$ and $d|mn$. Any division $d$ can be written in the form $d=d_1d_2$ such that $d_1|m$ and $d_2|n$. Since $(d_1,d_2)=1$, by Corollary 8, $$\phi(d)=\phi(d_1d_2)=\phi(d_1)\phi(d_2).$$
\begin{align*}f(mn)&=\sum_{d|mn}\phi(d)\\&=\sum_{d_1|m}\sum_{d_2|n}\phi(d_1)\phi(d_2)\\&=\sum_{d_1|m}\phi(d_1)\left(\sum_{d_2|n}\phi(d_2)\right)\\&=\sum_{d_1|m}\phi(d_1)f(n)\\&=f(n)\sum_{d_1|m}\phi(d_1)\\&=f(m)f(n).\end{align*} Let $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_r^{\alpha_r}$ be the prime factorization of $n$. Then
$$f(n)=f(p_1^{\alpha_1})f(p_2^{\alpha_2})\cdots f(p_r^{\alpha_r})$$ and for each $i=1,\cdots, r$, \begin{align*}f(p_i^{\alpha_i})&=\sum_{d|p_i^{\alpha_i}}\phi(d)\\&=\sum_{j=0}^{\alpha_i}\phi(p^j)\\&=\sum_{j=1}^{\alpha_i}p^j\left(1-\frac{1}{p}\right)+1\\&=\sum_{j=1}^{\alpha_i}(p^j-p^{j-1})+1\\&=p^{\alpha_i}.\end{align*} Therefore, $$f(n)=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_r^{\alpha_r}=n.$$