Lecture 16

It would be good to know something that reassures us that there will be a solution in some family of cases, and here’s a result, named after its discovery by the ancient Chinese:

Theorem

[Chinese Remainder Theorem] Let $m_1$ and $m_2$ be coprime, and let $a_1$ and $a_2$ be any integers. The simultaneous congruences $\begin{aligned} x &\equiv a_1\pmod{m_1}\\ x &\equiv a_2\pmod{m_2}\end{aligned}$ have a solution modulo $m_1m_2$.

Proving the Chinese Remainder Theorem

Proof

Suppose given $m_1$ and $m_2$ coprime.

We’ll solve two of the easiest imaginable pairs of simultaneous congruences first, and then we’ll observe that, in fact, that’s enough work to do anything.

The first easy pair of simultaneous congruences is $\begin{aligned} y &\equiv 1\pmod{m_1}\\ y &\equiv 0\pmod{m_2}.\end{aligned}$ The first equation says that $y=1-km_1$ for some $k$, and the second says that $y$ is a multiple of $m_2$. In other words, we have $m_2\mid 1-km_1$, so $km_1\equiv 1\pmod{m_2}.$ But $m_1$ and $m_2$ are coprime, so we know we can solve this.

Another easy pair of simultaneous congruences are $\begin{aligned} z &\equiv 0\pmod{m_1}\\ z &\equiv 1\pmod{m_2}.\end{aligned}$ This looks exactly the same, but the other way around: the second says that $z$ is of the form $z=1-lm_2$ for some $l$, and the first says that $z$ is a multiple of $m_1$. In other words, we need $lm_2\equiv 1\pmod{m_1}.$ We know we can do this.

In fact, instead of going through the method twice, the same process does both these pairs of congruences: if we use Euclid’s algorithm to give a solution to $rm_1 + sm_2 = 1,$ in fact taking $z=rm_1$ and $y=sm_2$ gives us what we want:

What then of our original equations $\begin{aligned} x &\equiv a_1\pmod{m_1}\\ x &\equiv a_2\pmod{m_2}?\end{aligned}$ I claim that if we take $x = a_1y + a_2z$, we have what we need.

Indeed, since $y\equiv 1\pmod{m_1}$ and $z\equiv 0\pmod{m_1}$, we have $x = a_1y + a_2z \equiv a_1\pmod{m_1},$ while, since $y\equiv 0\pmod{m_2}$ and $z\equiv 1\pmod{m_2}$, we have $x = a_1y + a_2z \equiv a_2\pmod{m_1}.$ Both of those are exactly what we needed.

This gives us a new way of finding solutions, which I’ll show off:

What are the solutions to: $\begin{aligned} x &\equiv 11 \pmod{14}\\ x &\equiv 10 \pmod{17}?\end{aligned}$

We’ll use our “building blocks” from the proof of the Chinese Remainder Theorem. In order to find this, we discovered we needed to invert $17$ mod $14$: we need to solve $14r+17s=1.$ This has a solution $5\times 17-6\times 14=1$.

As a result $5\times 17=85$ is congruent to $1$ mod $14$ and $0$ modulo $17$, and $-6\times14=-84$ is congruent to $0$ mod $14$ and $1$ modulo $17$.

Hence our solution is $11\times 85 + 10\times(-84) \equiv 95\pmod{238}.$

The bit in the statement which says that the moduli have to be coprime is definitely important!

Consider the following: $\begin{aligned} x &\equiv 2 \pmod{5}\\ x &\equiv 2 \pmod{5}\end{aligned}$ These have a solution, but it’s $x \equiv 2 \pmod{5}$, and not modulo 25.

On the other hand, these $\begin{aligned} x &\equiv 2 \pmod{5}\\ x &\equiv 3 \pmod{5}\end{aligned}$ don’t have any solution, modulo anything.

Similarly, you can check that $\begin{aligned} x &\equiv 17 \pmod{30}\\ x &\equiv 7 \pmod{20}\end{aligned}$ have a solution, which is $x\equiv 47\pmod{60}$.

We won’t prove it, but the rules are this:

• You can check whether two congruences with moduli $m_1$ and $m_2$ agree by looking what they say modulo $\gcd(m_1,m_2)$ (for example, the two above agree, because they both say $x\equiv 7\pmod{10}$);

• If two congruences agree, they have a common solution modulo $\operatorname{lcm}(m_1,m_2)$.

More calculations modulo primes

Earlier, we pointed out that modular arithmetic modulo primes is very well-behaved: every nonzero residue is invertible.

We’ll going to go on and use that.

The first thing we’ll talk about is exponentiation in modular arithmetic.

In integer arithmetic, it’s usually stupid to try to calculate very large powers: for example, $3^{1234}$ has a huge number of digits ($589$ of them, to be precise).

But, in modular arithmetic there are no large numbers. For example $3^{1234}$ will be congruent to something between $0$ and $9$ modulo $10$, and it’s reasonable to ask what.

One very stupid way of working it out would be to do the multiplication in the integers, then divide by $10$ and find the remainder.

We can do better, by doing our arithmetic all modulo $10$ in the first place. So: $\begin{aligned} {}3^2 &= 3\times 3\equiv 9\pmod{10},\\ {}3^3 &= 3\times 3^2\equiv 3\times 9\equiv 7\pmod{10},\\ {}3^4 &= 3\times 3^3\equiv 3\times 7\equiv 1\pmod{10}\ldots\end{aligned}$ That’s still going to be a lot of multiplication, if we keep multiplying by $3$ (modulo $10$) more than a thousand times!

There are considerably more intelligent ways. For example, we can square modulo $10$ quite quickly.

That lets us do some exponents by repeated squaring. For example, $3^8 = 3^{2\times 4} = {(3^2)}^4 = {(3^2)}^{2\times 2} = {\left({(3^2)}^2\right)}^2.$

1234 isn’t quite as nice, but we can do $\begin{aligned} 3^{1234} {}\equiv 3^{2\times 617} {}\equiv (3^2)^{617} {}\equiv 9^{617} {}\equiv 9^{2\times 308+1} {}\equiv (9^2)^{308}9\end{aligned}$ and end up getting the answer.

We’ll end up only multiplying about twenty times if we do it this way: that’s much less!

But, in fact, there’s a method that’s even faster still for this situation. We’ve just computed that $3^4\equiv 1\pmod{10}.$ That does almost all the work for us, as $3^{4k} {}= (3^4)^k {}\equiv 1^k {}= 1\pmod{10}.$ Hence $3^{1234} {}= 3^{4\times 308 + 2} {}= 3^{4\times 308}3^2{}= (3^4)^{308}3^2 {}\equiv 1^{308}3^2 {}\equiv 3^2 {}\equiv 9\pmod{10}.$ That makes the whole thing easy.

The relevant observation here was really that there was some integer $n$ such that $3^n\equiv 1\pmod{10}$. So two obvious questions are:

1. When does there exist such an $n$?

2. When it does exist, can we compute it?

Our answer to the first is not too difficult: