The Chinese Remainder Theorem

Sun Tsu Suan-Ching (4th century AD):

There are certain things whose number is unknown. Repeatedly divided by 3, the remainder is 2; by 5 the remainder is 3; and by 7 the remainder is 2. What will be the number?

In modern notation, Sun is asking us to solve the following system of equations:

$$x \equiv 2 \pmod{3}$$

$$x \equiv 3 \pmod{5}$$

$$x \equiv 2 \pmod{7}$$

The Chinese Remainder Theorem asserts that a solution to Sun's question exists, and the proof gives a method to find a solution.

Theorem 2.1 (The Chinese Remainder Theorem)   Let $a, b\in\mathbb{Z}$ and $n,m\in\mathbb{N}$ such that $\gcd(n,m)=1$. Then there exists $x\in\mathbb{Z}$ such that

$$x \equiv a\pmod{m}$$

$$x \equiv b\pmod{n}$$

Proof. The equation

$$tm \equiv b-a\pmod{n}$$

has a solution $t$ since $\gcd(m,n)=1$. Set $x=a+tm$. We next verify that $x$ is a solution to the two equations. Then

$$x \equiv a + (b-a) \equiv b \pmod{n},$$

and

$$x = a+tm \equiv a\pmod{m}.$$

$\qedsymbol$

Now we can solve Sun's problem:

$$x \equiv 2 \pmod{3}$$

$$x \equiv 3 \pmod{5}$$

$$x \equiv 2 \pmod{7}.$$

First, we use the theorem to find a solution to the pair of equations

$$x \equiv 2 \pmod{3}$$

$$x \equiv 3 \pmod{5}.$$

Set $a=2$, $b=3$, $m=3$, $n=5$. Step 1 is to find a solution to $t\cdot 3 \equiv 3-2\pmod{5}$. A solution is $t=2$. Then $x=a+tm=2+2\cdot 3 = 8$. Since any $x'$ with $x'\equiv x\pmod{15}$ is also a solution to those two equations, we can solve all three equations by finding a solution to the pair of equations

$$x \equiv 8 \pmod{15}$$

$$x \equiv 2 \pmod{7}.$$

Again, we find a solution to $t\cdot 15 \equiv 2-8\pmod{7}$. A solution is $t = 1$, so

$$x=a+tm=8+15=23.$$

Note that there are other solutions. Any $x'\equiv x\pmod{3\cdot 5\cdot 7}$ is also a solution; e.g., $23+3\cdot 5\cdot 7 = 128$.

We can also solve Sun's problem in PARI:

? chinese(Mod(2,3),Mod(3,5))
%13 = Mod(8, 15)
? chinese(Mod(8,15),Mod(2,7))
%14 = Mod(23, 105)