The Chinese remainder theorem is about solutions to congruences so let’s review solutions to linear congruences.

$$ \begin{align*} ax \equiv b \pmod{m} \end{align*} $$

This is solvable if and only \(b\) is divisible by \((a,m)\). Euclid’s algorithm can show this is sufficient. For example \(6x \equiv 7 \pmod{45}\) has no solutions since \((6,45) = 3\) but \(3 \not\mid 7\). On the other hand, \(6x \equiv 3 \pmod{45}\) has a solution since \((6,45) \mid 3\). How do we find the solution? \(6x \equiv 3 \pmod{45}\) implies that

$$ \begin{align*} 6x &= 3 + 45y \\ 2x &= 1 + 15y \end{align*} $$

We can now use Euclide’s algorithm to see that

$$ \begin{align*} x \equiv 8 \pmod{15} \end{align*} $$

So now \(8\) is a solution to the original equation but so is \(8+15, 8+30\) since we’re still okay module \(45\). Only when \((a,m) = 1\)m then the solution is unique and always exists. If \((a,m) > 1\), the solution might not exist or we might have many solution.

Polynomial Congruences

Suppose we want to solve a polynomial congruence such as

$$ \begin{align*} f(x) \equiv 0 \pmod{m} \end{align*} $$

There are a few cases

  • \(m\) is prime. This is easier since since if \((a,m) = 1\), then \(a\) has an inverse module \(m\)
  • \(m=p^n\) is a prime power. This can be reduced to the first case (we will see this in a later lecture).
  • \(m=p_1^{n_1}p_2^{n_2}\cdots\). We will reduce this to the case \(m = p^n\) using the Chinese Remainder Theorem (CRT).

In this lecture, we will study the reduction from case 3 to case 2.

System of Two Linear Congruences

Suppose we want to solve the following system of linear congruences

$$ \begin{align*} x \equiv a_1 \pmod{m_1} \\ x \equiv a_2 \pmod{m_2} \end{align*} $$

In general, to have a solution to this system we want \(m_1\) and \(m_2\) to be coprime. This solution is even unique when \(m_1\) and \(m_2\) are coprime. If they were not co-prime, then it’s not guaranteed. We could or could not have a solution.

Example

As an example, suppose we want to solve the following system of linear congruences

$$ \begin{align*} x \equiv 1 \pmod{17} \\ x \equiv 3 \pmod{21} \end{align*} $$

Here, we see that \((17,21) = 1\). So then we have

$$ \begin{align*} x = 1 + 17y = 3 + 21z \end{align*} $$

which simplifes to

$$ \begin{align*} -21z + 17y = 2 \end{align*} $$

By Euclid’s algorithm, \(-21z + 17y = 1\) has the solution \(z = 4, y = 5\). So now we can just double to get

$$ \begin{align*} z = 8, y = 10 \end{align*} $$

System of Three Linear Congruences

Suppose now that we have more than two linear congruences. Then by CRT:

The following system of congruences $$ \begin{align*} x &\equiv a_1 \pmod{m_1} \\ x &\equiv a_2 \pmod{m_2} \\ \cdots \\ x &\equiv a_k \pmod{m_k} \end{align*} $$ has a unique solution module \(M = m_1m_2\cdots m_k\) if \(m_1, m_2, \cdots m_k\) are pairwise coprime

Proof

So let’s focus on the first two equations and instead of using Euclid’s algorithm, let’s define a map from the set of all numbers from \(0\) up to \(m_1m_2 - 1\) to the product set of the numbers from \(0\) to \(m_1\) and \(0\) to \(m_2\).

$$ \begin{align*} \{0,1,2,\cdots,m_1m_2 - 1\} &\longrightarrow \{0,1,2,\cdots,m_1\} \times \{0,1,2,\cdots,m_2\} \\ \end{align*} $$

Given by

$$ \begin{align*} f(x) &\longrightarrow (x \bmod m_1, x \bmod m_2) \end{align*} $$

We claim that this is is a bijection. First we will show that \(f\) is injective. So suppose that \(f(x) = f(y)\), then we want to show that \(x = y\). Since \(f(x) = f(y)\), then

$$ \begin{align*} (x \bmod m_1, \bmod m_2) = (y \bmod m_1, y \bmod m_2) \end{align*} $$

Therefore

$$ \begin{align*} x \equiv y &\pmod{m_1} \\ x \equiv y &\pmod{m_2} \end{align*} $$

But equations imply \(x - y\) is divisible by \(m_1\) and \(m_2\). But \(m_1\) and \(m_2\) are coprime so \(x-y\) must also be divisible by the product \(m_1m_2\) which means that

$$ \begin{align*} x \equiv y \pmod{m_1m_2} \end{align*} $$

This implies that \(f\) is injective. Now, since we have an injective map between two sets of equal size, then this map must be surjective as well. Therefore, \(f\) is a bijection.

So now since this map is a bijection, then given a pair \((a_1, a_2) \in \mathbb{Z}/m_1, \mathbb{Z}/m_2\), this pair must correspond to exactly one \(x\) in \(\mathbb{Z}/(m_1m_2)\) such that

$$ \begin{align*} x &\equiv a_1 \pmod{m_1} \\ x &\equiv a_2 \pmod{m_2} \end{align*} $$

which is what we wanted to show. Note that this argument actually fails when \(m_1\) and \(m_2\) are not coprime. The map will not be injective.

Example

Suppose we want to solve the following

$$ \begin{align*} x^2 \equiv x \pmod{10^n} \end{align*} $$

What does this equation mean? We want to find a number such that they have the same last digit if you square it. For example, 6^2 = 36. Note here that that \(6^2 \pmod{10^1} = 6\). Similarly, \(76^2 = 5776 \pmod{10^3} = 6\). Note that \(10^n = 2^n \cdot 5^n\) and that \((5^n, 2^n) = 1\). By CRT, this is the same as solving the following two equations

$$ \begin{align*} x^2 &\equiv x \pmod{5^n} \\ x^2 &\equiv x \pmod{2^n} \end{align*} $$

For \(x^2 \equiv x \pmod{5^n}\), observe that

$$ \begin{align*} x^2 &\equiv x \pmod{5^n} \\ x^2 -x &\equiv 0 \pmod{5^n} \\ x(x-1) &\equiv 0 \pmod{5^n} \\ \end{align*} $$

There are at least two solutions \(x \equiv 0 \pmod{5^n}\) and \(x \equiv 1 \pmod{5^n}\). Similarly, for \(x^5 \equiv x \pmod{2^n}\), we also get the same two solutions. So now by CRT, we can combine these solutions. For example. When \(x \equiv 1\) in both,

$$ \begin{align*} x \equiv 1 \pmod{2^n} \\ x \equiv 1 \pmod{5^n} \end{align*} $$

By CRT,

$$ \begin{align*} x \equiv 1 \pmod{10^n} \end{align*} $$

so yes \(1^2 \equiv 1 \pmod{10^n}\) though it’s not an interesting solution. Let’s try another combination

$$ \begin{align*} x \equiv 0 \pmod{2^n} \\ x \equiv 1 \pmod{5^n} \end{align*} $$

From the first equation, \(x\) is divisible by \(2^n\) so \(x = 2^n \cdot t\). Plugging this in into the second equation

$$ \begin{align*} 2^nt &\equiv 1 \pmod{5^n} \end{align*} $$

Suppose \(n = 1\), then

$$ \begin{align*} 2t &\equiv 1 \pmod{5} \end{align*} $$

The smallest \(t\) satisfying this is \(t = 3\) since \(6 \equiv 1 \pmod{5}\). Then \(x = 2^n \cdot t = 2^1 \cdot 3 = 6\) so \(x = 6\) is a solution. If we try \(n = 2\), then

$$ \begin{align*} 4t &\equiv 1 \pmod{25} \end{align*} $$

The smallest \(t\) satisfying this is \(t = 19\). Then \(x = 2^2 \cdot 19 = 76\) so \(x = 76\) is a solution.

References