Chinese Remainder Theorem
Theorem
Let $a, b \in \Z$.
Let $r$ and $s$ be coprime integers.
Then:
- $a \equiv b \pmod {r s}$ if and only if $a \equiv b \pmod r$ and $a \equiv b \pmod s$
where $a \equiv b \pmod r$ denotes that $a$ is congruent modulo $r$ to $b$.
General Result
Let $b_1, b_2, \ldots, b_r \in \Z$.
Let $n_1, n_2, \ldots, n_r$ be pairwise coprime positive integers.
Let $\ds N = \prod_{i \mathop = 1}^r n_i$.
Then the system of linear congruences:
\(\ds x\) | \(\equiv\) | \(\ds b_1\) | \(\ds \pmod {n_1}\) | |||||||||||
\(\ds x\) | \(\equiv\) | \(\ds b_2\) | \(\ds \pmod {n_2}\) | |||||||||||
\(\ds \) | \(\vdots\) | \(\ds \) | ||||||||||||
\(\ds x\) | \(\equiv\) | \(\ds b_r\) | \(\ds \pmod {n_r}\) |
has a simultaneous solution which is unique modulo $N$.
Ultrageneral Result
Let $n_1, n_2, \ldots, n_k$ be positive integers.
Let $b_1, b_2, \ldots, b_k$ be integers such that:
- $\forall i \ne j: \gcd \set {n_i, n_j} \divides b_i - b_j$
Then the system of linear congruences:
\(\ds x\) | \(\equiv\) | \(\ds b_1\) | \(\ds \pmod {n_1}\) | |||||||||||
\(\ds x\) | \(\equiv\) | \(\ds b_2\) | \(\ds \pmod {n_2}\) | |||||||||||
\(\ds \) | \(\vdots\) | \(\ds \) | ||||||||||||
\(\ds x\) | \(\equiv\) | \(\ds b_k\) | \(\ds \pmod {n_k}\) |
has a simultaneous solution which is unique modulo $\lcm \set {n_1, \ldots, n_k}$.
Proof
Necessary Condition
This is proved in Congruence by Divisor of Modulus.
Note that for this result it is not required that $r \perp s$.
$\Box$
Sufficient Condition
Now suppose that $a \equiv b \pmod r$ and $a \equiv b \pmod s$.
We have by definition of congruence:
- $a \equiv b \pmod r \implies \exists k \in \Z: a - b = k r$
From $a \equiv b \pmod s$ we also have that:
- $k r \equiv 0 \pmod s$
As $r \perp s$, we have from Common Factor Cancelling in Congruence:
- $k \equiv 0 \pmod s$
So:
- $\exists q \in \Z: a - b = q s r$
Hence by definition of congruence:
- $a \equiv b \pmod {r s}$
$\blacksquare$
Examples
$n \equiv 7 \pmod {12}$ implies $n \equiv 3 \pmod 4$
Let $n \equiv 7 \pmod {12}$.
Then:
- $n \equiv 3 \pmod 4$
Warning
Let $r$ not be coprime to $s$.
Then it is not necessarily the case that:
- $a \equiv b \pmod {r s}$ if and only if $a \equiv b \pmod r$ and $a \equiv b \pmod s$
where $a \equiv b \pmod r$ denotes that $a$ is congruent modulo $r$ to $b$.
Historical Note
The Chinese Remainder Theorem was found in the Sun Tzu Suan Ching of Sun Tzu, who was active in China sometime between $200$ and $500$ C.E (nobody knows exactly when).
The principle is believed to have been used by Chinese calendar makers as far back as $100$ C.E.
Sources
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): remainder theorem: 2.
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Sun Tsu Suan Ching: $69$
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.4$: Integer Functions and Elementary Number Theory: Law $\text{D}$
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $18$