Linear congruence theorem

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In modular arithmetic, the question of when a linear congruence can be solved is answered by the linear congruence theorem. If a and b are any integers and n is a positive integer, then the congruence

ax \equiv b \pmod {n}

has a solution for x if and only if b is divisible by the greatest common divisor d of a and n (denoted by gcd(a,n)). When this is the case, and x0 is one solution of (1), then the set of all solutions is given by

\{x_0+k\frac{n}{d}\mid k\in\Bbb{Z}\}.

In particular, there will be exactly d = gcd(a,n) solutions in the set of residues {0,1,2,...,n − 1}. The result is a simple consequence of Bézout's identity.

Example[edit]

For example, examining the equation ax ≡ 2 (mod 6) with different values of a yields

3x \equiv 2 \pmod 6\

Here d = gcd(3,6) = 3 but since 3 does not divide 2, there is no solution.

5x \equiv 2 \pmod 6\

Here d = gcd(5,6) = 1, which divides any b, and so there is just one solution in {0,1,2,3,4,5}: x = 4.

4x \equiv 2 \pmod {6}\

Here d = gcd(4,6) = 2, which does divide 2, and so there are exactly two solutions in {0,1,2,3,4,5}: x = 2 and x = 5.

Solving a linear congruence[edit]

In general solving equations of the form:

ax \equiv b \pmod {n}

If the greatest common divisor d = gcd(a, n) divides b, then we can find a solution x to the congruence as follows: the extended Euclidean algorithm yields integers r and s such ra + sn = d. Then x = rb/d is a solution. The other solutions are the numbers congruent to x modulo n/d.

For example, the congruence

12x \equiv 20 \pmod {28}\

has 4 solutions since gcd(12, 28) = 4 divides 20. The extended Euclidean algorithm gives (−2)·12 + 1·28 = 4, i.e. r = −2 and s = 1. Therefore, one solution is x = −2·20/4 = −10, and −10 = 4 modulo 7. All other solutions will also be congruent to 4 modulo 7. Since the original equation uses modulo 28, the entire solution set in the range from 0 to 27 is {4, 11, 18, 25}.

System of linear congruences[edit]

By repeatedly using the linear congruence theorem, one can also solve systems of linear congruences, as in the following example: find all numbers x such that

2x \equiv 2 \pmod {6}
3x \equiv 2 \pmod {7}
2x \equiv 4 \pmod {8}

By solving the first congruence using the method explained above, we find x \equiv 1 \pmod{3}, which can also be written as x = 3k + 1. Substituting this into the second congruence and simplifying, we get

9k \equiv - 1 \pmod{7}.

Solving this congruence yields k \equiv 3 \pmod{7}, or k = 7l + 3. It then follows that x = 3 (7l + 3) + 1 = 21l + 10. Substituting this into the third congruence and simplifying, we get

42l \equiv - 16 \pmod{8}

which has the solution l \equiv 0 \pmod{4}, or l = 4m. This yields x = 21(4m) + 10 = 84m + 10, or

\equiv 10 \pmod{84}

which describes all solutions to the system.

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