Linear Congruences ax ≡bmod m
Theorem 1. If (a, m) = 1, then the congruence ax ≡bmod mphas exactly one solution
modulo m.
Constructive. Solve the linear system
sa +tm = 1.
Then
sba +tbm =b.
So
sba ≡b(mod m)
gives the solution x=sb.
If u1and u2are solutions, then
au1≡b(mod m) and au2≡b(mod m)
=⇒au1≡au2(mod m)
=⇒u1≡u2(mod m) since (a, m) = 1.
So there is only one solution.
Example 1. 3x≡50 (mod 113)
Note that ax ≡b(mod m) implies ax =b+qm for some integer q. So a common divisor of
a, m also divides b.
Example 2. 5x≡1 (mod 15) is not solvable.
Theorem 2. Consider the congruence ax ≡b(mod m).
1. The congruence has a solution if and only if (a, m)|b.
2. If u0is any particular solution, then a complete set of solutions is:
u0, u0+m
g, u0+2m
g, . . . , u0+(g−1)m
g
where g= (a, m). Thus there are gsolutions.
3. A particular solution u0can be obtained by solving the congruence
a
gx≡b
g(mod m
g)
This is possible since a
g,m
g= 1. (See last theorem.)
Example 3. 42x≡12 (mod 78)
