The document discusses linear congruences, defining congruence and linear congruence, establishing theorems about solutions, and demonstrating processes for solving specific examples. Key concepts include conditions for unique solutions, residue classes, and the existence of incongruent solutions based on the greatest common divisor of coefficients. Additionally, it outlines properties of integers in modular arithmetic and provides examples relating to finding solutions to congruences.

1. Linear Congruences, reduced
residue systems
Name of the unit: Congruences
Semester - 5, paper -7 , unit-4
Sri. B.Seethannaidu
Lecturer in Mathematics
SVLNS Government Degree College,
Bheemunipatnam

2. Congruence, Linear Congruence
• Congruence: Let 𝑚 be a fixed positive integer
and 𝑎, 𝑏 ∈ 𝑍, 𝑎 is said to be congruent to 𝑏 modulo
m if 𝑚 ∣ 𝑄 = 𝑏
• Linear Congruence: A polynomial congruence
of first degree is called a linear congruence.
• → Any linear congruence can be put in the form
𝑎𝑥 ≡ 𝑏(mod𝑚) where 𝑎 ≠ 0(mod𝑚)

3. Theorem: If (𝑎, 𝑚) = 1 then the linear
congruence 𝑎𝑥 ≡ 𝑏(mod𝑚) has a
unique solution.
proof:Given (𝑎, 𝑚) = 1
claim:The linear congruence 𝑎𝑥 ≡ 𝑏(mod𝑚) has a unique solution.
Since (𝑎, 𝑚) = 1
⇒ ∃𝑞, 𝑟 ∈ 𝑧 ⇒ 𝑎𝑞 + 𝑚𝑟 = 1
⇒ 𝑎𝑞𝑏 + 𝑚𝑟𝑏 = 𝑏
⇒ 𝑎(𝑞𝑏) − 𝑏 = 𝑚(−𝑟𝑏)
⇒ 𝑎𝑥0 − 𝑏 = 𝑘𝑚 where 𝑥0 = 𝑞𝑏, 𝑘 = −𝑟𝑏
⇒ 𝑚 ∣ 𝑎𝑥0 − 𝑏 ⇒ 𝑎𝑥0 ≡ 𝑏 mod𝑚 −−−−− −(1)
∴ 𝑥0 is a Solution of 𝑎𝑥 ≡ 𝑏(modm)
suppose if possible let 𝑥1 be any other Solution of 𝑎𝑥 ≡ 𝑏(modm)
⇒ 𝑎𝑥1 ≡ 𝑏(mod𝑚)
𝑏 ≡ 𝑎𝑥1 mod𝑚 −−−−−−−−−−−−− −(2)

4. From (1) &(2), 𝑎𝑥0 ≡ 𝑎𝑥1(modm)(𝐵𝑦 Transitivity)
⇒ 𝑚/𝑎𝑥0 − 𝑎𝑥1
⇒ 𝑚/𝑎 𝑥0 − 𝑥1
⇒ 𝑚/𝑥0 − 𝑥1 (∵ (𝑎, 𝑏) = 1)
𝑥0 ≡ 𝑥1(mod𝑚)
∴ 𝑎𝑥 ≡ 𝑏(modm) has a unique solution if (𝑎, 𝑚) = 1
Question :Solve the linear congruence, 16𝑥 ≡ 25 (mod 19)
Solution : Given linear congruence 16𝑥 ≡
25(mod19)
Comparing with 𝑎𝑥 ≡ 𝑏 (mod m)
𝑎 = 16, 𝑏 = 25, 𝑚 = 19
To Solve 16𝑥 ≡ 25
We add suitable Congruence or Congruences and reduce
it to the form

5. 𝑥 ≡ 𝑥0 (mod 19) which is the required Solution.
We have 247 ≡ 0(mod19)
0 ≡ 247(mad19) (Transitive property)
Adding with the given congruence
0 + 16𝑥 ≡ 247 + 25(mod19)
16𝑥 ≡ 272(mod19)
16𝑥 ≡ 16.17(mod19)
⇒ 𝑥 ≡ 17(mod19),
𝑥 = 17 + 19𝑡 where 𝑡 ∈ 𝑍
is the Set of congruent solutions of the Congruence

6. Solve the linear congruence
39𝑥 ≡ 65(mod52)
Solution:
Given linear congruence
39𝑥 ≡ 65(mod52)
comparing with 𝑎𝑥 ≡ 𝑏(mod𝑚)
𝑎 = 39, 𝑏 = 65, 𝑚 = 52
𝑑 = (𝑎, 𝑚) = (39,52) = 13.
Since 𝑑/𝑏 i. e 13 ∣ 65
By known theorem the
congruence
39𝑥 ≡ 65(mod52) has 13
incongruent solutions
The give Congruence
39𝑥 ≡ 65(mod52)
13.3𝑥 ≡ 13.5(mod52)

7. We have
𝑎𝑥 ≡ 𝑎𝑦 modm ⇔
𝑥 ≡ 𝑦 mod
𝑚
(𝑎, 𝑚)
3𝑥 ≡ 5 mod
52
13
⇒ 3𝑥 ≡ 5(mod4)
⇒ we have 4 ≡ 0(mod4)
0 ≡ 4(𝑚𝑜𝑑4) (Transitive
property)
Adding with 3𝑥 ≡ 5(mod4)
⇒ 3𝑥 ≡ 9(mod4)
⇒ 𝑥 ≡ 3 mod4
(∵ (3; 4) = 1. )
{ If 𝑎𝑏 ≡ 𝑎𝑐(mod𝑚) and
(𝑎, 𝑚) = 1
⇒ 𝑏 ≡ 𝑐(mod𝑚)}
By substituting in the given
congruence
we can see that
𝑥 = 3 is a solution
By known theorem
the 13 incongruent Solutions are
given 𝑥 = 3 + 4𝑡(mod52)
where 𝑡 = 0,1,2, ⋯ 12,
{∵ 𝑥 = 𝑥0 + 𝑡
𝑚
𝑎, 𝑚
}
ie 𝑥 =
3,7,11,15,19,23,27,31,35,39,43,
47,51(mod52)

8. Theorem: The linear congruence 𝑎𝑥 ≡
𝑏(mod𝑚)ℎ𝑎𝑠 a solution if (𝑎, 𝑚) ∣ 𝑏
Necessary part.
Suppose 𝑎𝑥 ≡ 𝑏(mod𝑚) has a
solution
claim: (𝑎, 𝑚)/𝑏
Let 𝑥0 be 𝑎 solution. of 𝑎𝑥 ≡
𝑏(mod𝑚)
𝑎𝑥0 ≡ 𝑏(mod𝑚)
⇒ 𝑚 ∣ 𝑎𝑥0 − 𝑏
⇒ 𝑎𝑥0 − 𝑏 = 𝑚𝑞 Where 𝑞 ∈ 𝑧
𝑎𝑥0 + 𝑚(−𝑞) = 𝑏
⇒ 𝑏 = 𝑎𝑥0 + 𝑚𝑦0 where 𝑦0 = −𝑞
if 𝑑 = (𝑎, 𝑚) then
d is the least positive integer in the
form 𝑎𝑥 + 𝑚𝑦
⇒ 𝑑 ∣ 𝑏
⇒ (𝑎, 𝑚) ∣ 𝑏
Sufficient part
Conversely suppose that (𝑎, 𝑚)/𝑏
claim: 𝑎𝑥 ≡ 𝑏(modm) has 𝑎 solution
Let (𝑎, 𝑚) = 𝑑 such that 𝑑 ∣ 𝑏
Since 𝑑 = (𝑎, 𝑚)
⇒ 𝑑 = 𝑎𝑥1 + 𝑚𝑦1 where 𝑥1, ∈ 𝑧,
𝑑)𝑏 ⇒ ∃𝑞1 ∈ 𝑧 ⇒ 𝑏 = 𝑑𝑞,
⇒ 𝑏 = 𝑞1 𝑎𝑥1 + 𝑚𝑦1
𝑏 − 𝑎 𝑥1, 𝑞1 = 𝑚 𝑞1𝑦1
⇒ 𝑚 ∣ 𝑏 − 𝑎 𝑥1𝑞
⇒ 𝑎 𝑥1, 𝑞1 ≡ 𝑏(mod𝑚)
∴ 𝑥1𝑞1 is 𝑎 solution of 𝑎𝑥 ≡ 𝑏(modm)

9. Residue classes:
If 𝑚 is a fixed tee integer then 𝑥0, 𝑥1, 𝑥2, … 𝑥𝑚−1 or
0, 1, 2, … 𝑚 − 1 are the residue classes modulo 𝑚.
Eg. If 𝑚 = 6 then
the residue classes modulo 6 are
0 = {⋯ , −12, −6,0,6,12, ⋯ } = {69: 9 ∈ 𝑧}
𝑇 = {⋯ − 11, −5,1,7,13, ⋯ } = {6𝑞 + 1: 𝑞 ∈ 𝑧}
𝑧 = {⋯ , 4,2,8,14, ⋯ } = {69 + 2: 9 ∈ 𝑧}
3 = {⋯ − 9 − 3,3,9,15, ⋯ } = {6𝑞 + 3: 𝑞 ∈ 𝑧}
4 = {⋯ − 8, −2,4,10,16, ⋯ } = {69 + 4: 𝑞 ∈ 𝑧}
5 = {⋯ − 7, −1,5,11,17, ⋯ } = {6𝑞 + 5; 𝑞 ∈ 𝑧}

10. Theorem: Let m be a positive integer and
𝑠 = {0,1,2, … 𝑚 − 1} then prove that no
two integers of 𝑠 are congruent modulo 𝑚,
proof: Given that 𝑚 is a tee integer and 𝑠 =
{0,1,2, … 𝑚 − 1}
Let 𝑎, 𝑏 ∈ 𝑠 and 𝑎 > 𝑏
0 ≤ 𝑎 < 𝑚, 0 ≤ 𝑏 < 𝑚
0 < a − 𝑏 < 𝑚
⇒ 𝑚 ∤ 𝑎 − 𝑏
⇒ 𝑎 ≡ 𝑏(mad 𝑚)
Hence no two integers of are congruent modulo m

11. Theorem: Let 𝑚 be a positive integer and
𝑠 = {0,1,2, ⋯ 𝑚 − 1}
then prove that every 𝑥 ∈ 𝑍 is congruent modulo 𝑚 to
ore of the integers of 𝑆.
proof. Given that 𝑚 is a tee integer and 𝑠 = {0,1,2, … 𝑚 − 1}
claim: 𝑥 ≡ 𝑟(mod𝑚)∀𝑥 ∈ 𝑧
By division algorithm for 𝑥 ∈ 𝑍
There exists unique integers 𝑞, 𝑟 such that 𝑥 = 𝑞𝑚 + 𝑟 where 0 ≤ 𝑟 <
𝑚
⇒ 𝑥 − 𝑟 = q𝑚, 𝑟 ∈ 𝑠
⇒ 𝑚/𝑥 − 𝑟
⇒ 𝑥 ≡ 𝑟(mod𝑚)
Hence for 𝑥 ∈ 𝑧 there exits one and only one integer r ∈ 𝑠 such that 𝑥 ≡
𝑟(mod𝑚)

12. • Set of Least positive residues modulo 𝐦
The remainder 𝑟, upon division of 𝑥 by 𝑚 is called the residue of 𝑥
mod m
The set of integers 𝑧𝑚 = {0,1,2, … 𝑚 − 1} is callers the least positive
residues modulo m
E.g: {0,1,2,3,4,5,6} is the set of least positive residue modulo 7 ,
Equivalence class or 𝜸-residue
class
The equivalence class 𝛾 is the
Set {𝑥 ∈ 𝑧: 𝑥 ≡ 𝑟(modm}
→ The set of 𝑚 equivalence
classes is denotes by 𝑧𝑚 =
0 , 1 , 2 ⋅ ⋯,m − 1 .}
complete set of residue classes
modulo m
Complete set of residue classes
modulo 𝑚 is the collection
0 , 1 , 2 ⋅ ⋯,m − 1 .}
A Set of integers
𝑥0, 𝑥1, … 𝑥𝑚−1 is called a
Complete set of residues,
if. 𝑥0, 𝑥1, … 𝑥𝑚−1 is the
complete Set of residue classes.

13. THANK YOU