This document defines and provides examples of linear Diophantine equations. It begins with a brief biography of Diophantus of Alexandria, who studied algebraic equations, including what are now called Diophantine equations. It then defines a linear Diophantine equation as a first-degree polynomial equation with integer solutions. The document provides a method to find the initial solution using the Euclidean algorithm and proves when a solution exists. It includes examples solving linear Diophantine equations in two variables and finding their integer solutions.

1. Linear Diophantine
Equations
By: MARY HARLENE BANATE
CHRISTY MAE JOY D. DOLFO
MATH 413

2. Diophantus of alexandria
BIOGRAPHY
- (Born c. AD 200- c. 214; died c.
AD 284- c. 298) was a Greek
Mathematician, who was the author
of a series of books called
Arithmetica, many of which are now
lost. His texts deal with solving
algebraic equations.

3. BIOGRAPHY
Title page of the original
1621 edition of the latin
translation by Claude
Gaspard Bachet de Mezziriac
of Diophantus Arithmetica
Diophantine equations,
Diophantine geometry, and
Diophantine approximations
are subareas of Number
Theory that are named after
him.

4. What is Linear Diophantine Equation?
 A Diophantine Equation is a polynomial equation whose
solutions are restricted to integers.
 A Linear Diophantine Equation is a first-degree equation
of this type. Diophantine equations are important when a
problem requires a solution in whole amounts.
 We can write ax + by = c where a, b, c ∈ Z

5. Method for computing the initial solution
to a Linear Diophantine Equation in Two
Variables
Given an equation ax + by = c
 Use the Euclidean algorithm to compute (a,b)=d,
taking care to record all steps.
 Determine if d | n. If not, then there are no solutions.

6. Method for computing the initial solution
to a Linear Diophantine Equation in Two
Variables

7. Theorem
Let a,b, and c be integers with a and b not both zero.
The linear diophantine equation
ax + by = c
has a solution if and only if d =(a, b) divides c.
Proof
Suppose that x 0 and y 0 is a solution. Then ax 0 + by 0 = c. Since d|a
and d|b, we get that d|ax 0 + by 0 and d|c.

8. 6x+9y=113
a=6 b=9 c=113
Using Euclidean
Algorithm
9 = 6(1) + 3
6 = 3(2) + 0
CDG(6,9)=3
3 does not divide 113
∴has no solution
10x+15y=33
a=10 b=15 c=33
Using Euclidean
Algorithm
15 = 10(1) + 5
10 = 5(2) + 0
GCD(10,15)=5
5 does not divide 33
∴has no solution
Example 1 Example 2

9. 54x+21y=3
Using Euclidean
Algorithm
54 = 21(2) +12
21 = 12(1) +9
12 = 9(1) +3
9 = 3(3) +0
CDG(54,21)=3
3 divides 3
∴has many solution
Example 3
3 = 12 – 9
3 = 12 – (21 -12)
= 12 – 21 + 12
3 = 12(2) – 21
3 = [54 – 21(2)] (2) – 21
3 = 54(2) – 21(4) – 21
Factor +21
3 = 54(2) + 21(-5)
c = ax + by
∴ x = 2 y = -5

10. 18x + 5y = 48
Using Euclidean
Algorithm
18 = 5(3) + 3
5 = 3(1) + 2
3 = 2(1) + 1
2 = 2(1) + 0
CDG(18,5)=1
1 divides 48
∴has many solution
1 = 3 – 2
1 = 3 – (5 -3)
= 3 – 5 + 3
1 = 3(2) – 5
1 = [18 – 5(3)] (2) – 5
1 = 18(2) – 5(6) – 5
Factor +5
1 = 18(2) + 5(-7)
c = ax + by
∴ x = 2 y = -7
Example 4

11. Practice
16x + 10y = 17
Using Euclidean
Algorithm
16 = 10(1) + 6
10 = 6(1) + 4
6 = 4(1) + 2
4 = 2(2) + 0
CDG(16,10)=2
2 does not divide 17
∴has no solution
20x + 6y = 55
Using Euclidean
Algorithm
20 = 6(3) + 2
6 = 2(3) + 0
CDG(20,6)=2
2 does not divide 55
∴has no solution

12. 45x + 33y = 9
Using Euclidean
Algorithm
45 = 33(1) + 12
33 = 12(2) + 9
12 = 9(1) + 3
9 = 3(3) + 0
CDG(45,33)=3
3 divides 9
∴has many solution
3 = 12 – 9
3 = 12 – [33 – 12(2)]
= 12 – 33 + 12(2)
Factor 12
3 = 12(3) – 33
3 = (45 – 33)(3) – 33
= 45(3) – 33(3) - 33
Factor +33
3 = 45(3) + 33(-4)
c = ax + by
∴ x = 3 y = -4
Practice

13. 15x + 11y = 3
Using Euclidean
Algorithm
15 = 11(1) + 4
11 = 4(2) + 3
4 = 3(1) + 1
3 = 1(3) + 0
CDG(15,11)=1
1 divides 3
∴has many solution
1 = 4 – 3
1 = 4 – [11 –4(2)]
= 4 – 11 + 4(2)
Factor 4
1 = 4(3) – 11
1 = (15 – 11)(3) – 11
= 15(3) – 11(3) - 11
Factor +11
1 = 15(3) + 11(-4)
c = ax + by
∴ x = 3 y = -4
Practice