Solving Linear Diophantine Equation with the Euclidean Algorithm

Solving Linear Diophantine Equation with the Euclidean Algorithm
Explore the fascinating world of solving linear Diophantine equations using the powerful Euclidean Algorithm! Join us on a journey through the step-by-step process of finding integer solutions to these intriguing mathematical puzzles. Unlock the secrets of Diophantine equations and elevate your problem-solving skills with this comprehensive tutorial. #lineardiophantineequation #euclideanalgorithm #mathematics #mathtutorial #problemsolving #algebra
Topics Covered:
1. Understanding the basics of Linear Diophantine equation in integers.
2. Analyzing the unique method of Euclidean Algorithm.
3. Step-by-step approach to solving equation for integers.
4. Tips and tricks for solving linear Diophantine equations
5. Particular and general solutions.
Timestamps:
0:00 Introduction
0:45 Steps of linear Diophantine equation
1:48 GCD
3:02 Euclidean algorithm
6:44 Particular solution
8:20 General solutions
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Thanks for Watching!
@infyGyan

[@mohammedsaysrashid3587]

Super wonderful explanation...thanks for sharing

[@infyGyan]

Thanks for watching 🙏

[@alibhukoo5400]

(x,y) =(3,-7),(12,-29),(21,-51),(30,-73),(39,-95),(48,-117)= any combo like x number(first number of x is 3 then always add 9) and for value of( y is then first number is -7 then always add -22)

[@alibhukoo5400]

x=(9n-6) and y=(15-22n) where n is the number of terms

[@mircoceccarelli6689]

👍👍👍
66x + 27y = 9
( x , y ) = ( - 6 + 9k , 15 - 22k )
= ( - 6 , 15 ) + k ( 9 , - 22 )
k € Z
😊🤪👋

[@allozovsky]

But why haven't you divided the whole equation by gcd(66, 27) = 3 right off at the very beginning of the solution? That would have given us a much simpler (but equivalent) form of the original equation: *22x + 9y = 3,* and we would have gotten exactly the same set of solutions anyway.

[@allozovsky]

But the presented solution is absolutely correct up to every tiny detail, of course. 👍

[@infyGyan]

Yes we can take that way too with gcd 1.
Thanks 👍

[@user-ji5su2uq9m]

another approach
66x + 27y = 9 => 22x + 9y = 3 => x ≡ 0 (mod 3)
let x = 3k, 66k + 9y = 3 => 22k + 3y = 1 => k ≡ 1 (mod 3)
let k = 3n +1, 22(3n +1) + 3y = 1, 22n + y + 7 = 0 => y = -7 - 22n
22*3k*(3n + 1) + 3*(-7 - 22n ) = 1
66kn + 22k - 21 - 66n = 1 => 3kn + k - 3n - 1 = (k - 1)*(3n + 1) = 0
so k = 1, x = 3k*(3n + 1) = 3 + 9n, y = -7 - 22n
(x,y) =(3 + 9n, -7 - 22n) which is same as (-6 +9k, 15 - 22k) with n = k -1

[@infyGyan]

Thanks for sharing!!