Linear Diophantine Equations with 3 Variables - 3 Different Methods

We want to solve the linear Diophantine equation with 3 variables:

35x+55y+77z=1
for integer solutions in

Three methods are discussed:
1. Split the equation into two linear equation each of which has two variables.
2. Parameterize with canonical form
3. Particular solution and general solutions of homogeneous equations

Check out the videos on the GCD, Euclidean algorithms here

Two Basic Theorems on gcd (Greatest Common Divisors) of Two Integers (Bezout's Identity)

An Example of GCD, and Extended Euclidean Algorithm In Finding the Bezout Coefficients

GCD, Euclidean Algorithm and Bezout Coefficients

00:00 Introduction
00:21 Review of Equation with 2 Variables
02:24 Method 1: Split Into 2 Equations
06:00 Method 2: Parameterize with Canonical Form
13:30 Method 3: Particular solution and General solutions

[oscarmontesan]

I think you made a mistake in the first method, as to get the general solution for Z you should do (Z1 - A/gcd * S) being Z1 the particular solution for Z, A the coeficient of Y and S a random integer. Well your soultion is Z=-2-11S but it should be Z=-2-5S as 5 is the coeficient of Y in the original equation (and it is divided by gdc=1 but that does not change anything)

Besides that, first method works great.

thanks for sharing!

[bradgilbertvalencia]

hello. for method 1, are the values of s and t always equal?

[onegreengoat]

I understand this was posted over a year ago, but I'm certain that your method 1 solution is incorrect. If t=0 and s=1, you would have x=5, y=160 and z=-208, which gives a large negative number, not 1. I think it came from a typo. What you're looking for is the following:

x=-5+11t
y=(3+7s)(16-35t)
z=(-2-5s)(16-35t)

[asharhaque]

Hi, this is one of the best videos outlining, in brief, the different methods to solve a Diophantine Equation.

Which book can I find questions related to "parameterizing with canonical form". I will be really grateful if you can help out because I need it for a project. Thank you for all your help in advance.

[dannkod]

My solution involves doing some modulo 5, 7 and 11 since they all are divisors of exact two of the numbers.
Modulo 5: z = 3 => z = 5c + 3
Modulo 7: y = 6 => y = 7b + 6
Modulo 11: x = 6 => x = 11a + 6

35(11a+6) + 55(7b+6) + 77(5c+3)=1
a+b+c = -2
(a, b, c) ∈ {(a, b, -(a+b+2) | a, b ∈ Z}
(x, y, z) ∈ {(11a+6, 7b+6, -5a-5b-7) | a, b ∈ Z}

[misfortune]

please explain thoroughly, i don't understand the flow of your solution. Some part is just to0 fast, can't keep up

[brunovuletic]

US and NATO planes and drones fly from the Black Sea every day for 3 years. What exactly is your problem?