Finding a solution to the Diophantine equation 6x + 10y + 45z =1

[MidgetDwarf]

Homework Statement

Find a solution to the Diophantine equation 6x + 10y + 45z = 1.

Relevant Equations

The following hint is given:

First express gcd(6,10) as a linear combination of 6 and 10. Then, express 1 as a linear combination of 45 and gcd(6,10).

I know from a previous result, that gcd of two nonzero integers a and b, can be written as
aX + bY = d. Where d is the gcd.

For the first sentence of the hint.

6x + 10y = 2. Hence, 6(2) + 10 (-1) = 2

45X + 2Y = 1. Where (1, -22) is an integral solution.

f

 

[fresh_42]

Homework Statement:: Find a solution to the Diophantine equation 6x + 10y + 45z = 1.
Relevant Equations:: The following hint is given:

First express gcd(6,10) as a linear combination of 6 and 10. Then, express 1 as a linear combination of 45 and gcd(6,10).

I know from a previous result, that gcd of two nonzero integers a and b, can be written as
aX + bY = d. Where d is the gcd.

For the first sentence of the hint.

6x + 10y = 2. Hence, 6(2) + 10 (-1) = 2

45X + 2Y = 1. Where (1, -22) is an integral solution.

f

So what?

 

[MidgetDwarf]

I actually found the solution out. I posted this thread on accident (working on two computers).

Solve 45a + 2b = 1.
So (1, -22) is an integral solution.
Then let z= 1 in the equation 6x + 10y + 45z = 1.
Hence, 6x + 10y = -44. Which simplifies to 3x +5y = -22.

Solving 3x + 5y = 1 . We find that (2,-1) is an integral solution. Then multiply (2,-1) by -22. To get the solution
(-44,22) for 6x + 10y = -44.

Putting this all together, (-44, 22, 1) is an integral solution of the original Diophantine equation.

Not sure if this method is valid in solving Diophantine equations of the form ax+by+cz= 1.
I am assuming that we need to impose the condition that a, b, and c are relatively prime in order to ensure existence of solution.

Yes, just proved that it is required the a, b , and c be relatively prime.