Here is the equation
$$ 6a+9b+20c=16 $$ To solve this, i follow the below steps :
$\gcd(6,9)(2a+3b)+20c = 16$ let, $w = 2a+3b$ So, $3w+20c =16$ then, specific solution of $w = 112+20n$, $c = -16-3n$ ; where $n \in \mathbb{Z}$.
After that, $w = 2a+3b$ So, $2a+3b = 112+20n$, I need to solve this equ. To find the solution of $a$ and $b$.
Finally I got, $$ A = -(112+20n)+3m \\ B = 112+20n-2m \\ C = -16-3n $$
I tried to do above by Euclidean algorithm.
Now, my question is, is it ok or the solving is right? and how do i got the positive ans for $a,b,c$ ? If there's any wrong in solving process , please explain me that.
Consider $6a + 9b + 20c = 16$ as the equation of a plane, in $a,b,c$.
Once that you know three points in it, with integer coordinates, ${\bf v}_{\bf 0},{\bf v}_{\bf 1},{\bf v}_{\bf 2} $, from ${\bf v}_{\bf 1}-{\bf v}_{\bf 0}$ and ${\bf v}_{\bf 2}-{\bf v}_{\bf 0}$ you can determine two vectors ${\bf u}_{\bf 1},{\bf u}_{\bf 2} $ having components with $gcd=1$. Then all the solutions will be given by $$ {\bf v} = {\bf v}_{\bf 0} + n\,{\bf u}_{\bf 1} + m\,{\bf u}_2 \quad \left| {\;{\rm integers}\;n,m} \right. $$ The ${\bf u}$ vectors can be more easily determined considering the parallel plane through the origin (the "homogeneous" solution). Therefore consider $$ 6a + 9b + 20c = 0 $$ here ${\bf v}_0 = \left( {0,0,0} \right)$ and ${\bf u}_{\bf 1} = \left( {3, - 2,0} \right)\quad {\bf u}_{\bf 2} = \left( {10,0, - 3} \right)$ follow immediately by putting one of the variables at zero.
Then we can easily find a "particular" solution by putting, e.g., $b=0$ and achieving $$ 6a + 9b + 20c = 16\quad \to \quad 6a + 0 + 20c = 16\quad \to \quad 3a + 0 + 10c = 8\quad \to \quad {\bf v}_{\bf p} = \left( {6,0, - 1} \right) $$ The solution is therefore
${\bf v} = {\bf v}_{\bf p} + n\,{\bf u}_{\bf 1} + m\,{\bf u}_2\quad$ that is: $\quad \left\{ \matrix{ a = 6 + 3n + 10m \hfill \cr b = - 2n \hfill \cr c = - 1 - 3m \hfill \cr} \right.$
In fact: $\quad16 = 6a + 9b + 20c = 16 + 0n + 0m$