Find all the solutions of $6a+9b+20c=-2$

Question

Find all integer solutions of $$6a+9b+20c=-2.$$

I did learn how to show there is no solution. First I find the solution set must also satisfy $$6a\equiv-2\mod 20$$ and so on to find a contradiction. But how to find all the solutions, if they exist?

Answer 1

We search for integer $a,b,c$ such that $6a+9b+20c=-2$. This means that $$a=-\frac{20c+9b+2}{6}$$ i.e $$20c+9b\equiv 4 (\mod 6)$$. Now we consider two cases:

1) $b$ is even $\Rightarrow 9b\equiv 0 (\mod 6)\Rightarrow 20c\equiv 4 (\mod 6)\Leftrightarrow 18c+2c\equiv 4 (\mod 6)$ $\Rightarrow 2c\equiv 4(\mod 6)\Rightarrow 2c=6k+4$ for $k\in\mathbb Z\Rightarrow c=\frac{6k+4}{2}$. Therefore the solution in this case is

$b-$ even, $c=\frac{6k+4}{2},\quad k=\pm 1, \pm 2, ..., a=-\frac{20c+9b+2}{6}$.

2) $b$ is odd. Then plugging in the equation $6a+9b+20c=-2$, we see that LHS is odd and RHS is even, which is impossible.

Finally the solution set is described in the first case:

$$b=2l,\quad l\in\mathbb Z,\quad c=\frac{6k+4}{2}=3k+2,\quad k\in\mathbb Z$$ $$a=-\frac{20c+9b+2}{6}=-\frac{10(6k+4)+18l+2}{6}=-(10k+3l+7)$$.

Answer 2

Changing notation, $6x+9y+20z=-2$ implies $y=2y_1$ which goes to the equation $3x+9y_1+10z=-1$.

Let $A=-1-3x$ be an auxiliary notation so we have the equation with two unknowns $9y_1+10z=A$ which has the particular solution $(y_1,z)=(-A,A)$. Hence the general solution is $y_1=-A+10t=1+3x+10t$; $z=A-9t=-1-3x-9t$. Thus the general parameterization with $s,t\in\mathbb Z$ $$\begin{cases}x=s\\y=2+6s+20t\\z=-1-3s-9t\end{cases}$$