According to my notes: $$ \begin{align} & \text{ Let } a,b \text{ not both } 0. \\ & ax+by \text{ has a solution iff } (a,b) \mid c \\ & \text{ If } d:=(a,b) \mid c \text{ and } a=d \cdot a_1 , b=d \cdot b_1 \text{ and } x_1,y_1 \text{ is a solution of } \\ & ax+by=c \text{ , then all the solutions are given by the formulas: } \\ & x=x_1+k \cdot b_1, y=y_1-k \cdot a_1, k \in \mathbb{Z} \end{align} $$
I am looking at the solution of this exercise:
Let $a,b \geq 0 , (a,b)=1$ . Show that the diophantine equation $ax-by=c$ has infinite solutions with $x,y>0$.
That is the solution:
Since $(a,b)=1$ , $\exists x_0,y_0 \text{ such that } ax_0+by_0=1 \Rightarrow a(cx_0)-b(-cy_0)=c$
Therefore, $\displaystyle x_1=cx_0 , y_1=-cy_0 \text{ is a solution of } ax-by=c$
The solutions of $ax-by=c$ are described by the formulas:
$$x=cx_0+kb , y=-cy_0+ka, k\in \mathbb{Z}$$
And,as we want $x>0,y>0$, it must be $\displaystyle k> \max\left\{\frac{-cx_0}{b},\frac{cy_0}{a}\right\}$
But...why are these : $x=cx_0+kb , y=-cy_0+ka, k\in \mathbb{Z}$ the solutions of $ax-by=c$,and not $x=cx_0+kb, y=-cy_0-ka$. Doesn't it have to be at one of $x,y$ , minus and at the other one plus???
So that $ax-by = \langle\text{stuff with $c$s}\rangle +akb-bka$ and the last two terms cancel.