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???

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    $\begingroup$ They changed sign on you, are looking at $ax-by=c$. $\endgroup$ – André Nicolas Jun 13 '14 at 17:08
  • $\begingroup$ A ok...I understand...So,if we would have $-ax+by=c$,we would take $x=x_1-k \cdot b_1 \text{ and } y=y_1+k \cdot a_1$,right? $\endgroup$ – evinda Jun 13 '14 at 17:17
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    $\begingroup$ No, it would be similar to the case $ax-by$, you use the same sign to get the cancellation. $\endgroup$ – André Nicolas Jun 13 '14 at 17:19
  • $\begingroup$ A ok...I got it...thank you very much!!! $\endgroup$ – evinda Jun 13 '14 at 17:23
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    $\begingroup$ You are welcome. Minus signs give me trouble too. $\endgroup$ – André Nicolas Jun 13 '14 at 17:24

So that $ax-by = \langle\text{stuff with $c$s}\rangle +akb-bka$ and the last two terms cancel.

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  • $\begingroup$ I understand...thanks a lot!!! $\endgroup$ – evinda Jun 13 '14 at 17:24

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