Prove the solutions to $ax+by=c$

Question

I have this math question, that I'm kind of stuck on.

Consider the equation $a x + b y = c$, for some non-zero integers $a, b$ and $c$. Suppose that $x = x_1, y = y_1$ is an integer solution to the equation $ax + by = c$.

It is know that if $x, y$ is an integer solution to the homogeneous equation $a x + by = 0$, then $$ x = \frac{b}{\gcd(a,b)} k, y= - \frac{a}{\gcd(a,b)} k, \text{ for some integer } k. $$ Use this fact to show that if $x=x_2, y=y_2$ is another solution to the equation $a x + b y = c$, then $$ x_2 = x_1+ \frac{b}{\gcd(a,b)}k, y_2=y_1 - \frac{a}{\gcd(a,b)}k, \text{ for some integer } k. $$

I'm not sure how to start this. Would I just plug in $x_2$ and $y_2$ for $x$ and $y$? Thanks

Answer 1

Hint:

$$ ax_2+by_2=c=ax_1+by_1\Rightarrow a(x_2-x_1)+b(y_2-y_1)=0 .$$

Answer 2

$$ax_2 + by_2 = c= ax_1 + by_1 ⇒ a(x_2 −x_1 )+b(y_2 −y_1 )=0.$$

$$a(x_2-x_1)= -b(y_2-y_1)$$ Divide by $g=(a,b)$ to get $$\frac{a}{g}(x_2-x_1)=-\frac{b}{g}(y_2-y_1)$$

Since $(\frac{a}{g},\frac{b}{g})=1$, we have $$\frac{a}{g} \mid (y_2-y_1)$$

The rest is straightforward.