1. ## Diophantine question

Hi plz give me a help to find the answer for following

If $a \ne 0$, gcd(a,b)=g and $(x_0,y_0)$ is a solution of $ax+by=c$ then show that
$x=x_0+\frac{b}{g}t$, $y=y_0-\frac{a}{g}t$

2. What is $t$ ?

3. Originally Posted by dhammikai
Hi plz give me a help to find the answer for following

If $a \ne 0$, gcd(a,b)=g and $(x_0,y_0)$ is a solution of $ax+by=c$ then show that
$x=x_0+\frac{b}{g}t$, $y=y_0-\frac{a}{g}t$
Here we are dealing with integer solution of
$ax+by=c$

Since,
$(x_0,y_0)$ is a solution
$ax+by=ax_0+by_0$
$a(x-x_0)+b(y-y_0)=0$
Since there are infinite integer solution(why?), let us introduce an integer parameter m such that $x-x_0=m$
It becomes
$am+b(y-y_0)=0$
$y-y_0=\frac{am}{b}$
Now, to ensure that RHS is always an integer, make substitution $m=\frac{bt}{g}$....where t is new integer parameter.

4. (1) ax + by = c
(2) $ax_{0} + by_{0} = c$ is an integer solution.

Subtracting (2) from (1),
$a\left(x-x_{0}\right)+b\left(y-y_{0}\right)=0$.

Now
(*) $\frac{a}{g}\left(x-x_{0}\right)=-\frac{b}{g}\left(y-y_{0}\right)$.

Thus
$\frac{b}{g}\mid \frac{a}{g}\left(x-x_{0}\right)$. But $\left(\frac{b}{g},\frac{a}{g}\right)=1$ therefore,

$\frac{b}{g}\mid \left(x-x_{0}\right)$. Hence it follows that there is a $t \in \mathbb{Z}$ such that

(3) $x-x_{0} = t \frac{b}{g}$. Substituting this in (*),

$\frac{a}{g} \cdot t\frac{b}{g} = -\frac{b}{g}\left(y-y_{0}\right)$. Hence $y-y_{0}=-t\frac{a}{g}$

Therefore, from above and (3),
$x = x_{0} + t\frac{b}{g}$

$y=y_{0}-t\frac{a}{g}$
$QED$

5. OK Thanks, I got the idea for the answer, thanks again