Problem :
Find the general solution and positive integral solutions of 775x -711y =1
My approach :
For the equation ax -by =c , the general solution can be given as :
Let h,k be a solution of ax -by =c, then ah -bk =c.
$ax -by =ah -bk ;$
$a(x-h)=b(y-k); $
$\frac{x-h}{b} = \frac{y-k}{a} =t,$ an integer
$x =h+bt, y =k +at;$ which is the general solution , therefore
We can compare the same with $775x -711y =1$
Let h,k be a solution of $775x -711y =1$ therefore , $775x -711y =1,$
$\Rightarrow 775h -711k =1 $
$\Rightarrow 775h -711 k = 775x -711y $
$\Rightarrow \frac{x-h}{711}=t; \frac{y-k}{775} =t$
Now please suggest how to take this further and get the integral solution of this.
I'd use the Euclidean algotithm. Let $a=775,b=711$.
$775=711=64$, so $64=775-711=a-b$.
$711=11*64+7$, so $7=711-11*64=b-11(a-b)=12b-11a$.
$64=9*7+1$, so $1=64-9*7=(a-b)-9(12b-11a)=100a-109b$.
So $x=x_0=100,y=y_0=109$ is one solution of the equation $775x-711y=1$. In parametric form, the general solution is $x=100+711t, y=109+775t,x\in\mathbb R$. If you want $x,y$ to be integers, the general solution is $x=100+711t, y=109+775t,x\in\mathbb Z$. If you want $x,y$ to be positive integers, you have to satisfy the inequalities $100+711t\gt0$ and $109+775t\gt0$, i.e., $t=0,1,2,3,\dots$. The resulting solutions are $(100,109),(811,884),(1522,1659),(2233,2434),\dots$.