The Diophantine equation $ax+by = b+c$ is solvable in integer $x , y$ iff $ax+by =c$ is solvable.

Question

Let $a,b,c \in \Bbb Z$. Show that the Diophantine equation $ax+by = b+c$ is solvable in integer $x , y$ iff $ax+by =c$ is solvable.

We know that a Diophantine equation $ax+by =c$ is solvable iff $d|c$ where $d=gcd(a,b)$. From here how can we do the above proof??

Answer 1

As you have said we use the following theorem :

A Diophantine equation $ax+by =c$ is solvable iff $d|c$ where $d=gcd(a,b)$.

$\Rightarrow$ Let us assume that $ax+by =b+c$ is solvable in integer $x,y$. Thus $d|b+c$ where $d=gcd(a,b)$ and since $d|b$ so $d|c$ thus $ax+by =c$ is solvable in integer $x,y$ using the above theorem.

$\Leftarrow$ can be done similarly!!

Answer 2

$(x,y)=(m,n)$ satisfies $ax+by=c$ iff $(x,y)=(m,n+1)$ satisfies $ax+by=b+c$.
Thus $\exists (x,y)(ax+by=c)\iff\exists (x,y)(ax+by=b+c)$.

Answer 3

Can't see why anyone would use complicated stuff like gcd.

If ax + by = b + c has a solution (x, y), then (x, y-1) is a solution of ax + by = c.

And if ax + by = c has a solution (x, y), then (x, y+1) is a solution of ax + by = b + c.

Answer 4

Hint $\,\ f(x,y)=0\ $ is solvable $\iff f(x,y\!-\!1)=0\ $ is solvable