Bezout's Identity in proving a pair of integers satisfies the linear integer combination [closed]

Question

The problem states:

Let $a,b,c \in \mathbb{Z}$.

Prove that if a pair of integers $(x_0,y_0)$ solves the equation $ax+by=c$, then the pair $(x_0+k \cdot b,y_0-k \cdot a)$ is also an integer solution of $ax+by=c$ (where $k$ is an arbitrary integer).

This shows that if $ax+by=c$ has an integer solution, then it has infinitely many solutions.

Answer 1

I guess the straight-forward calculation gives you the desired result easily:

We are given $ax+by=c$ where $a$, $b$ and $c$ are integers. Also we know that there exists a pair of integers $(x_{0},y_{0})$ which satisfies the equation, i.e $ax_{0}+by_{0}=c$. We need to show that $(x_{0}+k \cdot b, y_{0}-k \cdot a)$ is also an integer solution of our equation. Thus, we need to show two things:

1- $x_{0}+k \cdot b$ and $ y_{0}-k \cdot a$ are really integers

2- $a[x_{0}+k \cdot b]+b(y_{0}-k \cdot a]=c$.

(1) is clear since sum and product of integers are again an integer. For the (2) observe that $a[x_{0}+k \cdot b]+b(y_{0}-k \cdot a]= ax_{0}+akb+by_{0}-bka=ax_{0}+by_{0}=c$. Thus, $(x_{0}+k \cdot b, y_{0}-k \cdot a)$ is indeed a solution. Since for each integer value of k we have an integer solution to our equation, there are infinitely many solutions!

Let's look at a concrete example as well:

Suppose $a=2$, $b=3$ and $c=17$. Also, somehow, we found $(4,3)$ is a solution to our equation, i.e $x_{0}=4$ and $y_{0}=3$. Then, choosing

$k=1$ gives $(x_{0}+kb, y_{0}-ka= (7,1)$ is also a solution, surely $2 \cdot 7 + 3 \cdot 1 =17$. Thus, it is indeed a solution. $k=-2$ gives $(x_{0}+kb, y_{0}-ka= (-2,7)$ is another solution, since $2 \cdot (-2) + 7 \cdot 3 = 17$. I hope you get the idea.

Answer 2

$a(x_0+kb)+b(y_0-ka)=ax_0+akb+by_0-bka=ax_0+by_0$ that is equal to $c$ since $(x_0,y_0)$ solves the equation. Of course they are all integers.