0

I am not sure how to prove this theorem.

It states: Let $a,b,c$ be integers with $a$ and $b$ both no equal to $0$. If $x=x_0$ and $y=y_0$ is an integer solution to the equation $ax+by=c$, that is, $ax_0+by_0=c$, then for every integer $k$, the numbers $$x=x_0+\frac{kb}{(a,b)}\,\,\,\,\&\,\,\, y=y_0-\frac{ka}{(a,b)}$$ are integers that also satisfy the linear Diophantine equation $ax+by=c$. Moreover, every solution to the linear Diophantine equation $ax+by$ is of this form.

I understand how all solutions for a particular equation can be found using this equation, but I do not know how I can go about proving this theorem.

Example: Find all solutions to the equation $24x+9y=33$. The obvious solution to the equation is $x_0=1$ and $y_0=1$. Then, $x=1+\frac{9k}{3}$ and $y=1-\frac{24k}{3}$, so we have, $x=1+3k$ and $y=1-8k$ hence for all integer values of $k$, we have corresponding $x$ and $y$ values which satisfy the equation.

Bill Dubuque
  • 282,220
Obsessive Integer
  • 421

1 Answers1

1

Suppose that we have a solution of $ax+by=c$ given by $(x_0, y_0)$. Let us assume that all the solutions are of the form $(x_k, y_k)=(x_0+s_k, y_0+t_k)$. Our goal is to find $s_k$ and $t_k$.

Since $(x_k, y_k)$ satisfy the original equation, we have:

$ax_k+by_k=c$

$=> ax_0 + as_k + by_0+bt_k=c$

$ => as_k +bt_k=0$ $(\because ax_0+by_0=c)$

Now, we have:

$ s_k = \frac{-bt_k}{a}$

Consider the $\frac{-bt_k}{a}$. For integer solutions, $a$ must divide $-bt_k$. Now let the irreducible of $\frac{b}{a}$ be given by $\frac{p}{q}$. Then we know that $p=\frac{b}{gcd(a,b)}$ and $q=\frac{a}{gcd(a,b)}$. Therefore, we are now required to solve for $\frac{-pt_k}{q}$.

As we know that $gcd(p,q)=1$ (since the fraction is irreducible), $q$ must divide $t_k$.

And so, $t_k= k q = \frac{ka}{gcd(a,b)}$

Substituting, $t_k$ we have:

$s_k = \frac{-pt_k}{q} = \frac{-p(kq)}{q} = -pk = \frac{-kb}{gcd(a,b)}$

So, the final solution is given by:

$(x_k, y_k) = (x_k, y_k)=(x_0+s_k, y_0+t_k) = (x_0+ \frac{-kb}{gcd(a,b)}, y_0 +\frac{ka}{gcd(a,b)})$

SMIT PATEL
  • 76
  • This is a special case of a much more general result - see the linked dupes. Please strive not to add more dupe answers to dupes of FAQs. – Bill Dubuque Feb 11 '21 at 05:48