🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

$⟹$ Let $g:=gcd(a,b)$ and suppose that $x_0,y_0$ is a solution so that $a{x}_0+b{y}_0=c$ since $g\mid a,b$ we know that $g\mid a{x}_0+b{y}_0$ therefore $g\mid c$ as needed.

$⟸$ Recall that we have some $x_0,y_0\in \mathbb{Z}$ such that $gcd(a,b)=a{x}_0+b{y}_0$ since $g\mid c$ so $c=gk$ for some $k\in \mathbb{Z}$ thus multiplying our previous equation by $k$ we obtain $${\displaystyle c=gk=a\left(k{x}_0\right)+b\left(k{y}_0\right)}$$

Taking $s\in S$ then we see that $${\displaystyle s=(x_0+t\frac{b}{d},y_0-t\frac{a}{d})}$$ Now, let's confirm that this is a solution, so we have to make sure that $a{s}_1+b{s}_2=c$. $${\displaystyle a(x_0+t\frac{b}{d})+b(y_0-t\frac{a}{d})=a{x}_0+b{y}_0+0=c}$$ therefore every element in $S$ is a solution, but let's now show that it contains all solutions.

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We already know that $(x_0,y_0)$ is a solution, so let $x_1,y_1$ be a different solution, then we have $${\displaystyle a{x}_1+b{y}_1=c=a{x}_0+b{y}_0⟺a(x_1-x_0)=-b(y_1-y_0)}$$ dividing through by $g:=gcd(a,b)$ we see that $\frac{a}{g}(x_1-x_0)=\frac{-b}{g}(y_1-y_0)$ but then recall that $gcd(\frac{a}{g},\frac{b}{g})=1$ thus by forced division, we see that $\frac{a}{g}\mid y_1-y_0$so that there is some $k\in \mathbb{Z}$ such that $y_1-y_0=\frac{a}{g}k$ thus $${\displaystyle \frac{a}{g}(x_1-x_0)=\frac{-b}{g}(y_1-y_0)⟺\frac{a}{g}(x_1-x_0)=\frac{-b}{g}\left(\frac{a}{g}k\right)}$$ therefore by cancellation we have that $x_1-x_0=-\frac{b}{g}k$ and symmetrically $y_1-y_0=\frac{a}{g}k$ therefore $${\displaystyle (x_1,y_1)=(x_0-k\frac{b}{g},y_0+k\frac{a}{g})}$$ so therefore $(x_1,y_1)\in S$ as needed.

For notational ease, let $g:=gcd(a,b)$. Recall that $gcd(\frac{a}{g},\frac{b}{g})=1$ iff there are some $x,y\in \mathbb{Z}$ such that $\frac{a}{g}x+\frac{b}{g}y=1$ if and only if $a\left(dx\right)+b\left(dy\right)=dg$

This observation tells us that any integer multiple of $g$ is in $S$. Now supose we have $ax+by\in S$, we want to show this is a multiple of $g$, equivalently we want to show that $g\mid ax+by$, but this is certainly true becuase $\frac{ax+by}{g}=\left(\frac{a}{g}\right)x+\left(\frac{b}{g}\right)y\in \mathbb{Z}$ so it does