Solve $k_1 a = k_2 b + c$ Find all $k_1, k_2$ that satisfy $k_1  a = k_2 b + c$ where everything are integers. It feels like there should be some easy way to describe this in terms of congruence and gcd. 
 A: This is the simplest of the Diophantine equation i.e. linear Diophantine equation with 2 variables.$$ax+by=c$$
The condition for solvability is - $ax+by=c$ admits a solution if and only if $gcd(a,b)|c$ .
And if $(x_0,y_0)$ is any particular solution of this equation , then all other solutions are given by$$x=x_0+\frac{b}{d}t\quad \quad y=y_0-\frac{a}{b}t$$
For example consider the linear Diophantine equation$$172x+20y=1000$$
So applying Euclid's algorithms to find the gcd.
$$\begin{align*}172&=8.20+12 \\
20&=1.12+8\\
12&=1.8+4\\
8&=2.4\end{align*}$$
So the $\text{gcd}(172,20)=4$.And since $4|1000$ ,a solution to this equation exists.So working backward.
$$\begin{align*}4&=12-8\\ 
&=12-(20-12)\\
&=2.12-20\\
&=2(172-8.20)-20\\
&=2.172+(-17)20
\end{align*}$$
Multiplying by $250$ we get$$1000=500.172+(-4250)20$$
So $x=500 \text{ and }y=-4250$. And then putting these value in above formula you can get the general solution.
A: Let $d=\gcd(a,b)$. If $d$ does not divide $c$, there is no solution. So assume from now on that $d$ divides $c$. 
Suppose that we have found one particular solution $(x_0,y_0)$ of the equation $ax=by+c$. Then all solutions $(x,y)$ are given by 
$$x=x_0 +\frac{b}{d}t, \qquad  y=y_0+\frac{a}{d}t,\tag{$1$}$$
where $t$ ranges over the integers, positive, negative, and $0$.
So now look for a particular solution $(x_0,y_0)$. In "small" cases, a particular solution can be found by experimentation. In other cases, use the Extended Euclidean Algorithm to find integers $s$ and $t$ such that $as=bt+d$. Then a particular solution $(x_0,y_0)$ of our original equation is given by
$$x_0=\frac{c}{d}s,\qquad y_0=\frac{c}{d}t.$$
Now using $(1)$ we can generate all the solutions. 
