Is there a shortcut for solving $ax-by=c$ when a,b,c are relatively prime?

Question

I know how to find solutions for linear Diophantine equations like $ax-by=c$, but is there a shortcut when I already know that (a,b,c) are mutually prime?

Even if there isn't a conceptual shortcut, is there a nicer or briefer way to describe the algorithm in this special case?

No, because the general problem is trivially reducible to the case when $a,b$ are coprime. — Bill Dubuque, Mar 19 '17 at 22:56
Does the fact that I know c is also mutually prime help matters at all? — Joe Slater, Mar 20 '17 at 01:39
Generally it is solvable iff $,\gcd(a,b)\mid c.\ $ So your case is solvable iff $,\gcd(a,b) = 1,,$ and the same method of solution works for all $,c\neq 0.\ $ — Bill Dubuque, Mar 20 '17 at 02:48

score 2 · Accepted Answer · answered Mar 19 '17 at 00:15

Only $a$ and $b$ have to be coprime to ensure there are solutions. You begin with solving $u_oa+v_0b=1$ (Bézout's identity) with the extended Euclidean algorithm. Then $$x_0=cu_0,\quad y_0=-cv_0$$ is a solution.

As all solution to Bézout's identity are $$x=x_0+kb,\quad y=y_0-ka\quad(k\in\mathbf Z),$$ all solutions of the given equation have the form $$x=x_0+kb, \quad y=y_0+ka.$$

Is there a shortcut for solving $ax-by=c$ when a,b,c are relatively prime?

1 Answers1