2
$\begingroup$

I'm trying to find a method to find the set of solutions in non zero integers for this diophantine equation. I thought of using a generalized formula of the solutions of $ax +by = c$ but i don't know what is the best way to approach it. Any help is appreciated. Thanks in advance! EDIT: the solutions should be in terms of a,b,d regarding them as constants

$\endgroup$
2
  • $\begingroup$ I have not solved a diophantine equation in a looong time, but you can rewrite it $ab +1 = c(de +1)$ now $de +1$ is on the same form as $ab+1$, hmm. $\endgroup$ – mathreadler Oct 7 '17 at 11:53
  • 1
    $\begingroup$ If $a$ and $b$ are both constants, why not replace $ab$ by a single constant? $\endgroup$ – Adam Bailey Oct 7 '17 at 15:36
1
$\begingroup$

With $A=ab, C=c$ and $D=de$ we have the solutions $A=C(D+1)-1$ for arbitrary non-zero integers $C$ and $D$. For any $A$, we can find all ways to write it as a product $A=ab$, e.g., by computing the prime factorisation of $A$. For, say, $C=D=5$ we obtain $A=29$, which we can only write as $1\cdot 29$, or $29\cdot 1$ up to sign.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.