# Solutions to $ab+1=cde+c$

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

• 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. – mathreadler Oct 7 '17 at 11:53
• If $a$ and $b$ are both constants, why not replace $ab$ by a single constant? – Adam Bailey Oct 7 '17 at 15:36

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.