Diophantine Equation: $xy+ax+by+c=0$

Question

How to find integer solutions $x,y$ of $xy+ax+by+c=0$ for given $a,b,c \in \mathbb{Z}$?

Is there somewhere a treatise on this kind of equations?

Answer 1

Remark: We will assume that the equation is $xy+ax+by+c=0$. If it really is $xy+(a+b)x+c=0$, then it can be done in the same way as the problem below, but simpler. The equation then becomes $x(y+a+b)=-c$, and we are looking for factors of $-c$.

Hint: Rewrite as $(x+b)(y+a)=ab-c$.

We can find the solutions in a straightforward way once we find all divisors of $ab-c$. This includes negative divisors.

Answer 2

(I'm guessing that what was intended was $xy+ax+by+c=0$.)

\begin{align} u & = x+y \\ v & = x-y \\[8pt] \text{so that }x & = \frac{u+v}{2} \\[8pt] y & = \frac{u-v}{2} \end{align}

Then $xy = \dfrac{u^2-v^2}{4}$. So $$ xy + ax + by + c =0 $$ becomes $$ \frac{u^2-v^2}{4} + a\frac{u+v}{2} + b \frac{u-v}{2} + c =0, $$ and then multiplying both sides by $4$, we get $$ u^2 - v^2 + 2(a+b)u + 2(a-b)v + c = 0. $$ Then one can complete the square, getting $$ (u+a+b)^2 - (v-a+b)^2 = (a+b)^2+(a-b)^2 - c = 2a^2+2b^2-c. $$