Solving $\,axy+bx+cy=d\,$ [Completing a product or rectangle]

Question

How many integer solutions does $xy+9(x+y)=2006$ have? Here $x$ and $y$ are both integers.

My trying: I have tried to solve this problem. But I have no idea to solve this. Please help

Answer 1

Add $81$ to both sides to give $$xy + 9x + 9y + 81 = (x+9)(y+9) = 2087,$$ and then consider the divisors of the RHS.

Answer 2

Key Idea $ $ completing a $\rm\color{#0af}{square}$ generalizes to completing a $\rm\color{#0a0}{\text{product (rectangle)}},\,$ viz.

$$\begin{eqnarray} x^2\:\! &+&\ \ 2bx &=& \color{#0af}{(x + b)^2} - b^2\\ \iff\ x\color{#c00}x &+& bx\!+\!b\color{#c00}x &=\,& (x+b)(\color{#c00}x+b)-b^2\\ \iff \ x\color{#c00}y &+& bx\!+\!b\color{#c00}y\ &=& (x+b)(\color{#c00}y+b)-b^2\\ {\rm generally}\quad {xy}&+&bx\!+\!cy &=& \color{#0a0}{(x+c)(y+b)} - bc \end{eqnarray}\quad\ \ \,$$

Remark $\ $ The AC-method extends it to non-monics (lead coef $\,a\neq 1)$ as follows

$$\begin{eqnarray} && \ \ \ a\ x\ y &+& b\ x&+&c\ y &=&\ \ d\\ \smash{ \overset{\large \times\ a}\iff} && \ \ ax\,ay &+& b\,ax &+& c\,ay &=& ad\\ \iff && \ \ \ {X\ Y} &+& {b\,X} &+& {c\,Y} &=& ad,\quad X = ax,\ \ Y = ay\\ \iff && \ \color{#0a0}{(X\!+\!c)}&&\!\!\!\!\! \color{#0a0}{(Y\!+\!b)}&\color{#0a0}-&\, \color{0a0}{bc} &=\,& ad,\quad {\rm by\ \ \color{#0a0}{monic\ \ case}\ \ above}\\ \iff && (ax\!+\!c)\!\!&&\!\!\!\!\!\!(ay\!+\!b)\!\! && &=& ad\!+\!bc \end{eqnarray}$$

Summarizing: if $\,a\,$ is cancellable (e.g. $\,a\neq 0\,$ in $\Bbb Z)\,$ then

$$\bbox[1px,border:3px solid #c60]{\bbox[8px,border:1px solid #c00]{\begin{align} axy + bx + cy &\,=\, d\\[.2em] \!\!\!\iff (ax+c)(ay+b) &\,=\, ad+bc\end{align}}}\qquad\qquad\qquad$$

In this form it is clear that the solution of the equation reduces to a finite process - test which factorizations of $\,ad+bc\,$ have form $\,(ax+c)(ay+b)$.

This is one case of Lagrange's solution of the general quadratic binary Diophantine equation.

Note $ $ The special monic case $(a = 1)$ is sometimes called Simon's Favorite Factoring Trick in some problem solving / contest communities, but this name is not in wide use, so I recommend using the more descriptive name - completing a product (or rectangle).

Answer 3

Just to give another way to solve.

$$p+9s=2006\Rightarrow p\equiv8\pmod9\Rightarrow p=8+9n$$Then $s=t-n$ for some $t$ which is deduced from $$(8+9n)+9(t-n)=2006\Rightarrow t=222$$ so we have the general solution for the sum and the product $$\begin{cases}p=8+9n\\s=222-n\end{cases}$$ Now we have for values of $x$ and $y$ $$X^2-(222-n)X+(8+9n)=0$$ where for confort we put $2n$ instead of $n$ so we have the equation $$X^2-2(111-n)X+(8+18n)=0\Rightarrow X=111-n\pm\sqrt{n^2-240n+12313}$$ We need $$n^2-240n+12313=Y^2\\(n-120)^2+2087=Y^2\\2087=(Y+n-120)(Y-n+120)$$ Since $2087$ is prime we have an easy factorization giving setting values of $n$.

(By the way, $2087$ is the same integer reached in the concise and beautiful proof by @heropup).