Integer solutions of $xy+9(x+y)=2006$ [Completing a product / rectangle] 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 
 A: Add $81$ to both sides to give $$xy + 9x + 9y + 81 = (x+9)(y+9) = 2087,$$ and then consider the divisors of the RHS.
A: Key Idea $\ $ Completing a square $ $ generalizes to $ $ completing a product (rectangle)
$$\begin{eqnarray} x^2\:\! &+&\ \ \,2bx  &=& (x + b)^2 - b^2\\
\iff\ x\color{#c00}x  &+& bx\!+\!b\color{#c00}x &=\,&  (x+b)(\color{#c00}x+b)-b^2\\
\iff \ x\color{#c00}y &+&  bx\!+\!b\color{#c00}y\ &=&  (x+b)(\color{#c00}y+b)-b^2\\
{\rm generally}\quad  {xy}&+&bx\!+\!cy &=& \color{#0a0}{(x+c)(y+b) - bc}
\end{eqnarray}\quad\ \ \,$$
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 reduces to a finite process, namely, test which of the factorizations of $ad+bc$ can be written in the 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).
A: 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).
