Show that there are exactly 16 pairs of integers $(x,y)$ such that $11x+8y+17=xy$. Original problem
Show that there are exactly 16 pairs of integers $(x,y)$ such that $11x+8y+17=xy$.
My work
From case by case analysis I come to know that the equation will hold if and only if $x$ is odd and $y$ is even.
Also I found that $(8-x)|(11x+17)$ and $(11-y)|(8y+17)$
This is all what I have found.
Please see that my work are right or not.
This is a new kind of question which I have encountered so please help me is solving this problem.
 A: Given $a,b,c$ three integers, the idea when you have an equation of the form:
$$xy=ax+by+c$$
to solve if for unknown integers $x,y$ is to do the following factorization:
$$(x-b)(y-a)=c+ab$$
and hence the number of solutions is the number of divisors of $c+ab$.
A: This type of diophantine equation is solvable by a generalization of completing the square. Namely, completing a square generalizes to completing a product as follows:
$$\begin{eqnarray} &&axy + bx + cy\, =\, d,\ \ a\ne 0\\
\overset{\times\,a}\iff\, &&\!\! (ax+c)(ay+b)\, =\, ad+bc\end{eqnarray}\qquad\qquad$$
Here we deduce $\ xy - 11x - 8y = 17 \iff (x-8)(y-11) =\, \ldots$
So by uniqueness of prime factorizations the problem reduces to counting the divisors of $\,\ldots$
A: 16 ordered pairs indeed.  
$$\begin{align}
11x+8y+17&=xy \\
xy-11x-8y&=17 \\
xy-11x-8y+\color{red}{88}&=17+\color{red}{88} \\
(x-8)(y-11)&=105=3 \cdot 5 \cdot 7=d_1 \cdot d_2
\end{align}$$
Setting
$$\begin{cases} x-8=d_1\\ y-11=d_2 \end{cases}$$
Implies that
$$\begin{cases} x=8+d_1\\ y=11+d_2 \end{cases} \text{such that} \ \ (d_1,d_2) \in \lbrace \text{divisor-pairs of 105} \rbrace$$
For any divisor-pair $(d_1,d_2)$ that produces one solution, $(d_2,d_1)$ will produce another.  
Additionally, for any $(d_1,d_2)$ that produces a solution, $(-d_1,-d_2)$ will produce another as well.  
So for every divisor-pair, $(d_1,d_2)$, there will be four total solutions:
$$\begin{align} (d_1,d_2) &\to (d_1,d_2) \\
&\to (d_2,d_1) \\
&\to (-d_1,-d_2) \\
&\to (-d_2,-d_1)
\end{align}$$
This is due to a lack of symmetry in the expressions for $x$ and $y$ (for comparison, this situation is different).
The problem therefore reduces to finding exactly $4$ divisor-pairs of $105$.
Since $$105=3^{\color{red}{1}} \cdot 5^{\color{red}{1}} \cdot 7^{\color{red}{1}} \implies 105 \ \text{has} \ (1+{\color{red}{1}})(1+{\color{red}{1}})(1+{\color{red}{1}})=8 \ \ \text{divisors},$$
there are indeed four divisor-pairs:
$$(d_1,d_2) \in \bigg{\{} (1,105),(3,35),(5,21),(7,15) \bigg{\}}$$
As an example, setting $d_1=1$ and $d_2=105$ 
$$\begin{align} (1,105) &\to (1,105) && \implies (x,y)=(9,116)\\
&\to (105,1) && \implies (x,y)=(113,12)\\
&\to (-1,-105) && \implies (x,y)=(7,-94)\\
&\to (-105,-1) && \implies (x,y)=(-97,10)
\end{align}$$
That's four solutions from one divisor-pair alone. 
