Find all positive integer roots of $\,5xy=19x+96y$ 
Find all positive integer roots of : $5xy=19x+96y$

I tried using decomposition technique but no success...,it seems suitable  factorization of this equation is IMPOSSIBLE!!
Handy calculations show that it has as many as 6 answers.
 A: As you suggested, let us try to factor this somehow. We have the following equivalent conditions:
\begin{align*}
5xy-19x-96y&=0\\
25xy-5\cdot19x-5\cdot96y&=0\\
(5x-96)(5y-19)&=19\cdot96
\end{align*}
There is only finitely many possibilities how to write $96\cdot19=2^5\cdot3\cdot19$ as a product of integers. 
Moreover, we only have to check the possibilities where one of the factors is congruent to $1$ and the other one congruent to $4$ modulo $5$. (Because we want these factors to be equal to $5x-96 \equiv 4 \pmod5$ and $5y-19\equiv1\pmod5$.)
By going through all possibilities (all divisors of this number) we should be able to find all possibilities for $x$ and $y$. (So we have to check $24$ divisors, but many of them will not fulfill the conditions modulo $5$.)
A: Try to solve it as a diophantine equation 
${19x + 96y = D\\
\gcd ( 19, 96 ) = 1}$
${ 
\text {One   possible  solution of }\\
19x + 96y = 1\\
\text{is} \ (\bar x, \bar y) = (-5, 1)\\
\text{thus all the solution of}\\
19x + 96y = D\\
\text{are } (x,y) = (-5D - 96k, D + 19k) \text{ where } k \in \mathbb{Z}}$
${
\text{in our case } D = 5xy \text{ so }\\
\{
\begin{array}{1}
x = -25xy -96k\\y = 5xy + 19k
\end{array}\\
\text{namely}\\\{
\begin{array}{1}
x(1+25y) \equiv 0 \pmod{96}\\y(1-5x) \equiv 0\pmod{19}
\end{array}\\
\text{Then you can solve the system of congruences.} }$
A: HINT:
Like Ross Millikan,
$y=\dfrac{19x}{5x-96}$
As $(5x-96,5)=1,$
$(5x-96)|19x\iff(5x-96)\mid5\cdot19x=19(5x-96)+19\cdot96\iff(5x-6)|19\cdot96$
whose divisors are $?$
As $y>0,$  
$(i)$ either $x>0,5x-96>0\implies x>\dfrac{96}5=19.2$
$(ii)$ or $x<0,5x-96<0\implies x<0;$ but $x>0$
Can you take it from here?
A: We have $19x=(5x-96)y$, so $y=0\bmod19$ or $x=4\bmod19$.
If $y=19n$, then we have $\frac{96}{x}=5-\frac{1}{n}$. So $x\le24$, and $x(5n-1)=96n\ (*)$. Checking $n=1,2,3,4,5$ we find that $n=1$ works giving solution $(24,19)$. $n=2,3,4$ do not work. $n=5$ gives the solution $(20,95)$. If $n>5$ then since $5n-1,n$ are coprime we must have $5n-1|96$ and hence $5n-1=48,96$, neither of which work.
If $x=4\bmod19$, then $(5n-4)y=19n+4$. If $n=1$ we have $y=23$, giving the solution $(23,23)$. If $n\ge2$, then we have $5=\frac{96}{x}+\frac{19}{y}=\frac{1}{y}\left(\frac{96}{5n-4}+19\right)\le\frac{35}{y}$, so $y\le7$. If $y=7$, we get the solution $(42,7)$. If $y=6$ we get $11n=28$ which fails. If $y=5$ we get the solution $(80,5)$. If $y=4$ we get the solution $(384,4)$. If $y\le3$ we get $(19-5y)n=-4-4y$ which fails.
So in total we have the six solutions $(20,95),(23,23),(24,19),(42,7),(80,5),(384,4)$.
A: HINT.-You have $$5xy\equiv y\pmod{19}\Rightarrow 5x=1\pmod{19}\text{ discarding }\space y=0$$ It follows in $\mathbb N$ $$5x-19M=1$$ whose solutions is (since $(x,M)=(4,1)$ is solution) $$\begin{cases}x=4-19t\\M=5t+1\end{cases}$$ Hence $$5x=19(5t+1)+1\Rightarrow x=19t+4$$ Now trying with $t=1$ we test $x=23$ and for $t=2$ we test $x=42$ which are right and gives $$5\cdot23y=19\cdot23+96y\Rightarrow y=23\\5\cdot42y=19\cdot42+96y\Rightarrow y=7$$ so $$(x,y)=(23,23)\text{ and }(42,7)$$  are solutions, etc.  
A: Write it as $y=\frac {19x}{5x-96}$ and $x=\frac{96y}{5y-19}$  We must have $x \ge 20, y \ge 4$.  $x$ and $y$ move in opposite directions.  Plugging in $x=20$ gives $y=95$, so $4 \le y \le 95$.  One can make a spreadsheet to check all these.  I find six solutions.  This is more cases than factoring, but copy down makes it easy.
A: Hint $\ $ Apply the  AC-method to reduce to the monic case, which follows easily by  $\rm\,\color{#0a0}{\text{completing a product,}}$ in a similar way to completing a square, i.e.
$$\begin{eqnarray}  && \ \ \ a\ x\ y &+& b\ x&+&c\ y &=&\ \  d\\
\smash{\large \overset{\times\ a}\iff} && \ \ ax\,ay &+& b\,ax &+& c\,ay &\,=& ad\\
\iff  && \ \ \ {X\ Y} &+& {b\,X} &+& {c\,Y} &=& ad,\quad X = ax,\ \ Y = ay\\
\iff &&  \color{#0a0}{(X\!+\!c)}\!\!&&\!\!\!\!\!\! \color{#0a0}{(Y\!+\!b)}&\color{#0a0}-& \color{#0a0}{bc} &=\,& ad,\quad {\rm by\ \ \color{#0a0}{completing\  the\  product}}\\
\iff && (ax\!+\!c)\!\!&&\!\!\!\!\!\!(ay\!+\!b)\!\! && &=& ad\!+\!bc
\end{eqnarray}$$
