Positive pairs of integral values satisfying $2xy − 4x^2 +12x − 5y = 11$ The number of positive pairs of integral values of $(x, y)$ that solves
$2xy − 4x^2 +12x − 5y = 11$ is?
I rearranged it to $(2x-5)(y+1-2x)=6$, which took quite a bit of time.
So it can be $2*3$ , $3*2$, $6*1$ or $1*6$ which gives us 2 possible positive integral pairs. Answer: 2.
Is there a faster way to do similar problems?
 A: First note that:
$$y=\frac{11 - 12 x + 4 x^2}{2 x-5}$$
Applying division:
$$y=\frac{(2x-1)(2x-5)+6}{2x-5}=(2x-1)+\frac{6}
{2x-5}$$
To be an integer $2x-5$ must divide $6$ so $2x-5$ must be $-1,-3,1,2$ or $3$ or $6$. Check then by the positive integral solutions.
A: First of all, $6=\pm1\cdot\pm6=\pm1\cdot\pm6=\pm2\cdot\pm3=\pm3\cdot\pm2=\pm6\cdot\pm1$
We've got the following equations:
$$\begin{cases} 2x = 5+a\\-2x+y = -1+b\end{cases}$$
We can rearrange it to:
$$\begin{cases} x = 2.5+\frac{a}{2}\\y = 4+a+b\end{cases}$$
We want $x$ and $y$ to be integer, so $a$ must be an odd number and $b$ must be then an integer number. Also we want $ab=6$.
There are 4 pairs $[a,b]$ satisfying these conditions: $[\pm 1, \pm 6]$ and $[\pm 3, \pm2]$
We have then 4 different pairs $[x,y]$:
$$\left\{[2.5\pm \frac{1}{2}, 4 \pm 1 \pm 6], [2.5\pm \frac{3}{2}, 4 \pm 3 \pm 2] \right\} =\\
=\left\{[3,11],[2,-3],[4,9],[1,-1]\right\}$$
We also want $[x,y]$ to be a pair of positive integers, so from our pairs we pick two satisfying this condition:
$$\left\{[3,11],[4,9]\right\}$$
A: Hint
$$4x^2-2x(6+y)+5y+11=0$$
Once $x \in \Bbb N^*$ then:
$$\Delta=4(y^2-8y-8)=4k^2 \rightarrow (y-4)^2-k^2=24 \rightarrow (y-k-4)(y+k-4)=24$$
Now split $24$ as a product of two integer with the same parity and find all values for $y$ and $k$. After that you can find $x$ using:
$$x=\frac{2(6+y)\pm2|k|}{8}\quad (1)$$
Can you finish?
