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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?

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    Do you mean "integer values"? – 2017-01-12
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    Yes, in this context, "integral values" means "integer values". – 2017-01-12
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    Also, there are $4$ solutions, using $(-1)(-6)$ and $(-3)(-2)$. – 2017-01-12

3 Answers 3

5

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.

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    There are two more possibilities : -1 and -3 – 2017-01-12
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    actually, the denominator is $2x-5$, right? BTW, nice approach!+1 – 2017-01-12
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    @Arnaldo yes thanks! typo – 2017-01-12
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    $x$ and $y$ does have to be positive, but $2x-5$ don't have to be. – 2017-01-12
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    @JaroslawMatlak thanks man, fixed. – 2017-01-12
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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: $$\left\{[3,11],[4,9]\right\}$$

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    This does not answer the question: "Is there a faster way to do similar problems?" – 2017-01-12
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    @Ennar No, it doesn't. But it shows the right way to solve it. Author of question seemed to not solving it properly after the rearrangation. – 2017-01-13
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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?