Finding solutions

**Solaris123** · March 16th 2010, 07:45 PM

How would you be able to find the number of solutions on the next equation?

x^2-4xy+5y^2+2y-4=0

Given only this, how could you know the number of possible integer pairs of (x,y)? Please show the steps and explain.

**Solaris123** · March 16th 2010, 08:00 PM

Just a clarification for those that want to help me solve this, I will not have a calculator at hand when doing this type of problem so the most important thing is not the answer but rather the logic and procedure you used to get to the answer as that can help me in the future solve other similar problems. Thank you in advance for all your work.

**redsoxfan325** · March 16th 2010, 08:16 PM

Originally Posted by Solaris123

How would you be able to find the number of solutions on the next equation?

$x^2-4xy+5y^2+2y-4=0$

Given only this, how could you know the number of possible integer pairs of (x,y)? Please show the steps and explain.

This equation equals

$(x^2-4xy+4y^2)+(y^2+2y+1)-5=0\implies(x-2y)^2+(y+1)^2=5$

Therefore either

$x-2y=\pm2$ and $y+1=\pm1$

or

$x-2y=\pm1$ and $y+1=\pm2$

Solving those equations will get you the $(x,y)$ pairs that are integer solutions.

**Solaris123** · March 16th 2010, 08:21 PM

Ok, that makes sense but before I can say Im comfortable with this, how did you make those squares? I mean, you are right, this makes sense but is there any sort of rule I must follow to find them or I just have to play with values until I make those squares appear?

**Prove It** · March 16th 2010, 08:21 PM

Originally Posted by Solaris123

How would you be able to find the number of solutions on the next equation?

x^2-4xy+5y^2+2y-4=0

Given only this, how could you know the number of possible integer pairs of (x,y)? Please show the steps and explain.

$x^2 - 4xy + 5y^2 + 2y - 4 = 0$

You will want to get the equation to be $y$ in terms of $x$ , so move everything without a $y$ to the other side.

$5y^2 + (2 - 4x)y = 4 - x^2$ .

Now you will need to complete the square:

$y^2 + \frac{2(1 - 2x)}{5}\,y = \frac{4 - x^2}{5}$

$y^2 + \frac{2(1 - 2x)}{5}\,y + \left(\frac{1 - 2x}{5}\right)^2 = \frac{4 - x^2}{5} + \left(\frac{1 - 2x}{5}\right)^2$

$\left(y + \frac{1 - 2x}{5}\right)^2 = \frac{(4 - x^2)^2}{25} + \frac{(1 - 2x)^2}{25}$

$\left(y + \frac{1 - 2x}{5}\right)^2 = \frac{16 - 8x^2 + x^4 + 1 - 4x + 4x^2}{25}$

$\left(y + \frac{1 - 2x}{5}\right)^2 = \frac{x^4 - 4x^2 - 4x + 17}{25}$

$y + \frac{1 - 2x}{5} = \pm \frac{\sqrt{x^4 - 4x^2 - 4x + 17}}{5}$

$y = \frac{1 - 2x \pm \sqrt{x^4 - 4x^2 - 4x + 17}}{5}$ .

Now you need to determine what values of $x$ are going to give you a multiple of $5$ in the numerator.

**redsoxfan325** · March 16th 2010, 08:28 PM

Originally Posted by Solaris123

Ok, that makes sense but before I can say Im comfortable with this, how did you make those squares? I mean, you are right, this makes sense but is there any sort of rule I must follow to find them or I just have to play with values until I make those squares appear?

I hate to rain on your parade, but there's no general algorithm for solving Diophantine equations. (This was Hilbert's Tenth Problem.) You just have to play with each one as best as you can to try to devise a solution. Looking for squares is always a good place to start though.

**11rdc11** · March 16th 2010, 08:34 PM

Just wondering is this not an ellipses with a rotation of axes?

**Solaris123** · March 17th 2010, 04:41 AM

Just passed it throught Wolfram Alpha and indeed it's an ellipse with it's two foci slanted, does that make a difference or help solve the problem in any way?
PS: Anyone know how to use the fact that it is an ellipse to solve it?

**Solaris123** · March 17th 2010, 03:15 PM

Bump

**11rdc11** · March 17th 2010, 07:50 PM

Originally Posted by Solaris123

Just passed it throught Wolfram Alpha and indeed it's an ellipse with it's two foci slanted, does that make a difference or help solve the problem in any way?
PS: Anyone know how to use the fact that it is an ellipse to solve it?

The equation would not be anything easier to solve

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