The equation is

$\displaystyle

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

What is the total number of pair of integers (x,y) that satisfy the above equation?

How do I solve it?

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- Jul 29th 2009, 11:09 PMshrrikeshnumber of pair of integers (x,y) satisying an equation
The equation is

$\displaystyle

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

What is the total number of pair of integers (x,y) that satisfy the above equation?

How do I solve it? - Jul 29th 2009, 11:12 PMred_dog
$\displaystyle (x^2-4xy+4y^2)+(y^2+2y+1)-5=0\Rightarrow(x-2y)^2+(y+1)^2=5$

But $\displaystyle 5=1+4=4+1$

Can you continue? - Jul 30th 2009, 12:04 AMshrrikesh
Do I have to test pairs like (2,0), (0,2), (6,2), etc. or there is some other methods?

- Jul 30th 2009, 12:19 AMred_dog
$\displaystyle \left\{\begin{array}{ll}(x-2y)^2=1\\(y+1)^2=4\end{array}\right.$

From here you have to solve four systems:

$\displaystyle \left\{\begin{array}{ll}x-2y=1\\y+1=2\end{array}\right., \ \left\{\begin{array}{ll}x-2y=-1\\y+1=2\end{array}\right., \ \left\{\begin{array}{ll}x-2y=1\\y+1=-2\end{array}\right., \ \left\{\begin{array}{ll}x-2y=-1\\y+1=-2\end{array}\right.$

and to keep the integer solutions.

Similarly for $\displaystyle \left\{\begin{array}{ll}(x-2y)^2=4\\(y+1)^2=1\end{array}\right.$