The total number of pairs of integers (x, y) which satisfy the equation x^2-4xy+5y^2+2y-4=0 is?

Interesting and not very easy question.
First, we have to complete the squares for x and y and rewrite this equation in "standard" form.



x and y are integers and we know that if two numbers a and b are integers then (a+b), (a-b), a*b are integers too, so (x-2y) and (y+1) from our equation, are integers.

In equation, we add two squares and get 5 as result. This is possible only when we have:

1² + 2² = 1 + 4 = 5
or
2² + 1² = 4 + 1 = 5

If so, there will be:

1.



For y = 1



we have two pairs (3,1) and (1,1).

For y = -3



we have two pairs (-5,-3) and (-7,-3).

2.



For y = 0



we have two pairs (2,0) and (-2,0)

For y = -2



we have two pairs (-2,-2) and (-6,-2).

So the total number of pairs of integers which satisfy the equation is 8.