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

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### 1 Answer

First, you'll have to provide an equation. You've provided an expression, for the moment. We'll transform the expression into an equation:

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

We'll create the following groups;

(x^2-4xy+4y^2) + (y^2+2y + 1)- 1 - 4 = 0

We notice that we've made some changes within the expression you've provided, namely:

5y^2 = 4y^2 + y^2

- we've added and subtracted the value of 1

We notice that creating the groups above, we've created, in fact, two perfect squares:

(x-2y)^2 + (y+1)^2 - 5 = 0

Now, we'll have to determine what are the integers which added to give 5?

1 + 4, 2 + 3, 3 + 2, 4 + 1

Let x - 2y = 1 and y + 1 = 2

x = 2y + 1

y = 1 => x = 3

The first integer pair which satisfies the given equation is (3,1).

Let x - 2y = sqrt2 and y + 1 = sqrt3

Since x and y are not integer, we'll move to the next possibility:

x - 2y = 2 and y + 1 = 1 => y = 0

x = 2

The next possible pair of integer numbers, that makes the equality to hold is (2;0).

**Therefore, the pairs of integer numbers that satisfy the given equation are: (3,1) and (2;0).**