X and Y are in Roots!

We noticr that $4x + y + 4\sqrt{xy} - 36\sqrt{x} - 18\sqrt{y} + 80 = (2\sqrt{x} + \sqrt{y})^2 - 18(2\sqrt{x} + \sqrt{y}) + 80 =0$ .

Let $z = 2\sqrt{x} + \sqrt{y}$ , then the equation becomes $z^2 - 18z + 80 =0$ .

$(z-8)(z-10) =0 \Rightarrow z_{1,2} = (8,10)$ .

We have $2\sqrt{x} + \sqrt{y} = 8 \quad ...(1)$ and $2\sqrt{x} + \sqrt{y} = 10 \quad ...(2)$ .

Since $x$ and $y$ are integers, then $x$ and $y$ are both quadratic numbers.

Start with equation $(1)$

If $x=0, \quad y = 64$ ;

If $x = 1, \quad y =36$ ;

If $x = 4, \quad y = 16$ ;

If $x = 9, \quad y = 4$ ;

If $x = 16, \quad y=0$ .

From the equation $(1)$ , we have $5$ pairs of $(x,y)$ .

Now, equation $(2)$

If $x=0, \quad y=100$ ;

If $x= 1, \quad y= 64$ ;

If $x = 4, \quad y= 36$ ;

If $x = 9, \quad y = 16$ ;

If $x=16, \quad y = 4$ ;

If $x= 25, \quad y=0$ .

From equation $(2)$ , we have $6$ pairs of $(x,y)$ .

Hence, the total number of pairs of integers of $(x,y)$ is $\boxed{\color{#3D99F6}11}$ .