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The expression is equivalent to :

$\frac{x^2 + y^2 + x + y}{xy}$

Since it is an integer ,

$xy | x^2 + y^2 + x + y$

$\Rightarrow x | y(y + 1) , y | x(x + 1)$

Now, since $y , (y + 1)$ are co-prime. Therefore , either $x | y$ or $x | (y + 1)$ and is co -prime to the other.

Therefore , one of the following holds $x | y , y | x$ or $x | y + 1 , y | x + 1$ .

The former case leads to $x = y$ , using this in the expression forces $x | 2(x + 1) \Rightarrow x = 1 , 2$

Hence , two solutions is $\boxed{(x , y) = (1 , 1) ; (2 , 2)}$

The latter case $\Rightarrow x , y | x + y + 1$ .Since $x , y$ are co-prime , therefore $xy | x + y +1$

$\Rightarrow xy \leq x + y + 1$

$\Rightarrow (x - 1)(y - 1) \leq 2$

This leads to three pairs $\boxed{(x , y) = (2 , 3) ; (3 , 2) ; (2 , 2)}$ and hence we get four solutions namely, $\boxed{(x , y) = (1 ,1) ; (2 , 2) ; (2 , 3) ; (3 , 2)} \Rightarrow \boxed{k = 3 , 4}$

When x>y or y>x is impossible, so x=y

2+(2/x) is integer so x=y=2 and x=y=1

The answer is 4+3=7