Conceptual Equation

Set $\displaystyle x=u+1,y=v+1$ which leads to the equation $\displaystyle (uv+1)(u+v+4)=5$

Now here the factors are either equal to $5$ or $-5$ , which leads to four set of simultaneous equations of which only two are solvabe and they are,

$\displaystyle \begin{cases} u+v=1 \\ uv=0 \end{cases}$ & $\displaystyle \begin{cases} u+v=-5 \\ uv=-6 \end{cases}$

Hence $\displaystyle (x,y)=(u+1,v+1)=(1,2),(2,1),(-5,2),(2,-5)$ & the answer is $\boxed{0}$

First, use distributive and factor into: $x^2 y + y^2 x=x^2+y^2+1 \to xy(x+y)=(x+y)^2 -2xy + 1 \to xy(x+y+2)=(x+y)^2+1$ . Let $s=x+y$ and $t=xy$ we get $t(s+2)=s^2+1$ . Now, since neither $x=0$ nor $y=0$ give integer solutions, $t \neq 0$ . So we see that $s+2 | s^2 + 1=(s+2-2)^2+1=(s+2)^2 -4(s+2)+5 \to s+2 | 5$ . The last equation gives four answers, since 5 is prime, that are $\{3,-3,-1,-7\}$ . Now we have the pairs $(s,t) \in \{(3,2),(-3,-10),(-1,2),(-7,-10)\}$ . Notice that x and y must solve the equation $z^2-sz+t=0$ , meaning $\Delta=s^2-4t \geq 0$ and, since they are integers, $\Delta$ must be a perfect square, which the least two pairs don't do. So, we must have $(s,t) \in \{(3,2),(-3,-10)\}$ , therefore the sum in the problem must be $0$ . We should've doubled the result, accounting for the symmetry on the variables, but it's the same. For completeness, $(x,y) \in \{(1,2),(2,1),(2,-5),(-5,2)\}$