Fun with numbers

Note that $x^3+y^3-(x+y)^2=(x+y)(x^2-(y+1)x+y^2-y)=0$ $x+y=0$ gives infinite number of solutions in the form $(x,y)=(a, -a)$ . For getting integer solutions from $x^2-(y+1)x+y^2-y=0$ , which is a quadratic in terms of $x$ , discriminant of quadratic must be non-negative: $\Delta=-3y^2+6y+1 \ge 0 \\ y=0, 1, 2$ Plugging the values of $y$ in the quadratic gives $(x,y)=(0, 0), (1, 0), (0, 1), (2, 1), (1, 2), (2, 2)$

(0,0) is a solution. If x + y .ne. 0, we can divide both sided by x + y. Then x^2 - xy + y^2 = x + y, Writing as a quadratic in x, x^2 - (y + 1)x + y^2 - y = 0, with solution 2x = (y + 1) +/- sqrt(6y -3y^2 + 1). Letting D^2 = 6y - 3y^2 + 1, 3(y - 1)^2 + D^2 = 4. So y < 3.If y = 0, x = 1. If y = 1, x =0 or 2. If y = 2,x + 1 or 2. . So the ordered pairs are (0,0), (0,1), (1,0), (2,1), (1,2), (2,2). Adding all components, sum = 12. It can be shown that x or y cannot be <0.