Stringent Diophantine

$x^2 + y^2 = 2(x + y) + xy$

if both x, y are odd, then left is even but right is odd; if one of x, y is odd and the other is even, then left is odd but right is even. So both x, y must be even.

Let $x = 2u$ , $y = 2v$ , then: $\\ 4u^2 + 4v^2 = 4(u + v) + 4uv \\ 2u^2 + 2v^2 = 2(u + v) + 2uv \\ (u - v)^2 + (u - 1)^2 + (v- 1)^2 = 2$

Obviously, one of left 3 terms must be 0 and the other 2 are 1, then we can get 6 solutions for $(u, v)$ : $(0, 0), (2, 2), (1, 0), (1, 2), (0, 1), (2, 1)$

i. e. for $(x, y)$ : $(0,0), (4,4), (2, 0), (2,4), (0,2), (4,2)$

$x^2+y^2-2x-2y-xy=0$ We treat the above equation as a quadratic in terms of $x$ . So, we have, $x^2-(y+2)x+y^2-2y$ $\begin{aligned} \implies x&=\frac{y+2\pm\sqrt{(y+2)^2-4(y^2-2y)}}{2} \\ &=\frac{y+2\pm\sqrt{-3y^2+12y+4}}{2} \end{aligned}$ Since $x$ is an integer, the discriminant of the quadratic, $-3y^2+12y+4$ is greater that $0$ . $-3y^2+12y+4>0 \implies 3y(y-4)<4$ Since $y$ is an integer that satisfies the above inequality, $y\in \{0,1,2,3,4\}$

Checking each of the four cases, the order pairs $(x,y)$ are, $(x,y)\in \{(0,0),(2,0),(0,2),(4,2),(2,4),(4,4)\}$ Therefore, there are $\boxed{6}$ order pairs $(x,y)$ that satisfy the given equation.

We have

$\begin{aligned} x^2-xy+y^2-2x-2y=&0 \\ x^2-2xy+y^2+x^2-4x+4+y^2-4y+4=&8 \\ (x-y)^2+(x-2)^2+(y-2)^2=&8 \end{aligned}$

which means that we have $((x-y)^2, (x-2)^2, (y-2)^2)\in \{(0,4,4), (4,0,4), (4,4,0)\}$ (since $0\leq (x-y)^2, (x-2)^2, (y-2)^2 \leq 8$ ) and by considering each case, we get that $\boxed{(x,y)=(0,0),(4,4),(2,0),(2,4),(0,2),(4,2)}.$