Weird Systems

From $x^2 + y^2 = 8x+8y+2$

$\begin{aligned} \Rightarrow x^2 - 8x + 16 + y^2 - 8y + 16 & = 2+16 +16 \\ (x-4)^2 + (y-4)^2 & = 34 \\ \Rightarrow (y-4)^2 & = 34 - (x-4)^2 \quad \quad \small \color{#3D99F6}{\text{For natural numbers } x \text{ and }y,} \\ & = \begin{cases} 34 - 0^2 = 34 & \small \color{#D61F06}{\text{not perfect square, rejected}} \\ 34 - 1^2 = 33 & \small \color{#D61F06}{\text{not perfect square, rejected}} \\ 34 - 2^2 = 30 & \small \color{#D61F06}{\text{not perfect square, rejected}} \\ 34 - 3^2 = 25 & \Rightarrow \color{#3D99F6}{x = 7, \quad y = 9} \\ 34 - 4^2 = 18 & \small \color{#D61F06}{\text{not perfect square, rejected}} \\ 34 - 5^2 = 9 & \Rightarrow \color{#3D99F6}{x = 9, \quad y = 7} \end{cases} \end{aligned}$

Substituting the values of $x$ and $y$ in $y^3+3y = x^3 + 3x^2 + 266$ ,

$\begin{aligned} x=7 \quad y=9 \quad \Rightarrow 9^3 + 3(9) & = 7^3 + 3(7^2) + 266 \\ \Rightarrow \color{#3D99F6}{756} & \color{#3D99F6}{= 756 \quad \text{accepted}} \\ x=9 \quad y=7 \quad \Rightarrow 7^3 + 3(7) & = 9^3 + 3(9^2) + 266 \\ \Rightarrow \color{#D61F06}{364} & \color{#D61F06}{\ne 1238 \quad \text{rejected}} \end{aligned}$

Since $xy = 7\times 9 = 63$ , and its sum of digits is $\boxed{9}$ .

the second equation is sufficient enough for the answer , only 2 points either (7,9) or (9,7) are the points with natural coordinates that satisfy the equation of circle , hence product 63 and sum is 9.