Let's Try Substitution First

Let $X=x+3$ and $Y=y-3$ . Then the given equation becomes ${(X+Y)}^{2}=XY$ .

So ${X}^{2}+XY+{Y}^{2}=0$ . However, ${X}^{2}, {Y}^{2}$ and $XY = ({X+Y})^{2}$ are non-negative.

Hence $X=Y=0$ ; so $x=-3$ and $y=3$ is the only solution.

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See here for a more difficult way to solve this problem.

Let $u = x+3$ and $v = x-3$ . Then we have:

$\begin{aligned} (x+y)^2 & = (x+3)(x-3) \\ (u+v)^2 & = uv \\ u^2 + uv + v^2 & = 0 \\ \Rightarrow u & = \frac{-v \pm \sqrt{v^2-4v^2}}{2} \\ & = \frac{-v \pm \sqrt{\color{#3D99F6}{-3v^2}}}{2} \end{aligned}$

We note that $\color{#3D99F6}{-3v^2} \le 0$ therefore, $u$ is only real when $\color{#3D99F6}{-3v^2} = 0$ , that is $v = 0$ and $u=0$ and that $x = -3$ and $y = 3$ . Therefore, there is only $\boxed{1}$ pair of $(x,y)$ that satisfies the equation.

$(x+y)^2=(x+3)(y-3)$ $x^2+y^2+2xy=xy-3x+3y-9$ $x^2+y^2+xy+3x-3y+9=0$ $(x^2+6x+9)+(y^2-6y+9)+(xy-3x+3y-9)=0$ $(x+3)^2+(y-3)^2+(x+3)(y-3)=0$ Since $(x+3)^2$ and $(y-3)^2$ can't be lower than zero, so both of them must be zero $$ $x+3=0$ and $y-3=0$ $$ $x=-3$ and $y=3$ $$ So the only solution is $x=-3$ and $y=3$ .