Relations and Functions

$x^2+y=y^2+x$

$x^2-y^2=x-y$

$(x-y)(x+y)=x-y$

$(x-y)(x+y-1)=0$

Case 1: $x=y$

$x^2+x=12$

$x^2+x-12=0 \Rightarrow y=x=3$ or $y=x=-4$

Case 2: $x+y=1$

$24=x^2+y^2+x+y=(x+y)^2-2xy+x+y=1^2-2xy+1=2-2xy \Rightarrow xy = -11$

So by Vieta's formulas x and y are the roots of the quadratic equation $x^2-x-11=0 \Rightarrow x=\frac {1+3\sqrt{5}}{2}$ and $y=\frac {1-3\sqrt{5}}{2}$ or $x=\frac {1-3\sqrt{5}}{2}$ and $y=\frac {1+3\sqrt{5}}{2}$ .

In summary we have

$x_1=y_1=3$

$x_2=y_2=-4$

$x_3=\frac {1+3\sqrt{5}}{2}$ , $y_3=\frac {1-3\sqrt{5}}{2}$

$x_4=\frac {1-3\sqrt{5}}{2}$ , $y_4=\frac {1+3\sqrt{5}}{2}$

So the answer is 4.

$y=12-x^2$

$(12-x^2)^2+x=12$

Solutions: $x=3, x=-4, x=\frac12(1+3\sqrt5), x=\frac12(1-3\sqrt5)$

Corresponding $y$ 's: $y=3, y=-4, y=\frac12(1-3\sqrt5), y=\frac12(1+-3\sqrt5)$

Relevant wiki: Rational Root Theorem - Basic

$\begin{cases} x^2 + y = 12 & ...(1) \\ y^2 + x = 12 & ...(2) \end{cases}$

From (1): $\quad y = 12 - x^2$

From (2):

$\begin{aligned} \left(12-x^2\right)^2 + x & = 12 \\ x^4 -24x^2 + x + 132 & = 0 & \small \color{#3D99F6} \text{Using rational root theorem, }x = -4, 3 \\ (x+4)(x-3)(x^2 - x -11) & = 0 \\ (x+4)(x-3)\left(x-\frac {1-3\sqrt 5}2\right)\left(x-\frac {1+3\sqrt 5}2\right) & = 0 \end{aligned}$

Since there are 4 roots for $x$ , there are $\boxed{4}$ ordered pairs $(x,y)$ and they are:

$\begin{cases} x = - 4 & y = - 4 \\ x = 3 & y = 3 \\ x = \frac {1-3\sqrt 5}2 & y = \frac {1+3\sqrt 5}2 \\ x = \frac {1+3\sqrt 5}2 & y = \frac {1-3\sqrt 5}2 \end{cases}$