Problem 1985

Let us write the given equations:
$$x + 4 = (y - 2)^2 \tag{1}$$
$$y + 4 = (x - 2)^2 \tag{2}$$
Expand the right sides:
From (1):
$$x + 4 = y^2 - 4y + 4$$
So
$$x = y^2 - 4y$$
From (2):
$$y + 4 = x^2 - 4x + 4$$
So
$$y = x^2 - 4x$$
Now substitute $x$ from above into the expression for $y$:
$$y = (y^2 - 4y)^2 - 4(y^2 - 4y)$$
Let $s = y$ for simplicity:
$$s = (s^2 - 4s)^2 - 4(s^2 - 4s)$$
Expand $(s^2 - 4s)^2$:
$$(s^2 - 4s)^2 = s^4 - 8s^3 + 16s^2$$
So
$$s = s^4 - 8s^3 + 16s^2 - 4s^2 + 16s$$
$$s = s^4 - 8s^3 + 12s^2 + 16s$$
Bring all terms to one side:
$$0 = s^4 - 8s^3 + 12s^2 + 15s$$
Factor $s$:
$$s(s^3 - 8s^2 + 12s + 15) = 0$$
Since $x$ and $y$ are distinct, $s = 0$ is a possible value, but let's factor the cubic:
Let us try rational roots for $s^3 - 8s^2 + 12s + 15 = 0$.
Try $s = -1$:
$$(-1)^3 - 8(-1)^2 + 12(-1) + 15 = -1 - 8 + (-12) + 15 = -1 - 8 - 12 + 15 = -21 + 15 = -6$$
Try $s = 1$:
$$1 - 8 + 12 + 15 = 1 - 8 + 12 + 15 = -7 + 12 + 15 = 5 + 15 = 20$$
Try $s = 3$:
$$27 - 72 + 36 + 15 = 27 - 72 + 36 + 15 = -45 + 36 + 15 = -9 + 15 = 6$$
Try $s = -3$:
$$(-3)^3 - 8(-3)^2 + 12(-3) + 15 = -27 - 72 - 36 + 15 = -27 - 72 - 36 + 15 = -99 - 36 + 15 = -135 + 15 = -120$$
Try $s = 5$:
$$125 - 200 + 60 + 15 = 125 - 200 + 60 + 15 = -75 + 60 + 15 = -15 + 15 = 0$$
So $s = 5$ is a root.
Divide $s^3 - 8s^2 + 12s + 15$ by $(s - 5)$:
Use synthetic division:
Coefficients: 1, -8, 12, 15
5 | 1 -8 12 15
5 -15 -15
1 -3 -3 0
So $s^3 - 8s^2 + 12s + 15 = (s - 5)(s^2 - 3s - 3)$
So the equation is:
$$s(s - 5)(s^2 - 3s - 3) = 0$$
So possible $y$ values are $0$, $5$, and the roots of $s^2 - 3s - 3 = 0$.
Solve $s^2 - 3s - 3 = 0$:
$$s = \frac{3 \pm \sqrt{9 + 12}}{2} = \frac{3 \pm \sqrt{21}}{2}$$
Now recall $x = y^2 - 4y$.
If $y = 0$:
$$x = 0^2 - 4 \cdot 0 = 0$$
But $x$ and $y$ are distinct, so $x \neq y$.
If $y = 5$:
$$x = 25 - 20 = 5$$
Again, $x = y$, not allowed.
Try $y = \frac{3 + \sqrt{21}}{2}$:
$$x = \left(\frac{3 + \sqrt{21}}{2}\right)^2 - 4 \left(\frac{3 + \sqrt{21}}{2}\right)$$
Compute $\left(\frac{3 + \sqrt{21}}{2}\right)^2$:
$$= \frac{(3 + \sqrt{21})^2}{4} = \frac{9 + 6\sqrt{21} + 21}{4} = \frac{30 + 6\sqrt{21}}{4} = \frac{15 + 3\sqrt{21}}{2}$$
So
$$x = \frac{15 + 3\sqrt{21}}{2} - 2(3 + \sqrt{21})$$
$$x = \frac{15 + 3\sqrt{21}}{2} - (6 + 2\sqrt{21})$$
$$x = \frac{15 + 3\sqrt{21} - 12 - 4\sqrt{21}}{2}$$
$$x = \frac{3 - \sqrt{21}}{2}$$
So for $y = \frac{3 + \sqrt{21}}{2}$, $x = \frac{3 - \sqrt{21}}{2}$.
Similarly, for $y = \frac{3 - \sqrt{21}}{2}$:
$$x = \frac{3 + \sqrt{21}}{2}$$
So the two distinct real solutions are:
$$x = \frac{3 - \sqrt{21}}{2}, \quad y = \frac{3 + \sqrt{21}}{2}$$
or
$$x = \frac{3 + \sqrt{21}}{2}, \quad y = \frac{3 - \sqrt{21}}{2}$$
Now compute $x^2 + y^2$:
Let $x = \frac{3 - \sqrt{21}}{2}$, $y = \frac{3 + \sqrt{21}}{2}$.
Then
$$x^2 + y^2 = \left(\frac{3 - \sqrt{21}}{2}\right)^2 + \left(\frac{3 + \sqrt{21}}{2}\right)^2$$
Compute each term:
$$\left(\frac{3 - \sqrt{21}}{2}\right)^2 = \frac{9 - 6\sqrt{21} + 21}{4} = \frac{30 - 6\sqrt{21}}{4}$$
$$\left(\frac{3 + \sqrt{21}}{2}\right)^2 = \frac{9 + 6\sqrt{21} + 21}{4} = \frac{30 + 6\sqrt{21}}{4}$$
Add:
$$x^2 + y^2 = \frac{30 - 6\sqrt{21}}{4} + \frac{30 + 6\sqrt{21}}{4} = \frac{60}{4} = 15$$
Answer: $x^2 + y^2 = 15$.