Problem 1233

We are trying to solve the equations
$$xy = x + y$$
$$1 = x^2 + y^2.$$

Squaring both sides of $xy = x + y$ and using $x^2 + y^2 = 1$ to simplify gives
$$x^2 y^2 = x^2 + 2xy + y^2 = 1 + 2xy \Rightarrow (xy)^2 - 2(xy) - 1 = 0.$$
This gives $xy = 1 \pm \sqrt{2}$ or $y = \frac{1\pm\sqrt{2}}{x}$.
Notice that from equation $xy = x + y$, we can write $y = \frac{x}{x-1}$.
Combining this with the previous result gives
$$x^2 - x(1 \pm \sqrt{2}) + 1 \pm \sqrt{2} = 0.$$
If $x^2 - x(1 + \sqrt{2}) + 1 + \sqrt{2} = 0$, then
$$\begin{aligned} x &= \frac{(1 + \sqrt{2}) \pm \sqrt{(1 + \sqrt{2})^2 - 4(1 + \sqrt{2})}}{2} \\ &= \frac{(1 + \sqrt{2}) \pm \sqrt{-1 - 2\sqrt{2}}}{2}. \end{aligned}$$
which is non-real.
If $x^2 - x(1 - \sqrt{2}) + 1 - \sqrt{2} = 0$, then
$$\begin{aligned} x &= \frac{(1 - \sqrt{2}) \pm \sqrt{(1 - \sqrt{2})^2 - 4(1 - \sqrt{2})}}{2} \\ &= \frac{(1 - \sqrt{2}) \pm \sqrt{2\sqrt{2} - 1}}{2}. \end{aligned}$$
This presents a viable solution. Notice that if $(x, y)$ is a solution, then $(y, x)$ is also a solution. The two roots of the above equation are the desired values (since they satisfy equation $xy = x + y$ by Vieta's formulas) and we can have $(x, y) = (y, x) = (r_1, r_2)$ where
$$r_1 = \frac{1 - \sqrt{2} + \sqrt{2\sqrt{2} - 1}}{2} \quad \text{and} \quad r_2 = \frac{1 - \sqrt{2} - \sqrt{2\sqrt{2} - 1}}{2}.$$