, 2021

The numbers we are looking for are $k \in \{0, 3, 4, 5, 8\}$.
A pair $(x, x) \in \mathbb{R}^2$ is auroral if and only if $x = x(3 - x)^2$, and it is easy to see that this cubic equation has the solution set $\{0, 2, 4\}$. These pairs give us 0, 4 and 8 as possible values for $k$.

Next we investigate the case $k = 3$. Here we can simplify the given equations as $x = yx^2$ and $y = xy^2$. The number $x$ cannot be zero in this case, since otherwise $y$ and $k$ would also be zero. We can conclude that $xy = 1$. The equations $x + y = 3$ and $xy = 1$, according to Vieta's Theorem, imply that $x$ and $y$ are the solutions of the equation $\lambda^2 - 3\lambda + 1 = 0$. Hence
$$(x, y) = \left( \frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2} \right) \quad \text{or} \quad (x, y) = \left( \frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2} \right)$$
and it is easy to verify that both pairs are actually auroral. These pairs give us 3 as a possible value for $k$.

A simple calculation shows that if a pair $(x, y) \in \mathbb{R}^2$ is auroral, then the pair $(4 - x, 4 - y)$ is also auroral. From the pairs we have just found, we can therefore construct auroral pairs
$$(x, y) = \left( \frac{5 - \sqrt{5}}{2}, \frac{5 + \sqrt{5}}{2} \right) \quad \text{and} \quad (x, y) = \left( \frac{5 + \sqrt{5}}{2}, \frac{5 - \sqrt{5}}{2} \right),$$
which give us 5 as a possible value for $k$.

Next we investigate the case $k = 4$, where we can simplify the first given equation as $x = (4 - x)(x - 1)^2$. We already know $x = 2$ must be a solution of this cubic equation, since the pair $(2, 2)$ is auroral. By polynomial division we calculate the quotient of $x - (4 - x)(x - 1)^2$ by $x - 2$ as $x^2 - 4x + 2$. The roots of this quadratic polynomial yield pairs
$$(x, y) = (2 + \sqrt{2}, 2 - \sqrt{2}) \quad \text{and} \quad (x, y) = (2 - \sqrt{2}, 2 + \sqrt{2})$$
and it is easy to verify that both pairs are actually auroral.

If $(x, y) \in \mathbb{R}^2$ is auroral, then $x$ satisfies the polynomial equation
$$x - x(3 - x)^2 [3 - x(3 - x)^2]^2 = 0$$
which is of degree 9. Since we have already found nine values for $x$ that must satisfy the equation, there are no more auroral pairs other than those already found. $\square$

Let $f(x) = x(3 - x)^2$. It is easy to check that if $x < 0$ then $f(x) < x$. In particular $f(f(x)) < f(x) < x$ in this case, so that the pair $(x, f(x))$ cannot be auroral. Similarly, $f(x) > x$ if $x > 4$, so the pair $(x, f(x))$ cannot be auroral in this case either.

Suppose that $(x, y) \in \mathbb{R}^2$ is auroral. According to the previous remark $x \in [0, 4]$, and similarly, $y \in [0, 4]$. Hence we may write $x = 2 + 2r$ and $y = 2 + 2s$ with $r, s \in [-1, 1]$. After substitution and simplification, the equation $x = y(3 - y)^2$ transforms into the equation $r = 4s^3 - 3s$. Recall the trigonometric identities for threefold angles. If $s = \cos(\alpha)$ for some $\alpha \in \mathbb{R}$, then $r = 4 \cos^3(\alpha) - 3 \cos(\alpha) = \cos(3\alpha)$. In the same way $s = 4r^3 - 3r = \cos(9\alpha)$.

We can deduce that $9\alpha = 2\pi m + \alpha$ or $9\alpha = 2\pi l - \alpha$ for some integers $m$ and $l$. In the former case we have $8\alpha = 2\pi m$, so that $m \in \{0, 1, 2, 3, 4\}$, and the corresponding possible auroral pairs can be found in Figure 3. In the former case we have $10\alpha = 2\pi l$, so that $l \in \{0, 1, 2, 3, 4, 5\}$, where $l = 0$ and $l = 5$ result in angles that we have already considered in the first case. We consider the other options in Figure 4 taking into account the well-known identities $\cos(\pi/5) = (1 + \sqrt{5})/4$ and $\cos(3\pi/5) = (1 - \sqrt{5})/4$. $\square$