Tripping triples

Write $a = x + \frac 1 x$ , $b = y + \frac 1 y$ and $c = z + \frac 1 z$ for complex $x, y, z$ . Then we have:

$b = a^3 - 3a = x^3 + \frac 1 {x^3}, \ c = y^3 + \frac 1 {y^3}, \ a = z^3 + \frac 1 {z^3}.$

Thus, this gives $x + \frac 1 x = x^{27} +\frac 1 {x^{27}}$ . Next, we use the following fact (whose proof we omit, since it's easy):

• if $\alpha + \frac 1 \alpha = \beta + \frac 1 \beta$ , then $\alpha = \beta, \frac 1 \beta$ .

Thus, we conclude that $x = x^{27}$ or $x = x^{-27}$ . Now $x \ne 0$ so $x$ is a 26-th or 28-th root of unity. So $a = x + \frac 1 x = 2 \text{Re}(x)$ and replacing $x$ by $x^{-1}$ if necessary, we shall assume $\text{Im}(x) \ge 0$ .

Now, if $x = \pm 1$ , then $a = b = c = \pm 2$ so this solution is not pairwise distinct.

Likewise, for $x = i$ , we have $a = b = c =0$ which is omitted.

For the remaining solutions, we have:

• $x = e^{2\pi i k/26}$ , where $1 \le k \le 12$ , or
• $x = e^{2\pi i k/28}$ , where $1 \le k \le 13$ and $k \ne 7$ .

This gives 24 solutions.

If $|a| > 2$ then $|a^2-3|>1$ , and so $|a^3-3a|>|a|$ . Thus $|b|>|a|>2$ , and hence $|c|>|b|>|a|$ and finally $|a|>|c|>|a|$ , which is impossible. Thus $|a| \le 2$ , so we can write $a = 2\cos\alpha$ . But then $b = 2\cos3\alpha$ and $c = 2\cos9\alpha$ , and we know that $2\cos\alpha = a = c^3 - 3c \; = \; 2\cos27\alpha$ so that $0 \; = \; \cos27\alpha - \cos\alpha \; = \; -2\sin14\alpha\sin13\alpha$ Thus possible values of $\alpha$ are $\tfrac{k\pi}{14}$ for $0 \le k \le 14$ and $\tfrac{m\pi}{13}$ for $0 \le m \le 13$ . Other values of $k$ and $m$ merely repeat the possible values of $a=2\cos\alpha$ .

We must have $a \neq b$ (if $a=c$ or $b=c$ the cyclic nature of the definition would force $a=b$ ). Thus we want $0 \neq \cos3\alpha - \cos\alpha \; = \; -2\sin2\alpha\sin\alpha$ and so we have to exclude $\alpha = 0,\tfrac{1}{2}\pi,\pi$ from the above lists. Thus possible values of $a$ are $2\cos\tfrac{k\pi}{14}$ for $1 \le k \le 6$ or $8 \le k \le 13$ and $2\cos\tfrac{m\pi}{13}$ for $1 \le m \le 12$ Since $13$ and $14$ are coprime, these values are distinct. Thus there are $13+12-1=24$ possible values of $a$ , each of which determine the corresponding values of $b$ and $c$ .