Brazilian Math Olympiad

a. First notice that, since $r$ is a root, then $r^3 + r^2 - 4r + 1 = 0 \iff r^2 + r - 3 = 1 - \frac{1}{r}$. So we need to prove that
$$\begin{aligned} & \left(1 - \frac{1}{r}\right)^3 + \left(1 - \frac{1}{r}\right)^2 - 4\left(1 - \frac{1}{r}\right) + 1 = 0 \\ \iff & 1 - \frac{3}{r} + \frac{3}{r^2} - \frac{1}{r^3} + 1 - \frac{2}{r} + \frac{1}{r^2} - 4 + \frac{4}{r} + 1 = 0 \\ \iff & -1 - \frac{1}{r} + \frac{4}{r^2} - \frac{1}{r^3} = 0 \\ \iff & r^3 + r^2 - 4r + 1 = 0, \end{aligned}$$
which is true.

b. Iterating $1 - \frac{1}{r}$, we get $1 - \frac{1}{1-\frac{1}{r}} = \frac{1}{1-r}$. It's not hard to see that $r, 1 - \frac{1}{r} = \frac{r-1}{r}$ and $\frac{1}{1-r}$ are all distinct. In fact, if $r = 1 - \frac{1}{r}$ then $r^2 - r + 1 = 0$, and $r^3 = -1$, so $r^2 - 4r = 0$, which is not true.
So there are two possible ways of computing $\frac{\alpha}{\beta} + \frac{\beta}{\gamma} + \frac{\gamma}{\alpha}$:
* $(\alpha, \beta, \gamma) = (r, \frac{r-1}{r}, \frac{1}{1-r})$:
$$\begin{aligned} \frac{\alpha}{\beta} + \frac{\beta}{\gamma} + \frac{\gamma}{\alpha} &= \frac{r^2}{r-1} - \frac{(r-1)^2}{r} - \frac{1}{r(r-1)} \\ &= \frac{r^3 - (r-1)^3 - 1}{r(r-1)} = \frac{3r(r-1)}{r(r-1)} = 3 \end{aligned}$$

* $(\alpha, \beta, \gamma) = (r, \frac{1}{1-r}, \frac{r-1}{r})$:
$$\begin{aligned} \frac{\alpha}{\beta} + \frac{\beta}{\gamma} + \frac{\gamma}{\alpha} &= -r(r-1) - \frac{r}{(r-1)^2} + \frac{r-1}{r^2} \\ &= \frac{-r^6 + 3r^5 - 3r^4 + r^3 - 3r^2 + 3r - 1}{r^4 - 2r^3 + r^2} \end{aligned}$$

Since $-r^6 + 3r^5 - 3r^4 + r^3 - 3r^2 + 3r - 1 = (r^3 + r^2 - 4r + 1) \cdot (-r^3 + 4r^2 - 11r + 29) - 80r^2 + 130r - 30 = -80r^2 + 130r - 30$ and $r^4 - 2r^3 + r^2 = (r^3 + r^2 - 4r - 1) \cdot (r - 3) + 8r^2 - 13r + 3 = 8r^2 - 13r + 3$. So
$$\frac{\alpha}{\beta} + \frac{\beta}{\gamma} + \frac{\gamma}{\alpha} = \frac{-80r^2 + 130r - 30}{8r^2 - 13r + 3} = -10.$$

Another way to solve this problem is realizing that if $(\alpha, \beta, \gamma) = (r, \frac{r-1}{r}, \frac{1}{1-r})$ then the other sum is actually $\frac{\beta}{\alpha} + \frac{\gamma}{\beta} + \frac{\alpha}{\gamma}$. By Vieta's formula, $\sigma_1 = \alpha+\beta+\gamma = -1$, $\sigma_2 = \alpha\beta + \beta\gamma + \gamma\alpha = -4$ and $\sigma_3 = \alpha\beta\gamma = -1$.
$$\begin{aligned} \frac{\alpha}{\beta} + \frac{\beta}{\gamma} + \frac{\gamma}{\alpha} + \frac{\beta}{\alpha} + \frac{\gamma}{\beta} + \frac{\alpha}{\gamma} &= \frac{\alpha^2\beta + \alpha^2\gamma + \beta^2\alpha + \beta^2\gamma + \gamma^2\alpha + \gamma^2\beta}{\alpha\beta\gamma} \\ &= \frac{\sigma_1\sigma_2 - 3\sigma_3}{\sigma_3} = \frac{(-1) \cdot (-4) - 3(-1)}{-1} = -7, \end{aligned}$$
so
$$$$
$\frac{\beta}{\alpha} + \frac{\gamma}{\beta} + \frac{\alpha}{\gamma}$ = -7 - 3 = -10.