3. The real numbers $a, b$ and $c$ satisfy that the polynomial $p(x)=x^{4}+a x^{3}+b x^{2}+a x+c$ has exactly three distinct real roots; these roots are equal to $\tan (y), \tan (2 y)$ and $\tan (3 y)$, for some real number $y$. Find all possible values of $y, 0 \leq y<\pi$.
Solution. Let $r, r, s, t$ be the three distinct real zeros of the polynomial $p(x)$. By the Cardano-Viète formulas, identifying the cubic and linear coefficients, we obtain

$$
2 r+s+t=r^{2} s+r^{2} t+2 r s t \Leftrightarrow 2 r(1-s t)=\left(r^{2}-1\right)(s+t)
$$

From the last equality, it follows that if $r^{2}-1=0$ then $1-s t=0$ and vice versa. Next, under the hypothesis $r^{2}=s t=1$ and considering $0 \leq y<\pi$, we will analyze the following cases:

- If $r=\tan y$ then $y \in\left\{\frac{\pi}{4}, \frac{3 \pi}{4}\right\}$ but for these values $s=\tan 2 y$ is not defined.
- If $r=\tan 2 y$ then $y \in\left\{\frac{\pi}{8}, \frac{3 \pi}{8}, \frac{5 \pi}{8}, \frac{7 \pi}{8}\right\}$. Setting $s=\tan y$ and $s=\tan 3 y$ we have that for all the above values $r s=\tan y \cdot \tan 3 y=1$, which is true if $y+3 y=4 y$ is an odd multiple of $\pi / 2$, and this occurs for all the above values.
- If $r=\tan 3 y$ then it must be that $r s=\tan y \cdot \tan 2 y=1$ and $\tan 3 y$ is not defined.

Therefore, only the cases where $r^{2}-1 \neq 0$ and $1-s t \neq 0$ remain to be analyzed. Dividing both sides of the equality $2 r(1-s t)=\left(r^{2}-1\right)(s+t)$ by $\left(r^{2}-1\right)(1-s t)$ results in

$$
\frac{2 r}{1-r^{2}}+\frac{s+t}{1-s t}=0
$$

These expressions separately would correspond to the formulas for the tangent of the double angle and the tangent of the sum. Now, using the last expression, we will distinguish the following cases:

- If $r=\tan y, s=\tan 2 y, t=\tan 3 y$ then $\tan 2 y+\tan 5 y=0$ which admits solutions that are multiples of $\frac{\pi}{2}$ less than $\frac{7 \pi}{2}$, so

$$
y \in\left\{\frac{\pi}{7}, \frac{2 \pi}{7}, \frac{3 \pi}{7}, \frac{4 \pi}{7}, \frac{5 \pi}{7}, \frac{6 \pi}{7}\right\}
$$

- If $r=\tan 2 y, s=\tan y, t=\tan 3 y$ then $\tan 4 y+\tan 4 y=0$ which admits solutions that are multiples of $\frac{\pi}{2}$ less than $4 \pi$, so

$$
y \in\left\{\frac{\pi}{8}, \frac{3 \pi}{8}, \frac{5 \pi}{8}, \frac{7 \pi}{8}\right\}
$$

and we have rejected $y=\frac{\pi}{2}$ since then $\tan y$ would not be a real number, and also $\frac{\pi}{4}$ and $\frac{3 \pi}{4}$ since then $\tan (2 y)$ would also not be a real number.

- If $r=\tan 3 y, s=\tan y, t=\tan 2 y$ then $\tan 6 y+\tan 3 y=0$ which admits solutions that are multiples of $\frac{\pi}{2}$ less than $\frac{9 \pi}{2}$, so

$$
y \in\left\{\frac{\pi}{9}, \frac{2 \pi}{9}, \frac{\pi}{3}, \frac{4 \pi}{9}, \frac{5 \pi}{9}, \frac{2 \pi}{3}, \frac{7 \pi}{9}, \frac{8 \pi}{9}\right\}
$$

These are all the values, and we are done.