Problem 1096

Solution: Since $x^{3}-3 x^{2}+\left(a^{2}+2\right) x-a^{2}=(x-1)\left(x^{2}-2 x+a^{2}\right)$, in order for there to be three distinct real roots it is necessary that $D=1-a^{2}>0$. Therefore $a^{2}\sqrt{1-a^{2}}>0$. The roots of our equation are $x_{1}=1, x_{2}=1+\sqrt{1-a^{2}}, x_{3}=1-\sqrt{1-a^{2}}$. It follows now that $x_{2}+x_{3}=2$ and $2 \geq x_{2}>1$ and $1>x_{3} \geq 0$.

There are two cases to consider:

1. The second term of the progression is $\sin \left(\frac{2 \pi}{3} x_{1}\right)$. Then

$$\begin{aligned} \sin \left(\frac{2 \pi}{3} x_{2}\right)+\sin \left(\frac{2 \pi}{3} x_{3}\right) & =2 \sin \left(\frac{2 \pi}{3}\right) \\ 2 \sin \left(\frac{2 \pi}{3}\left(\frac{x_{2}+x_{3}}{2}\right)\right) \cos \left(\frac{2 \pi}{3}\left(\frac{x_{2}-x_{3}}{2}\right)\right) & =2 \sin \left(\frac{2 \pi}{3}\right) \\ \cos \left(\frac{\pi}{3}\left(x_{2}-x_{3}\right)\right) & =1 \end{aligned}$$

But $\frac{\pi}{3}\left|x_{2}-x_{3}\right|=\frac{2 \pi}{3} \sqrt{1-a^{2}} \leq \frac{2 \pi}{3}$, and hence $\frac{\pi}{3}\left(x_{2}-x_{3}\right) \in$ $\left[-\frac{2 \pi}{3}, \frac{2 \pi}{3}\right]$. Therefore $\cos \left(\frac{\pi}{3}\left(x_{2}-x_{3}\right)\right)=1$ when $x_{2}=x_{3}$, which is impossible, since the roots are distinct.

2. The first or the third term of the progression is $\sin \left(\frac{2 \pi}{3} x_{1}\right)$. Then

$$\sin \left(\frac{2 \pi}{3}\right)+\sin \left(\frac{2 \pi}{3} x_{i}\right)=2 \sin \left(\frac{2 \pi}{3}\left(2-x_{i}\right)\right)$$

for $i=2$ or 3 . Hence

$$\sin \frac{2 \pi}{3}+\sin \left(\frac{2 \pi}{3} x_{i}\right)=2 \sin \frac{4 \pi}{3} \cos \left(\frac{2 \pi}{3} x_{i}\right)-2 \cos \frac{4 \pi}{3} \sin \left(\frac{2 \pi}{3} x_{i}\right)$$

After simple calculations we get $\cos \left(\frac{2 \pi}{3} x_{i}\right)=-\frac{1}{2}$. From the restrictions for $x_{2}$ and $x_{3}$ we obtain $x_{i}=1$ or $x_{i}=2$. In the first case $a^{2}=1$, which is impossible, and in the second case $x_{2}=2, x_{3}=0$ and $a^{2}=0$.

Thus $a$ has a unique value and it is $a=0$.