## SUBJECT IV

Determine the numbers $a, b, c \in[-2,2]$ with $a \leq b \leq c$ and $n \in \mathbb{N}^{*}$ knowing that $a+b+c=-3, a^{3}+b^{3}+c^{3}=-15$, $a \leq b \leq c$ and $a^{n}+b^{n}+c^{n}=8 n+1$.
Solution:

Consider the real numbers $x, y, z$ such that $\sin x=\frac{a}{2}, \sin y=\frac{b}{2}, \sin z=\frac{c}{2}$.

$a+b+c=-3 \Leftrightarrow \sin x+\sin y+\sin z=-\frac{3}{2}$

$a^{3}+b^{3}+c^{3}=-15 \Leftrightarrow \sin ^{3} x+\sin ^{3} y+\sin ^{3} z=-\frac{15}{8}$

$\sin 3 x+\sin 3 y+\sin 3 z=3(\sin x+\sin y+\sin z)-4\left(\sin ^{3} x+\sin ^{3} y+\sin ^{3} z\right)=3$ $1 p$

Since $\sin 3 x, \sin 3 y, \sin 3 z \in[-1,1]$, it follows that $\sin 3 x=\sin 3 y=\sin 3 z=1$ $1 p$

$\sin 3 x=1 \Leftrightarrow 4 \sin ^{3} x-3 \sin x+1=0$ with solutions $\sin x=-1, \sin x=\frac{1}{2}$

Thus $a, b, c \in\{-2,1\}$ and from the given conditions we have that $a=b=-2$ and $c=1$ $1 p$

$a^{n}+b^{n}+c^{n}=8 n+1 \Leftrightarrow(-2)^{n}=4 n$ with the unique solution $n=4$ $1 \mathrm{p}$