Determine all triples of real numbers $(a, b, c)$, that satisfy the following system of equations: $\left\{\begin{array}{l}a^{5}=5 b^{3}-4 c \\ b^{5}=5 c^{3}-4 a . \\ c^{5}=5 a^{3}-4 b\end{array}\right.$

## SOLUTION:
Without loss of generality, we can assume that $a=$ maximum $\{a, b, c\}$.

First case: $c \geq b$. Then $a^{5}+4 c \geq c^{5}+4 b$ and $b \geq a$. In this way, $a=b=c$.

Second case: $b \geq c$. Then $b^{5}+4 a \geq c^{5}+4 b$ and $c \geq a$. Therefore, $a=b=c$.

Thus, everything reduces to solving the equation $t^{5}-5 t^{3}+4 t=0$, where $t=a=b=c$.

Clearly, $t \in\{0,1,-1,2,-2\}$ and there are five possible triples that satisfy the given system.

4.