For real numbers $x,y,z$ solve the system of equations $$\begin{align} x^5 = 5y^3 − 4z, \\ y^5 = 5z^3 − 4x, \\ z^5 = 5x^3 − 4y.\end{align}$$
Source: polish olympiad for high schoolers. Marked as a 'hard' system of equations
No luck getting anywhere. Help would be apprciated.
We can WLOG $x = \max(x, y, z)$ by symmetry. Now, we have two cases to consider:
Case 1: $x \geq y \geq z$
This implies $y^5 \geq z^5, -5z^3 \geq -5x^3$, which we can add to get $y^5 - 5z^3 \geq z^5 - 5x^3$, so $-4x \geq -4y$, or $x \leq y$. Thus, $x = y$, which we can sub back to get $x^5 = 5x^3 - 4z = 5z^3 - 4x$, so $5(x^3 - z^3) + 4(x - z) = 0$. Since $x \geq z$, this equality only occurs when $x = z$, so $x = y = z$.
Case 2: $x \geq z \geq y$
This implies $x^5 \geq z^5, -5y^3 \geq -5x^3$, which we can add to get $x^5 - 5y^3 \geq z^5 - 5x^3$, so $-4z \geq -4y$, or $z \leq y$. Thus, $z = y$, which we can sub back to get $y^5 = 5y^3 - 4x = 5x^3 - 4y$, so $5(x^3 - y^3) + 4(x - y) = 0$. Since $x \geq y$, this equality only occurs when $x = y$, so $x = y = z$.
Thus, in both cases, $x = y = z$, so we have $x^5 = 5x^3 - 4x \implies x(x - 1)(x + 1)(x - 2)(x + 2) = 0 \implies x = 0, \pm 1, \pm 2$. Therefore, all solutions are
$$\boxed{(-2,-2,-2),(-1,-1,-1),(0,0,0),(1,1,1),(2,2,2)}.$$