288. Does the equation $x^{2}+y^{3}=z^{4}$ have solutions in prime numbers $x, y, z$?
288. Answer: does not have. From $y^{3}=\left(z^{2}-x\right)\left(z^{2}+x\right)$ and the simplicity of $y$, we get either $z^{2}-x=1, z^{2}+x=y^{3}$, or $z^{2}-x=y$, $z^{2}+x=y^{2}$. Both systems do not have solutions in prime numbers