12. If $x \in \mathbf{N}^{*}$, and $x^{5}+x+1$ is a prime number, then $x=$
12. $x=1$

Since $\omega^{5}+\omega+1=0,\left(\omega^{2}\right)^{5}+\omega^{2}+1=0$ (where $\omega$ is a cube root of 1), it follows that $x^{2}+x+1 \mid x^{5}+x+1$, so $x=1$