Let $p^2-p+1=x^3$ with $x \in \mathbb{N}^*$ $\Rightarrow p(p-1)=(x-1)(x^2+x+1)$ Since $p$ is prime, $p \mid x-1 \vee p \mid x^2+x+1$ If $p \mid x-1$, $p \leq x-1 \Rightarrow p-1 \leq x-2 <x^2+x+1$ $\Rightarrow x^2+x+1 \nmid p-1$. Therefore, $p \mid x^2+x+1 \Rightarrow x-1 \mid p-1$ Let $x^2+x+1=kp (k \in \mathbb{N}^*) \Rightarrow p-1=k(x-1)$ $\Rightarrow x^2+x+1=k[k(x-1)-1]$ $\Rightarrow x^2+(1-k^2)x+k^2-k+1=0$ (1) $\Delta =(k^2-1)^2-4(k^2-k+1)=k^4-6k^2+4k-3$ (1) has integer solution only if $\Delta =k^4-6k^2+4k-3$ is a perfect square. Plug $k=1,k=2$ we have $\Delta < 0$ (unsatisfied) When $k \geq 3$, $(k^2-3)^2 \leq \Delta < (k^2-2)^2$ $\Rightarrow \Delta =(k^2-3)^2 \Rightarrow k=3$ $\Rightarrow x^2-8x+7=0 \Rightarrow x=1 \vee x=7 $ When $x=1$, $p=0$ (unsatisfied) When $x=7,p=19$. So $p=19$ is the only number satisfied.