2. Find all prime numbers $p$ for which $p^{2}-p+1$ is a perfect cube.

(Albania)
2. The equation $p^{2}-p+1=b^{3}$ can be rewritten as $p(p-1)=(b-1)\left(b^{2}+b+1\right)$. Since $b>1$; furthermore, $k \geqslant 3$ because $b^{2}+b+1$ is odd. Now we have $p=k b-k+1$ and

$$
b^{2}+b+1-k p=b^{2}-\left(k^{2}-1\right) b+\left(k^{2}-k+1\right)=0
$$

which is a quadratic equation in $b$. Its discriminant $D=\left(k^{2}-1\right)^{2}-4\left(k^{2}-k+1\right)=$ $k^{4}-6 k^{2}+4 k-3$ must be a perfect square, and since $\left(k^{2}-3\right)^{2} \leqslant D<\left(k^{2}-2\right)^{2}$, it must be $D=\left(k^{2}-3\right)^{2}$, which gives us $k=3$. Now from (1) we get $b=7$ and $p=19$, which is the only solution.