<s>[INST] Find all primes $p$ such that $p^2-p+1$ is a perfect cube Find all primes $p$ such that $p^2-p+1$ is a perfect cube.
I found out that p is of the form $18n+1$ and $p=19$ is a solution but I am not getting anything further.
$p^2-p-(m^3-1)=0$
$1+4(m^3-1)=k^2$
$4m^3-3=k^2$
every square is $0,1,4 or 7(\mod9)$ and every cube is $0,1 or 8\mod9$ 
Using this fact we can conclude that $m^3\equiv{1\mod 9}$ 
Hence $k$ is either $1$ or $-1$ $\mod9$
But $k\equiv{-1\mod9}$ is not possible since $p$ is a prime. So putting  $k\equiv{1\mod9}$ in the quadratic formula we get $p\equiv{1\mod9}$ and since $p$ is odd(as $p=2$ is not a solution),  $p\equiv{1\mod18}$ 
 [/INST] you can work in $Z[1/2+\sqrt{-3}/2]$.we have unique factorization in this ring,and units are $1,-1,1/2+\sqrt{-3}/2,1/2-\sqrt{-3}/2$,and 2 is prime,yo can easily check this facts by the norm
$N(a+b\sqrt{-3})=a^2+3b^2$
we have
$4m^3=k^2+3=(k-\sqrt{-3})(k+\sqrt{-3})$,we can see $gcd((k-\sqrt{-3}),(k+\sqrt{-3})|2\sqrt{-3}$
and because$(k,3)=1$ we have $gcd((k-\sqrt{-3}),(k+\sqrt{-3})=2 or 1 $and we have$4(a/2+\sqrt{-3}b/2)\not=k+\sqrt{-3}$ so we have this cases:
$$1:k+\sqrt{-3}=2(a/2+\sqrt{-3}b/2)^3,k-\sqrt{-3}=2(a/2-\sqrt{-3}b/2)^3$$
$$2:k+\sqrt{-3}=(1+\sqrt{-3})(a/2+\sqrt{-3}b/2)^3,k-\sqrt{-3}=(1-\sqrt{-3})(a/2-\sqrt{-3}b/2)^3$$
$$3:k+\sqrt{-3}=(1-\sqrt{-3})(a/2+\sqrt{-3}b/2)^3,k-\sqrt{-3}=(1+\sqrt{-3})(a/2-\sqrt{-3}b/2)^3$$
in case 1 by take imaginary part we have:$2(-3b^3/8+3a^2b/8)=1$ so case 1 is impossible.
in case 2 by take real and imaginary part we have :
$$-3b^3/8+3a^2b/8=1-k=2p$$
$$a^3/8-9ab^2/8=k+3$$
because $p\not=3$ case 2 is impossible
case 3:we have
$$-3b^3/8+3a^2b/8=1+k=2-2p$$
$$a^3/8-9ab^2/8=k-3=-2-2p$$
$$-3b^3/8+3a^2b/8+a^3/8-9ab^2/8=-4p$$
i cant go further in this case
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