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),  $p\equiv{1\mod18}$ 

this is an easier answer:
$$$$we have$$p^2-p=p(p-1)=m^3-1=(m-1)(m^2+m+1)$$we can see $p>m-1$ so we have $$p-1|m-1\to p=1\, MODm-1$$and we have$$m^2+m+1=3\, MODm-1$$we know $$p|m^2+m+1$$ so we have$$m^2+m+1=3p\, or\, m^2+m+1\ge (m+2)p$$if $$m^2+m+1=3p$$ we can see p=19,we have $3(m-1)=p-1$ so we have $m^2+m+1=9m-6$ so m=1 or
,m=7,p=19,$$$$if $$m^2+m+1\ge (m+2)p$$ we have$$(m+2)(m-1)\ge p-1$$ so $$m^2+m-1\ge p=(m^2+m+1)/m+2$$because $m>1$ it is clearly impossible.
