I'm trying to prove that the primes with form $3k+1$ are not Eisenstein prime. This step: to find $a$ such that $a^2-a\equiv -1\pmod p$ when $p$ is a prime with $p\equiv 1\pmod 3$ is the only obstacle now, and I have thought of many methods, but none of them work.
I know that when proving prime of form $p=4n+1$ aren't Gaussian primes, we need $a$ such that $a^2 \equiv -1\pmod p$. This can be obtained by Wilson's theorem, that $(4n)!\equiv -1\pmod p$, which gives $(4n)!\equiv(1\times 2\times \cdots \times(2n))((-2n)\times\cdots\times(-2)\times(-1))=(1\times 2\times \cdots \times(2n))^2\pmod p$. And this constructs the $a$ we want.
But in this case, things are different. I've tried via mathematica, that the $(3k)!$ or $(6m)!$ cannot be factorized into $a(a-1)$. Such as $720=24\times30$. But when I minus some multiple of the primes, it becomes $24\times 23\equiv-1(\mod 7)$. This special case can't give me more details. So I wonder if there is some method to find $a$ or prove existence without finding it? Only hints are needed! Thanks!