I encountered a number theory problem when doing my research:
1.I want to know whether or not there are infinitely many primes $p$ satistying $gcd(\frac{p-1}{6},6)=1$, such that $6$ is a cubic residue mod $p$, but $2$ and $3$ are not cubic residues mod $p$? If there are, can we give a expression of $p$?
- I have deduced that if $p=1296k^2+36k+7$ is a prime, then $p$ satisfies the above conditions. Are there infinitely many such primes?
I have done the following:
Since $p\equiv 1\pmod{3}$, we have $p=\pi\bar{π}$, where $\pi$ is a primitive prime in the ring $\mathbb{Z}[\omega]$ and $\omega^2+\omega+1=0$. Assume $\pi=3m−1+3nω$, where $m,n∈\mathbb{Z}$. I have worked out that $p$ satisfies all the requirements (except $gcd(\frac{p-1}{6},6)=1$), if and only if
- $m$ is odd, $n$ is odd, $n\equiv 1\pmod{3} $
or
2)$m$ is even, $n$ is odd, $n\equiv 2\pmod{3} $
I just don't know whether or not there are infinite primes $p$ satisfying all the requirements (including $gcd(\frac{p-1}{6},6)=1$). $p=1296k^2+36k+7$ is a example of 1), and I have found many primes $p=1296k^2+36k+7$ , but I don't know if there are infinite primes like this.
