I wonder if we can assume the following statement to be true in general:
Let $p$ be a prime of the form $6k+1$ and $n<p$ a natural number less than $p$. If $3$ does not divide $n$ and none of the prime factors of $n$ is of the form $6k+1$, then $p$ is a cube modulo $n$.
Some ideas / observations:
If $p$ is a prime and $g$ a primitive root of $p$, then the least residues modulo $p$ of $g^1,g^2,g^3,\ldots,g^{p−1}$ is a permutation of $1,2,3,...,p−1$ (this is clear so far). Let $T_i$ ($i=1,2,\ldots,n$) denote the least residues modulo $p$ of $g^i,g^{i+n},g^{i+2n},\ldots,g^{i+(r−1)n}$ where $r=\frac{p-1}{n}$. This way we obtain a partitioning of the natural numbers $1,2,3,\ldots,p−1$ into $n$ sets. Finally, let $s_i$ denote the number of consecutive integers in $T_i$ (after sorting). If, for example, $x$, $x+1$ and $x+2$ are elements of a set, then the number of consecutive integers in this sequence is considered to be $2$.
For $p=157$ interestingly $s_2$, $s_3$ and $s_4$ are in arithmetic progression. In this case $r$ is even and $2^{2r}\equiv1\pmod{p}$. Similar $p$ values (among many others) are $2017$, $2281$, $2341$ and $3889$. For $p=2017$, $s_1=66$, $s_2=58$, $s_3=54$, $s_4=50$, $s_5=42$ and $s_6=65$. Reciprocity does not occur for $13$ (a prime of the form $6k+1$) or any product containing $13$.
There seems to be three possibilities for $n=6$: Either $r$ is even and $2^{2r}\equiv1\pmod{p}$, $r$ is even and $2^{2r}\not\equiv1\pmod{p}$ or $r$ is odd.
My question:
Is there a way to (maybe straight-forwardly) prove the above stated behavior? Or does there possibly already exist a theorem for it?
Comment: some experimental facts; $P=6k+1$ is prime if k is odd with some exceptions:
$(k, p)=(1, 7), (3, 19),(5, 31), (7, 43), (9, 55), (11, 67), (13, 79, (15, 91)\cdot\cdot\cdot$
$(19=2^3+11$; $11\neq3t$), ($31=2^3+23; 23\neq 3t$)$\cdot\cdot\cdot$
But:
$43=2^3+35; 35=5\times 7$ and $7=6\times 1+1$
However as can be seen $3\nmid(p-a^3)$ .