Let $\left(\dfrac{a}{p}\right)_n$ be the $n$th power residue symbol such that when $a$ is an $n$th power residue $\pmod p$, $\left(\dfrac{a}{p}\right)_n=1$ otherwise $\left(\dfrac{a}{p}\right)_n=\zeta$. Let $f_n(x)$ be a defining polynomial for the $n$th cyclotomic field $K=\mathbb{Q}(e^{2\pi i/n})$. Suppose that there exists some integer $m$ such that $f_n(m)=0 \pmod {n^2}$, and there are primes $p$ and $q$ such that $p=f_n(z)$ and $q=f_n(z+a)$. Then $\left(\dfrac{a}{p}\right)_n = \left(\dfrac{a}{q}\right)_n$. When $n=3$, this conjecture agrees with cubic reciprocity theorems. The same is true for $n=4$.
Here is an example of the conjecture for $n=5$:
$f_5(x)=x^4+5x^3+10x^2+25$ and $f_5(0)=0\pmod {5^2}$
$a=3$, $p=f_5(1)=41$ and $q=f_5(1+a)=f_5(4)=761$ are both prime, thus
$\left(\dfrac{3}{41}\right)_5 = \left(\dfrac{3}{761}\right)_5$.
Since $3$ is a quintic residue $\pmod {41}$, then it is also a quintic residue of $761$. In general, if $p=f_5(3a+1)$ is prime, then $3$ is a quintic residue $\pmod p$.