Question

For the solution of congruences $x^2\equiv p \pmod q$ and $x^2\equiv q \pmod p$, we have the law of quadratic reciprocity. Is there any solution for the above congruences?

Answer 1

Let me interpret the question as follows: is there a simple relationship between the solvability of $x^n \equiv p \bmod q$ and that of $x^n \equiv q \bmod p$ for primes $p \equiv q \equiv 1 \bmod n$?

Here the answer is yes if $n = 2$, and no otherwise. The reason is the following: the congruence $x^n \equiv p \bmod q$ for primes $q \equiv 1 \bmod n$ is solvable if and only if the primes above $q$ in the field $K$ of n-th roots of unity split completely in the Kummer extension $L = K(\sqrt[n]{p})$. This splitting behaviour depends only on the residue classes modulo primes above $p$ by class field theory, the reason being that $L/K$ is abelian. This gives rise to a reciprocity law, which basically states a very close connection between $(p/{\mathfrak q})_n$ and $({\mathfrak q}/p)_n$ (the second symbol has to be suitably interpreted, which is possible if $K$ has class number $h$ coprime to $n$).

What this means is that any reasonable connection between $(p/q)_n$ and $(q/p)_n$ will have to use information on the principal ideals ${\mathfrak p}^h$ and ${\mathfrak q}^h$ . In the simplest nontrivial case $n = 4$, the connection is provided by "Burde's" reciprocity law $(p/q)_4 (q/p)_4 = (ac-bd/p)$, where $p = a^2 + b^2$ and $q = c^2 + d^2$, with everything suitably normalized.

Moreover, you can't expect something similar to hold over the rationals, because the extension ${\mathbb Q}(\sqrt[n]{p})$ is not abelian over ${\mathbb Q}$ except when $n = 2$.

There is a similar (but already more complicated) law for $n = 3$; the case $n = 4$ is special because $i^2 = -1$ is rational.