MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
I'm guessing you are not really demanding the same $x$ work in both equations... – Will Jagy Jul 22 '10 at 19:29
I think it would be reasonable to interpret this question as asking for generalizations of quadratic reciprocity. The most down-to-earth reference I know for this is Ireland and Rosen's Classical Introduction to Modern Number Theory. – Qiaochu Yuan Jul 22 '10 at 19:39
@Will By the Chinese remainder theorem, it doesn't matter whether we take x to be the same in both cases or not. – David Speyer Jul 22 '10 at 20:08
There is no article on the higher reciprocity law on wikipedia, but at least the cubic case,, is elaborated in quite details. Then you can guess that in general one has to go to cyclotomic extensions and generalised Gauss sums. – Wadim Zudilin Jul 22 '10 at 23:45
I think what you're looking for is known as the Eisenstein Reciprocity Law. A good source for this and related material is Reciprocity Laws: From Euler to Eisenstein by Franz Lemmermeyer. – Kevin Ventullo Jul 23 '10 at 3:15

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.

share|cite|improve this answer
I oppose the word "simple" in the interpretation. – Dror Speiser Jul 24 '10 at 9:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.