Reciprocity laws

I've been reading about a new type of reciprocity than the usual quadratic reciprocity theorem. Instead of being about square roots mod p it is about divisors of the cyclotomic polynomial.

• A PRELUDE TO THE STUDY OF RECIPROCITY LAWS - JONAH SINICK
• WHAT IS A RECIPROCITY LAW? - B. F. WYMAN

An easy case of the theorem is that the following are equivalent, for all primes $p,q$:

The theorem in an easy case

• (I) exists $x$, $q | x^{p-1} + \cdots + x + 1$
• (II) $q = p$ or $q = 1 \,(\text{mod}\,p)$

In the I=>II direction we have $x^p \equiv 1 \,(\text{mod}\,q)$ so $x$ is either the identity or an order $p$ element of the group of size $q-1$. That implies that $x = 1 \,(\text{mod}\,q)$ (so $q | 1^{p-1} + \cdots + 1 + 1 = p$ hence $q = p$) or $p | q - 1$ (hence $q = 1 \,(\text{mod}\,p)$).

In the other direction II=>I the $q = p$ we have $x = 1$ and otherwise we get an element $x$ of order $p$ by Cauchy's theorem in group theory and it satisfies the equation.

Example

A nice example of this (taken from the first PDF) is that all prime factors other than 5 of $x^4 + x^3 + x^2 + x + 1$ end in 1:

? f(x) = x^4 + x^3 + x^2 + x + 1
%1 = (x)->x^4+x^3+x^2+x+1
? factor(f(51))
%2 = 
[  5 1]
[ 41 2]
[821 1]

? factor(f(781))
%3 = 
[       5 1]
[    5641 1]
[13207921 1]

? factor(f(7831))
%4 = 
[          5 1]
[      15551 1]
[48372221411 1]

? factor(f(7231))
%5 = 
[         5 1]
[        11 1]
[        31 1]
[      1061 1]
[1511519461 1]

Connection to quadratic reciprocity

First of all this holds much more generally than just for primes $p$, we can use any $k$ and it's cyclotomic polynomial. The proof for that is in the second PDF.

Second while this is a very elegant theorem, I'm interested in that the cyclotomic reciprocity law can be used to prove quadratic reciprocity. I don't have the proof yet though.