# Primes dividing the values of integer polynomials

Problem: Let $n$ be an integer and $p$ a prime dividing $5(n^2-n+\frac{3}{2})^2-\frac{1}{4}$. Prove that $p \equiv 1 \pmod{10}$.

The polynomial can be re-written as $(\sqrt{5}(n^2-n+\frac{3}{2})-\frac{1}{2})(\sqrt{5}(n^2-n+\frac{3}{2})+\frac{1}{2})$. If this vanishes mod $p$ then $5$ is a quadratic residue mod $p$, which shows that $p \equiv \pm 1 \pmod{5}$ (the primes 2 and 5 are easily ruled out). It feels like the problem should be solvable by understanding the splitting of primes in the splitting field of this polynomial, but I can't find an appropriate "reciprocity law".

The things I'm not sure about are:

1. How does one rule out the primes congruent to $-1$ mod $5$?
2. Under what circumstances is it the case that the set {rational primes that split in the ring of integers of some number field} is the union of arithmetic progressions? This a kind of generalized reciprocity law but I don't know in what generality they are known to hold.
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The only way that this situation could possibly arise is if the splitting field of $f(x)=5(x^2-x+3/2)^2-1/4$ is the cyclotomic field generated by the fifth roots of unity. To show this, it helps to get your hands dirty and actually solve the equation $f(x)=0$.

In general, number fields are determined up to isomorphism by their splitting laws. The splitting law will be determined by a congruence condition iff the extension is an Abelian extension of $\mathbb{Q}$ (by class field theory).

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I actually did solve the equation - it's just a matter of solving the two quadratic I indicated. I'm pretty sure it's indeed Abelian, but why are you saying it has to be the one generated by fifth roots of unity? The radical expressions contain things like sqrt(125-10 sqrt(5)), this sure doesn't look like Z[zeta] but I could be wrong... – Alon Amit Aug 11 '10 at 9:24
If you solve $x^4+x^3+x^2+x+1=0$ you'll find some expressions for $\zeta_5$ which will be reminiscent of the roots of $f$; the task is to express one lot in terms of the other lot :-) – Robin Chapman Aug 11 '10 at 9:31
I think i see what you're saying now - the primes that split in Z[zeta] where zeta is a 5th root of unity are the ones congruent to 1 mod 5, so if this question has the same answer it must be the same field. Neat. I'll see if I can prove my intuition wrong and show that this really is the 5 cyclotomic. – Alon Amit Aug 11 '10 at 9:34
If one root of $f$ is $a\zeta+b\zeta^2+c\zeta^3+d\zeta^4$ where $a,\ldots,d$ are rational, the others should be $a\zeta^2+b\zeta^4+c\zeta+d\zeta^3$ $a\zeta^3+b\zeta+c\zeta^4+d\zeta^2$ $a\zeta^4+b\zeta^3+c\zeta^2+d\zeta$ in some order. If you can compute the zeros of $f$ numerically and put them in the right order, then finding $a,\ldots,d$ is just linear algebra :-) – Robin Chapman Aug 11 '10 at 11:19
"In general, number fields are determined up to isomorphism by their splitting laws." I see that this is true for Galois extensions of $\mathbb{Q}$. For a non-Galois extension of $\mathbb{Q}$, the same primes split completely as split in the Galois closure, so by "splitting laws" you must mean more information than this. Could you say what, exactly? – Pete L. Clark Aug 11 '10 at 14:21

HINT Your polynomial $p(n)$ splits over ${\mathbb Q}(w), w = \zeta_5$, namely

$125 \; p(x) = 125 \; (5 x^4-10 x^3+20 x^2-15 x+11)$

$\quad\quad\quad\quad\quad\; = \;\; (5 x+3 w^3-4 w^2-w-3) (5 x+4 w^3+3 w^2+7 w+1)$

$\quad\quad\quad\quad\quad\quad\; * \; (5 x-3 w^3+4 w^2+w-2) (5 x-4 w^3-3 w^2-7 w-6)$

Regarding the other questions in the query and the comments: there has been much research on various ways of characterizing number fields by splitting behavior, norm sets, etc - going all the way back back to Kronecker. Searching on the terms "Kronecker equivalent" or "arithmetically equivalent" will find pertinent literature. E.g. below is one enlightening review

MR0485790 (58 #5595) 12A65 (12A75)
Gauthier, François
Ensembles de Kronecker et représentation des nombres premiers par une forme quadratique binaire.
Bull. Sci. Math. (2) 102 (1978), no. 2, 129--143.

L. Kronecker [Berlin Monatsber. 1880, 155--162; Jbuch 12, 65] first tried to characterize algebraic number fields by the decomposition behavior of primes. Recently, the Kronecker classes of algebraic number fields have been studied by W. Jehne [J. Number Theory 9 (1977), no. 2, 279--320; MR0447184 (56 #5499)] and others.

(a) When does the set of primes having a given splitting type in an algebraic number field contain (up to a finite set) an arithmetic progression?
(b) When is this set a union of arithmetic progressions?

If $K$ is an algebraic number field, let $\text{spl}(K)$ denote the set of rational primes which split completely in $K$ and let $\text{spl}^1(K)$ denote the set of rational primes which have at least one linear factor in $K$. Moreover, if $K/Q$ is a Galois extension with Galosis group $G$, let ${\text Art}_{K/Q}$ denote the Artin map which assigns a conjugacy class of $G$ to almost all rational primes $p$. If $C$ is a conjugacy class of $G$ then $\text{Art}_{K/Q}^{-1}(C)$ is the set of primes having Artin symbol $C$. Finally a set $S$ of rational primes is said to contain an arithmetic progression or to be the union of arithmetic progressions if the set of primes in the arithmetic progression(s) differs from $S$ by at most a finite set.

Let $G'$ denote the commutator subgroup of the Galois group $G$. Two results proved in the article are:

Theorem A. The following statements are equivalent:
(a) $|C|=|G'|$;
(b) $\text{Art}_{K/Q}^{-1}(C)$ is the union of arithmetic progressions;
(c) $\text{Art}_{K/Q}^{-1}(C)$ contains an arithmetic progression.

Theorem B. The following statements are equivalent:
(a) $K/Q$ is abelian;
(b) $\text{spl}(K)$ contains an arithmetic progression;
(c) $\text{spl}(K)$ is the union of arithmetic progressions;
(d) there exist a modulus $m$ and a subgroup ${r_1,\cdots,r_t}$ of the multiplicative group modulo $m$ such that $\text{spl}(K)$ is the union of the arithmetic progressions $mx+r_i\ (i=1,\cdots,t)$.

When $K/Q$ is a non-Galois extension it is well known that $\text{spl}(K)=\text{spl}(\overline K)$ where $\overline K$ denotes the normal closure of $K$. It follows from Theorem B that $\text{spl}(K)$ cannot contain an arithmetic progression. However, the author gives two conditions, one necessary and the other sufficient, for $\text{spl}^1(K)$ to be the union of arithmetic progressions when $K/Q$ is non-Galois. As a final application of his result the author gives a necessary and sufficient condition for the set of primes represented by a quadratic form to be the union of arithmetic progressions.

The proofs use class field theory, properties of the Artin map and the Čebotarev density theorem.

Reviewed by Charles J. Parry

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Thanks so much - this is just what I was looking for in my second question! – Alon Amit Aug 11 '10 at 18:16