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:
- How does one rule out the primes congruent to $-1$ mod $5$?
- 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.
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).