Research proposal

Beyond quadratic residues

Negative Pell equations and finer arithmetic information

xyx² − 2y² = −1

Question & scope

In the prime family d=5q below, what do the two quartic residue signs explain—and fail to explain—about negative Pell solubility?

The domain is prime q≡5 (mod 8), q≠5, with Legendre symbol (5/q)=1. The two rational quartic signs mean 5^((q−1)/4) mod q and q^((5−1)/4) mod 5, each interpreted as ±1. Continued-fraction parity already decides the integer equation; the goal is structural explanation in this family.

First calculation: a signature is a hypothesis to test

q≡5(mod8),(5q)=1,d=5q,x2−dy2=−1.q\equiv5\pmod8,\quad\left(\frac5q\right)=1,\quad d=5q,\quad x^2-dy^2=-1.

For prime q≤10000 in this family, record the two quartic signs, the exact period and an integer norm −1 witness when one exists. Separate examples with the same signs and opposite outcomes. This finite computation checks a proposed signature, not a density or an infinite-family theorem.

The next comparison is between the ordinary and narrow ideal class groups of Q(√d), and the class of the principal ideal (√d) in the narrow group. Write down its relation to the existence of a negative-norm unit before choosing a finer 2-primary invariant. Known quartic reciprocity and class-group criteria must be read first; no sufficiency assertion is made here.

Known boundaries

The 145/305 pair disproves only the displayed quadratic signature. Exact pilot calculations also give q=101 and 461 with the same quartic signs (+1,+1), but period lengths 4 and 1 for √505 and √2305. These are assistant-prepared test cases, not new theorems. Stevenhagen’s density problem has a resolution; current novelty of a restricted criterion must still be checked.

Stevenhagen’s density conjecture has already been resolved. This proposal concerns restricted decision problems and a path into the methods behind arithmetic statistics.

Matching coarse data, different answers

x² − dy² = −1   (x, y ∈ ℤ)
d = 145 = 5 · 29   :   12² − 145 = −1
d = 305 = 5 · 61   :   √305 = [17; 2, 6, 2, 34, …]

The period of √305 has even length, giving non-solubility by the classical period criterion. This is an exact seed calculation; a finite search alone would not establish non-solubility.

Supporting study notes

Classical background and worked examples for this proposal.

Work to undertake

  1. Compare the two named quartic signs in the fixed prime family, with exact integer certificates.
  2. Use 505/2305 as a same-quartic-signature sanity check, keeping finite and general conclusions separate.
  3. Derive the negative-norm-unit and narrow-class-group correspondence, then choose and literature-check a finer invariant.
Number theoryQuadratic fieldsArithmetic statistics
  • Continued fractions
  • Norms and units
  • Local methods
  • Narrow class groups

References

Books behind the question