# Two negative Pell equations that coarse data cannot separate

145 and 305: matching congruence signatures, opposite integral answers.

2026-10-17 · Study note · classical criterion

> Assistant-prepared seed examples for the arithmetic proposal. Continued-fraction periods and integer identities are checked exactly; no general density or new solvability criterion is claimed.

## How this example entered the record

The saved number-theory reading includes Pell equations, quadratic rings and p-adic viewpoints. The 145/305 contrast was selected with an assistant as a classical test case; there is no saved independent discovery narrative to claim.

## The next question I can actually test

The period criterion already answers solubility. The proposed next step fixes d=5q and asks what the two specified quartic signs explain, using exact counterexamples rather than calling another finite search a proof of non-solubility.

## Specify integral solubility

$$
x^2-dy^2=-1,\qquad x,y\in\mathbb Z,\quad d>1\text{ nonsquare}.
$$

Integers are essential here. A rational point on a conic and an integral unit of norm −1 are different requirements. A finite search that fails to find an integer pair does not decide the equation.

The two squarefree radicands below have two prime factors, each congruent to 5 modulo 8. The indicated Legendre symbols are both +1. Indeed 11² ≡ 5 (mod 29), and 26² ≡ 5 (mod 61).

$$
145=5\cdot29,\qquad305=5\cdot61,\qquad\left(\frac5{29}\right)=\left(\frac5{61}\right)=1.
$$

## The period is a complete test

For a positive nonsquare integer d, write the simple continued fraction of √d as [a₀; overline(a₁,…,aL)]. The classical negative Pell criterion says that an integer solution exists if and only if L is odd. This is a theorem about the period, not a bounded search heuristic.

$$
\sqrt{145}=[12;\overline{24}],\qquad\sqrt{305}=[17;\overline{2,6,2,34}].
$$

The first period has length one and gives (x,y) = (12,1). The second has length four and rules out every integer solution, even beyond any feasible search range.

$$
12^2-145\cdot1^2=-1.
$$

## Calculate with integers throughout

Start with m₀ = 0, q₀ = 1, a₀ = floor(√d). The complete-quotient recurrence gives the period without floating-point subtraction. Stop when the next coefficient equals 2a₀.

$$
\begin{aligned}m_{n+1}&=q_na_n-m_n,\\q_{n+1}&=\frac{d-m_{n+1}^2}{q_n},\\a_{n+1}&=\left\lfloor\frac{a_0+m_{n+1}}{q_{n+1}}\right\rfloor.\end{aligned}
$$

For an odd period, the convergent with coefficients a₀,…,aL₋₁ supplies a solution. The interactive calculation uses arbitrary-precision integers for the numerator and denominator, then verifies x² − dy² = −1 exactly. The input limit controls browser work; it is not a restriction of the theorem.

See the website for the interactive calculation; exact seed scripts and results are provided alongside it.

## What a stronger invariant must explain

These examples disprove only the sufficiency of the displayed coarse signature. They do not say that all local information is useless, nor that class groups automatically decide every chosen formulation. A next project must fix a family and a precise proposed invariant.

There is a unit-theoretic reformulation in Z[√d]: a solution is a unit x+y√d of norm −1. When d ≡ 1 modulo 4, the full ring of integers also contains half-integral elements; one must not silently replace the order Z[√d] by that larger ring without checking the relation between their units.

Stevenhagen’s density conjecture is already the subject of a resolution by Koymans–Pagano. The proposed work is a route into restricted decision questions and the methods behind them; it does not reopen a solved conjecture as an unknown problem.

## Sources

- [Koymans & Pagano — On Stevenhagen’s conjecture](https://arxiv.org/abs/2201.13424)