LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Convergents to $\sqrt{D}$ satisfy the norm identity

Statement

Let $D$ be a positive nonsquare integer, let $p_{n} / q_{n}$ be the convergents of $\sqrt{D}$ , and let $P_{n}, Q_{n}$ be the complete-quotient state variables from The complete quotients of $\sqrt{D}$ satisfy the $P_{n}, Q_{n}$ recurrence. Then for every $n \geq 0$ , $p_{n}^{2} - D q_{n}^{2} = (- 1)^{n + 1} Q_{n + 1} .$

Facts & Assumptions

Given: A positive nonsquare integer $D$ , the convergents $p_{n} / q_{n}$ of $\sqrt{D}$ , and the state variables $P_{n}, Q_{n}$ .

[F1]

The complete-quotient tail formula gives $\sqrt{D} = \frac{α_{n + 1} p_{n} + p_{n - 1}}{α_{n + 1} q_{n} + q_{n - 1}}$ for every $n \geq 0$ (Complete-quotient tail formula).

[F2]

Consecutive convergents satisfy $p_{n} q_{n - 1} - p_{n - 1} q_{n} = (- 1)^{n - 1} .$ (Determinant identity for consecutive convergents).

[F3]

The next complete quotient has the form $α_{n + 1} = \frac{\sqrt{D} + P_{n + 1}}{Q_{n + 1}}$ with $Q_{n + 1} > 0$ (The complete quotients of $\sqrt{D}$ satisfy the $P_{n}, Q_{n}$ recurrence).

Proof

technique · direct

1.1

Substitute the expression from [F3] into [F1] and clear denominators. One gets $\sqrt{D} ((\sqrt{D} + P_{n + 1}) q_{n} + Q_{n + 1} q_{n - 1}) = (\sqrt{D} + P_{n + 1}) p_{n} + Q_{n + 1} p_{n - 1} .$ Comparing the rational and irrational coefficients of $1$ and $\sqrt{D}$ yields $D q_{n} = P_{n + 1} p_{n} + Q_{n + 1} p_{n - 1}, p_{n} = P_{n + 1} q_{n} + Q_{n + 1} q_{n - 1} .$

F1F3algebra

2.1

Multiply the second identity of step 1.1 by $p_{n}$ , the first by $q_{n}$ , and subtract. Then $p_{n}^{2} - D q_{n}^{2} = Q_{n + 1} (p_{n} q_{n - 1} - p_{n - 1} q_{n}) = (- 1)^{n + 1} Q_{n + 1}$ by [F2].

F2step 1.1algebra∎

Depends on

Used by

The Pell equation for D=14 Example
Every Pell equation has a positive nontrivial integral solution Theorem
Negative Pell is soluble exactly for odd period length Theorem

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Peter Hackman, Elementary Number Theory (standard reference, not scraped)
MIT 18.781, Lecture 21: Brahmagupta-Pell Equation (standard reference, not scraped)

Convergents to DD​ satisfy the norm identity