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The large fundamental solution for $D = 61$

Example

A direct continued-fraction calculation gives $\sqrt{61} = [7; \overline{1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14}],$ so the period length is $11$ . The least positive solution of $x^{2} - 61 y^{2} = - 1$ is $29718 + 3805 \sqrt{61},$ and the least positive solution of $x^{2} - 61 y^{2} = 1$ is its square $1766319049 + 226153980 \sqrt{61} .$

Facts & Assumptions

Given: The positive nonsquare integer $D = 61$ .

[F1]

Negative Pell is soluble exactly for odd period length, and when the period length is odd the least positive negative-Pell solution is $p_{ℓ - 1} / q_{ℓ - 1}$ while the least positive norm-one solution is $p_{2 ℓ - 1} / q_{2 ℓ - 1}$ (Negative Pell is soluble exactly for odd period length).

Verification

technique · direct

1.1F1givenalgebra

The displayed continued fraction has odd period length $ℓ = 11$ . Running the convergent recurrence through one full period gives $p_{10} / q_{10} = 29718 / 3805.$ Fact [F1] therefore gives $29718^{2} - 61 \cdot 3805^{2} = - 1,$ so $29718 + 3805 \sqrt{61}$ is the least positive negative-Pell solution.

2.1F1step 1.1algebra∎

Squaring that element gives $(29718 + 3805 \sqrt{61})^{2} = 1766319049 + 226153980 \sqrt{61},$ and therefore $1766319049^{2} - 61 \cdot 226153980^{2} = 1.$ By [F1], this is the least positive solution of Pell's equation for $D = 61$ . The example shows that a finite continued-fraction algorithm can still produce a very large fundamental solution.

Depends on

Negative Pell is soluble exactly for odd period length

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

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

The large fundamental solution for D=61D=61