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Generalized Pell orbits for $x^{2} - 6 y^{2} = 3$

Example

For $x^{2} - 6 y^{2} = 3,$ the bounded representatives are exactly $\pm (3 + \sqrt{6}),$ and every integral solution is $\pm (3 + \sqrt{6}) (5 + 2 \sqrt{6})^{k}, k \in Z .$

Facts & Assumptions

Given: The equation $x^{2} - 6 y^{2} = 3$ .

[F1]

Every positive Pell solution is a power of the fundamental unit (All positive Pell solutions are powers of the fundamental solution).

[F2]

Every generalized Pell solution is Pell-equivalent to one in the explicit bounded rectangle given by the orbit theorem (Generalized Pell solutions fall into finitely many Pell orbits).

Verification

technique · direct

1.1

The element $5 + 2 \sqrt{6}$ has norm $1$ , and there is no positive norm-one solution with $y = 1$ because $x^{2} = 7$ is impossible; hence $5 + 2 \sqrt{6}$ is the fundamental Pell unit, so [F1] identifies the Pell powers with $(5 + 2 \sqrt{6})^{k}$ . The element $3 + \sqrt{6}$ has norm $3$ . For $N = 3$ and $ε_{6} = 5 + 2 \sqrt{6}$ , the bound of [F2] is $∣ x_{0} ∣ \leq \frac{\sqrt{3}}{2} (\sqrt{5 + 2 \sqrt{6}} + (5 + 2 \sqrt{6})^{- 1 / 2}) = 3,$ $∣ y_{0} ∣ \leq \frac{\sqrt{3}}{2 \sqrt{6}} (\sqrt{5 + 2 \sqrt{6}} + (5 + 2 \sqrt{6})^{- 1 / 2}) < 2.$

F1F2givenalgebra

2.1

The only integer solutions in that box are $(x, y) = (\pm 3, \pm 1)$ . Moreover, $(3 - \sqrt{6}) = (3 + \sqrt{6}) (5 - 2 \sqrt{6}) = (3 + \sqrt{6}) (5 + 2 \sqrt{6})^{- 1},$ so $(3, 1)$ and $(3, - 1)$ lie in the same Pell orbit, and multiplying by $- 1$ gives the negative orbit. Therefore every solution of $x^{2} - 6 y^{2} = 3$ is $\pm (3 + \sqrt{6}) (5 + 2 \sqrt{6})^{k} (k \in Z) .$

F2step 1.1algebra∎

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Sources

Keith Conrad, Pell's Equation, II (standard reference, not scraped)
Peter Hackman, Elementary Number Theory (standard reference, not scraped)

Generalized Pell orbits for x2−6y2=3x2−6y2=3