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Generalized Pell orbits for x2−6y2=3

Example

For x2−6y2=3, the bounded representatives are exactly ±(3+6), and every integral solution is ±(3+6)(5+26)k,k∈Z.

Facts & Assumptions

Given: The equation x2−6y2=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
1.1F1F2givenalgebra

The element 5+26 has norm 1, and there is no positive norm-one solution with y=1 because x2=7 is impossible; hence 5+26 is the fundamental Pell unit, so [F1] identifies the Pell powers with (5+26)k. The element 3+6 has norm 3. For N=3 and ε6=5+26, the bound of [F2] is ∣x0∣≤32(5+26+(5+26)−1/2)=3, ∣y0∣≤326(5+26+(5+26)−1/2)<2.

2.1F2step 1.1algebra∎

The only integer solutions in that box are (x,y)=(±3,±1). Moreover, (3−6)=(3+6)(5−26)=(3+6)(5+26)−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 x2−6y2=3 is ±(3+6)(5+26)k(k∈Z).

Depends on

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Sources