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Generalized Pell solutions fall into finitely many Pell orbits

Statement

Let D be a positive nonsquare integer, let NZ{0}, and let εD>1 be the fundamental Pell solution. Every integral solution x2Dy2=N is Pell-equivalent to a solution x0+y0D satisfying x0N2(εD+εD1/2),y0N2D(εD+εD1/2). Consequently the generalized Pell equation has only finitely many Pell-equivalence classes.

Facts & Assumptions

Given: A nonzero integer N, the fundamental Pell solution εD>1, and a solution α=x+yD of ND(α)=N.

[F1]

Two norm-N solutions are Pell-equivalent exactly when one is obtained from the other by multiplication by a power of εD (Pell-equivalence for generalized solutions).

[F2]

The Pell norm is multiplicative (The Pell norm is multiplicative).

[F3]

The fundamental solution satisfies εD>1 (The fundamental Pell solution).

Proof

technique · direct
1.1

Put t:=αN>0. Because εD>1, the half-open intervals (εDm1/2,εDm+1/2](mZ) partition (0,): their endpoints are strictly ordered, they tend to 0 as m, and they tend to + as m+. Hence there is a unique integer k with εD1/2<tεDkεD1/2. Put α0:=αεDk=x0+y0D. By [F1] and [F2], α0 is Pell-equivalent to α and still satisfies ND(α0)=N.

F1F2F3givenalgebra
2.1

Let β:=α0. Step 1.1 gives NεD1/2<βNεD1/2. Since α0=x0y0D=Nα0, one has α0=NβNεD1/2 and also α0>NεD1/2. Therefore 2x0=α0+α0β+α0N(εD+εD1/2), and similarly 2Dy0=α0α0β+α0N(εD+εD1/2). This is exactly the stated bound.

F2step 1.1algebra
3.1

The bounds of step 2.1 leave only finitely many integer pairs (x0,y0). Every solution is Pell-equivalent to one of them by step 1.1, so only finitely many Pell-equivalence classes occur.

step 1.1step 2.1

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