Problem: Find all integer solutions of ax^{2} + bxy + cy^{2} + dx + ey + f = 0. (1)

ν = (φ_{1} + ψ_{1}√Δ_{1})/2 gives the least positive solution of u^{2} – Δ_{1}v^{2} = 4.

κ = εν^{n} = (φ + ψ√Δ_{1})/2, where ε = ±1, n ∈ ℤ.

v_{11} = (φ – bψ)/2, v_{12} = -cψ, v_{21} = aψ, v_{22} = (φ + bψ)/2.

α = 2cd – be, β = 2ae – bd.

K_{1}(κ) = (α – (αv_{11} + βv_{12}))/Δ_{1} ,
K_{2}(κ) = (β – (αv_{21} + βv_{22}))/Δ_{1}

Case 1. κ = ν, K_{1}(ν) and K_{2}(ν) are both integers.

Case 2. κ = -ν, Case 1 does not hold, but K_{1}(-ν) and K_{2}(-ν) are both integers.

Case 3. κ = ν^{2}. Neither Case 1 nor Case 2 occurs. (We know K_{1}(ν^{2}) and K_{2}(ν^{2}) are both integers.)

Make a transformation Δ_{1}x = X + α, Δ_{1}y = Y + β.

equation (1) transforms to AX^{2} + BXY + CY^{2} = N, (2)

where A = a/gg, B = b/gg, C = c/gg, gg = gcd(a,b,c), and
N = -Δ_{1}(ae^{2} - bde + cd^{2} + fΔ_{1})/gg.

If equation (2) has no integer solutions, then neither does equation (1).

Suppose (2) has g fundamental solutions (X_{i}, Y_{i}), 1 ≤ i ≤ g.

Δ_{2} = B^{2} – 4AC.

μ = (u_{1} + v_{1}√Δ_{2})/2 gives the least positive solution of u^{2} – Δ_{2}v^{2} = 4.

U = [(u_{1} – Bv_{1})/2, -Cv_{1}]

[Av_{1}, (u_{1} + Bv_{1})/2]

Determine k ≥ 1 such that ν = μ^{k}.

In Case 3, test (X,Y)^{t} = εU^{j}(X_{i}, Y_{i})^{t}, 1 ≤ i ≤ g, 0 ≤ j ≤ 2k-1, ε = ±1

to see if (x_{0}, y_{0}) = (X + α)/Δ_{1}, (Y + β)/Δ_{1}) is an integer solution of (1).

This gives a corresponding family

(Δ_{1}x_{n}, Δ_{1}y_{n}) = εU^{kn + j}(X_{i}, Y_{i})^{t} + (α, β)^{t}, n ∈ ℤ

In Case 1, test (X,Y)^{t} = εU^{j}(X_{i}, Y_{i})^{t}, 1 ≤ i ≤ g, 0 ≤ j ≤ k-1, ε = ±1

In Case 2, test (X,Y)^{t} = ε(-U)^{j}(X_{i}, Y_{i})^{t}, 1 ≤ i ≤ g, 0 ≤ j ≤ k-1, ε = ±1

The families satisfy the recurrence relations

x_{n+1} = v_{11}x_{n} + v_{12}y_{n} + K_{1}(κ),

y_{n+1} = v_{21}x_{n} + v_{22}y_{n} + K_{2}(κ).

x_{n} = v_{22}x_{n+1} – v_{12}y_{n+1} + K_{1}(κ^{-1}),

y_{n} = -v_{21}x_{n+1} + v_{11}y_{n+1} + K_{2}(κ^{-1}).

*Last modified 18th September 2022*

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