Questions: Pell's Equation

Question 1

The fundamental solution to x² − 3y² = 1 is (x₁, y₁) = (2, 1). What is the next positive integer solution?

• A. There are no further solutions — Pell's equation has only finitely many
• B. (4, 2) — double each component of the fundamental solution
• C. (7, 4) — computed by expanding (2 + √3)² = 7 + 4√3
• D. (5, 3) — the next convergent of the continued fraction of √3

Answer: (C)

All solutions are generated by xₙ + yₙ√D = (x₁ + y₁√D)ⁿ. Here (2 + √3)² = 4 + 4√3 + 3 = 7 + 4√3, giving (7, 4). Check: 49 − 3·16 = 1. ✓ Option A is wrong — there are infinitely many solutions. Option B is wrong — solutions scale multiplicatively via the ring structure, not additively by doubling components. The repeated-multiplication structure is the key insight.

Question 2

Why do the convergents of the continued fraction expansion of √D yield solutions to x² − Dy² = 1?

• A. Because all convergents are integers, and integer pairs always satisfy some Diophantine equation
• B. Because convergents pₖ/qₖ are the best rational approximations to √D, making pₖ² − Dqₖ² land exactly on 1 at the right index
• C. Because the continued fraction of √D terminates, producing exact values
• D. Because Pell's equation was designed specifically to be solved by continued fractions

Answer: (B)

The connection is deep: p/q ≈ √D means p ≈ q√D, so p² ≈ Dq². The error in best rational approximations is so controlled by continued fraction theory that pₖ² − Dqₖ² lands on exactly ±1 (and eventually +1) at the period index. The continued fraction of √D does not terminate — it is periodic — so option C is wrong. Option D inverts the history.

Question 3

Pell's equation x² − Dy² = 1 has infinitely many positive integer solutions for every positive non-square integer D.

• T. True
• F. False

Answer: True

True. Once the fundamental solution (x₁, y₁) is found (which exists for all non-square D by the theory of continued fractions), the formula xₙ + yₙ√D = (x₁ + y₁√D)ⁿ generates a distinct positive integer solution for every positive integer n. Solutions grow exponentially, so the sequence is infinite.

Question 4

If (x₁, y₁) is the fundamental solution to x² − Dy² = 1, then subsequent solutions are obtained by the recurrence xₙ = n·x₁ and yₙ = n·y₁.

• T. True
• F. False

Answer: False

False. Solutions are generated multiplicatively, not additively. The rule is xₙ + yₙ√D = (x₁ + y₁√D)ⁿ, which corresponds to a matrix recurrence — solutions grow exponentially, not linearly. For D = 2, the fundamental solution is (3, 2); the next is (17, 12), not (6, 4). Multiplying components by n gives a different (and usually non-solution) pair.

Question 5

Explain the connection between the continued fraction expansion of √D and the fundamental solution to Pell's equation x² − Dy² = 1. Why does multiplication in the ring ℤ[√D] generate all solutions from just one?

Model answer: The continued fraction of √D is periodic; the convergents pₖ/qₖ at the end of each period satisfy pₖ² − Dqₖ² = 1, because convergents are best rational approximations and the approximation error is small enough to land exactly on 1. This gives the fundamental solution (x₁, y₁). All solutions come from powers because the norm form x² − Dy² = N(x + y√D) is multiplicative in ℤ[√D]: N(α)·N(β) = N(αβ). Since N(x₁ + y₁√D) = 1, every power also has norm 1, generating a new solution. Solutions are precisely the units (norm-1 elements) of the ring ℤ[√D].

The algebraic insight is that Pell's equation is asking for units of ℤ[√D]. Units form a group under multiplication, so the product of two solutions is again a solution. The group is generated by the fundamental solution — analogous to how ℤ* is generated by 1. The continued fraction connection comes from the theory of best approximations: the only integers p, q with |p² − Dq²| small arise as convergents, and the period of the continued fraction determines exactly when the first hit on +1 occurs.