The fundamental solution to x² − 3y² = 1 is (2, 1). What is the next-smallest positive integer solution?
A(4, 2) — double the fundamental solution
B(5, 3) — adding the fundamental solution to itself
C(7, 4) — computed by squaring ε₁ = 2 + √3 in ℚ(√3)
D(11, 6) — the next convergent of the continued fraction of √3
To generate new solutions from the fundamental solution (x₁, y₁) = (2, 1), form ε₁ = x₁ + y₁√3 = 2 + √3 and compute ε₁² = (2+√3)² = 4 + 4√3 + 3 = 7 + 4√3, giving (x₂, y₂) = (7, 4). Verify: 7² − 3·4² = 49 − 48 = 1 ✓. Simply doubling (option A) fails because solutions don't scale linearly — (4,2) gives 16 − 12 = 4 ≠ 1. The multiplication in ℚ(√D) is what generates solutions, not arithmetic on the integers.
Question 2 Multiple Choice
Why does the continued fraction expansion of √D (for non-square positive integer D) always yield solutions to x² − Dy² = 1?
AEvery sufficiently close rational approximation to √D automatically satisfies the Pell equation exactly
BThe continued fraction is infinite, so by the pigeonhole principle some convergent must work
C√D is irrational, so its continued fraction is eventually periodic, and convergents at the end of each complete period satisfy p² − Dq² = ±1, with full periods giving +1
DThe Pell equation is defined to have solutions wherever the continued fraction algorithm terminates
The periodicity of the continued fraction of √D is the key fact (proved by Lagrange). Because the expansion repeats, the convergents at the end of each period are not just good rational approximations to √D — they satisfy the Pell equation exactly. This is not a coincidence or a pigeonhole argument: the arithmetic of convergents and the structure of the equation are connected through the theory of quadratic irrationalities. Not every convergent works — only those at period-end — which is why knowing the period length is essential.
Question 3 True / False
The fundamental solution to x² − Dy² = 1 always appears as a convergent of the continued fraction expansion of √D.
TTrue
FFalse
Answer: True
This is the central result connecting continued fractions to Pell's equation. The fundamental solution (x₁, y₁) is the smallest positive-integer solution, and it always occurs at the convergent p_k/q_k corresponding to the end of the first complete period of the continued fraction of √D. Every subsequent solution is generated algebraically from this seed.
Question 4 True / False
If (x₁, y₁) is the fundamental solution to x² − Dy² = 1, then (2x₁, 2y₁) is also a solution.
TTrue
FFalse
Answer: False
Scaling a solution by a constant does not produce new solutions. Check: (2x₁)² − D(2y₁)² = 4x₁² − 4Dy₁² = 4(x₁² − Dy₁²) = 4·1 = 4 ≠ 1. Solutions grow through multiplication in ℚ(√D): the n-th solution comes from computing (x₁ + y₁√D)ⁿ = xₙ + yₙ√D, an operation that preserves the norm-1 property but is non-linear in the integers x and y.
Question 5 Short Answer
Explain why knowing the fundamental solution (x₁, y₁) to x² − Dy² = 1 guarantees infinitely many solutions, and describe how they are generated.
Think about your answer, then reveal below.
Model answer: Form ε₁ = x₁ + y₁√D in the ring ℤ[√D] ⊂ ℚ(√D). The equation x² − Dy² = 1 is equivalent to saying ε = x + y√D has norm 1 (i.e., ε·ε̄ = (x + y√D)(x − y√D) = 1). Because norms multiply — N(αβ) = N(α)·N(β) — every power ε₁ⁿ also has norm 1. Since ε₁ > 1, the powers ε₁ⁿ for n = 1, 2, 3, … are all distinct elements xₙ + yₙ√D with xₙ, yₙ positive integers satisfying the Pell equation. The continued fraction finds the seed; the multiplicative structure of ℚ(√D) generates the infinite family.
The algebraic key is that x² − Dy² = 1 defines a group under multiplication in ℚ(√D): the set of norm-1 elements is closed under multiplication and inversion. The fundamental solution is a generator of this group (it has infinite order because ε₁ > 1). So all solutions are exactly the integer powers of ε₁ — positive and negative — making the solution set infinite and completely described by one seed value.