The equation x²−2y²=1 asks for an integer point on a hyperbola. The pair (3, 2) works because nine minus eight is one. Multiplying 3+2√2 by itself gives 17+12√2, and 289−288 is again one. The fixed worked example in this page uses D=2. The live exploration has its own selectable values and should not be read as the source of those fixed numbers.
Predict
Predict whether repeated multiplication always preserves the equation. Define the conjugate of x+y√D as x−y√D and its norm as their product x²−Dy². Conjugation respects multiplication, so norms multiply. Products of norm-one expressions remain norm one with integer coefficients. If D is a positive square, the equation factors into ordinary integer factors and this theory changes; throughout this lesson D is a positive nonsquare integer.
Experiment
Live experiment: a good approximation can solve an exact equation
For a nonsquare positive integer D, build convergents p/q to √D. Predict when p²−Dq² will equal 1 or −1. Approximate closeness is not the certificate: the displayed integer residual checks the Diophantine equation exactly.
The continued fraction begins [3; 1, 1, 1, 1, 6]. This is a finite prefix, not a statement that the expansion terminates. We start p₋₂=0, p₋₁=1, q₋₂=1, q₋₁=0; each new digit aₖ gives pₖ=aₖpₖ₋₁+pₖ₋₂ and qₖ=aₖqₖ₋₁+qₖ₋₂.
The surd recurrence is mₖ₊₁=dₖaₖ−mₖ, dₖ₊₁=(D−mₖ₊₁²)/dₖ, aₖ₊₁=⌊(⌊√D⌋+mₖ₊₁)/dₖ₊₁⌋, starting m₀=0,d₀=1. The adjacent determinant pₖqₖ₋₁−pₖ₋₁qₖ alternates −1,+1, proving each convergent is reduced.
No +1 solution in the displayed prefix. The term budget has ended; this is not evidence that the equation has no solution.
The prefix also contains a −1 solution: 18²−13×5²=−1.
Try D=13 with six terms, then ten. Explain why the exact equation can change while every fraction remains a good approximation. Compare D=3 and D=2, and prove a modular obstruction to the negative equation when available. Arithmetic for convergents and certificates uses exact integers. The supported small surd states use exact safe integer operations, with a bounded square-root calculation only to find ⌊√D⌋. Settings are temporary.
Inspect exact convergents, their recurrence and their residuals in the exploration below. Change D and the number of terms. Look for +1 and −1 separately. A reported first occurrence means first within the displayed prefix; a prefix with no +1 is not an impossibility proof. The controls are temporary exploration, not saved mastery evidence.
Fixed D=2 convergent
Residual p²−2q²
Interpretation
1/1
−1
Negative Pell
3/2
1
Positive Pell
7/5
−1
Negative Pell
17/12
1
Positive Pell
Check the last row directly using integer squares.
Understand
Existence can be proved without assuming periodicity. Infinitely many convergents p/q to √D satisfy absolute error below 1/q². Their nonzero integer norms k=p²−Dq² are bounded in absolute value by a constant, since the norm equals (p/q−√D)q²(p/q+√D). Thus infinitely many share one nonzero k. Among these choose an infinite subsequence with the same p and q residues modulo the positive integer ∣k∣. For two distinct positive expressions α_i=p_i+q_i√D and α_j=p_j+q_j√D in this subsequence, their ratio is (p_i p_j−Dq_i q_j+(q_i p_j−p_i q_j)√D)/k. Both numerator coefficients are divisible by k by the residue agreement. The ratio therefore has integer coefficients and norm one. Choose α_i>α_j, giving a ratio greater than one and consequently positive coefficients x, y. This proves a nontrivial solution exists.
Apply
Choose the least positive x among solutions with y>0 and call its expression ε=x+y√D. This also minimizes the expression because y=√((x²−1)/D) increases with x. For any positive norm-one expression η>1, choose an integer n≥0 with 1≤η/ε^n<ε. The inverse of ε is its integer conjugate, so the quotient still has integer coefficients and norm one. If it exceeded one, writing it as z+w√D and using its reciprocal conjugate would give positive z, w and a smaller solution. Therefore it is one and η=ε^n. This classifies positive solutions once a fundamental unit is established.
Continue
Negative Pell behaves differently: x²−3y²=−1 is impossible modulo three, although positive Pell has solution (2, 1). Squaring a negative-norm solution gives positive norm, but the reverse implication fails. Continued-fraction periodicity gives powerful computational criteria; its general theorem is beyond the proof supplied here. Our existence and power classification are proved independently, while a bounded widget search remains a finite calculation. For implementation, retain exact integers because small residuals arise from subtracting large squares.
Drafts save locally as you type. Reveal worked solutions when useful; the site remembers that exposure. Automatic matches check the requested answer, while written arguments require review against the rubric.
Square 2+√3 and report the new rational coefficient.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Report the √3 coefficient in that square.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Prove multiplying norm-one expressions gives another integer solution.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Prove x²−3y²=−1 has no integer solutions.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Explain why every positive norm-one unit is a power of the least one greater than 1.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
A search of twelve convergents finds no norm+1. What is justified?
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
