Patterns,
Proofs, and Primes

The same data, read forward: 391/299 = [1; 3, 4]. These partial quotients generate convergents whose signed coefficients reproduce the Euclidean remainders.

The alternating trap: the 1st, 3rd, 5th, … convergents rise below the target; the 2nd, 4th, 6th, … converge downward from above.

The matrix spine: multiplying the digit matrices [[a, 1], [1, 0]] places consecutive convergents in its columns. Every product has determinant ±1, so it can be reversed over the integers.

Linear fractional transformations: a matrix M = [[a, b], [c, d]] with ad − bc ≠ 0 acts by T_M(t) = (at + b)/(ct + d). Composition is matrix multiplication, with the rightmost factor acting first. A repeating tail is a fixed point, giving ct² + (d − a)t − b = 0.

A prefix conjugates the period: if x = T_N(x) and y = T_M(x), then y = T_MNM⁻¹(y). The nontrivial period matrix stays nontrivial after conjugation, so an eventually periodic continued fraction is still a quadratic irrational. Chapter 3 works this out for √3.

Exactly which numbers repeat? Lagrange’s theorem says: precisely the real quadratic irrationals. Galois’s sharper criterion says the repetition starts immediately exactly when x > 1 and −1 < x′ < 0, where x′ is the other root of its quadratic. Thus √2 is eventually periodic, while 1 + √2 is purely periodic. The expanded book proves both directions using bounded integer states and unique reduced predecessors, with worked examples, exercises, and solutions. Sage Lab 4 turns these proofs into exact matrix and quadratic-field experiments.

A calendar is an approximation: the tropical year has about 0.2422 extra days. Its convergents begin 1/4, 7/29, 8/33, 31/128. The Julian rule is 1/4; after requiring a cycle of whole centuries, the convergent 97/4 to the per-century excess 100 × 0.2422 becomes the Gregorian rate 97/400.

Almost base 11: the Khinchin–Lévy law gives a typical precision factor of e^{π²/(6 log 2)} ≈ 10.731 per continued-fraction term—about 1.03 decimal digits.

π looks typical; e does not: for the first 100,000 partial quotients of π, the geometric mean is 2.67956 and q_n^2/n is 10.67835, close to the generic targets 2.68545 and 10.73102—but no limiting theorem is known for π. Euler’s patterned expansion of e instead makes both statistics diverge.

Continued fractions can calculate: Gosper’s homographic and bihomographic algorithms absorb input terms and emit certified output terms using only integer matrices. A modern interval-valued refinement prevents exact boundary cases—such as √2 × √2 = 2—from stalling forever.

Periodicity meets Pell: The identity p_k² − Dq_k² = (−1)^k+1Q_k+1 proves that a period endpoint gives norm ±1. The exact Pell lab shows the parity rule and the palindromic digits; Archimedes’ cattle add a divisibility condition to the resulting Pell equation.

CRT becomes interpolation: a selector is 1 at one remainder and 0 at the others. Replace integer moduli with X − c and the same construction gives Lagrange’s interpolation formula. Extended gcd supplies the denominator inverses over a prime field. Read the construction and worked examples → At roots of unity, these Lagrange selectors give exactly the inverse discrete Fourier transform. Connect interpolation, Fourier inversion, and Möbius inversion →

Why Fibonacci residues repeat: follow the pair (a, b) → (b, a + b). Its inverse (u, v) → (v − u, u) makes every step reversible, so finitely many pairs force a cycle from the start. Return to (0, 1), not just a zero. Prove periodicity now → Explore extension fields and exact periods later →

From Fermat to Euler: with two colors and nine positions, remove the 2³ words that repeat a three-bead block. The remaining 2⁹ − 2³ = 504 words form 56 orbits of size 9. Cancel the invertible factor 2³ to get 2⁶ ≡ 1 (mod 9). The same count works at every prime power; CRT combines the congruences into Euler’s theorem. Follow the proof, starting with length 9 →

Undo a count: the totient is a multiplicative function, and its sum over the divisors of n is n. Möbius inversion recovers the individual contributions. The same formula counts words of each exact period. Explore multiplicative functions and divisor sums → Bonus: posets, inclusion–exclusion, and partition counts →

Fermat already gives the sum formula: in the prime field, (a + b)^p = a + b = a^p + b^p. The power map is the identity there. The binomial proof makes this a polynomial identity valid in extensions, where Frobenius can move the new elements. See Frobenius act in a quadratic extension →

Look for it: blank cells are the coefficients divisible by 2. The repeating blocks are base-2 arithmetic made visible.

Sieve through 50: boxed numbers survive. Crossing out multiples of 2, 3, 5, and 7 is enough because every composite has a prime factor no larger than its square root.

Existence versus a witness: the classical argument with x = (√2)^√2 proves by cases that some pair of irrationals has a rational power, but does not say which pair. The explicit choice a = √2 and b = log_√23 gives a^b = 3; unique factorization proves that b is irrational.

The palindromic part: remove the final 2a₀; the remaining digits read the same from either end. Matching colors mark reflected positions. Galois’s criterion and reversal of the period of √D + a₀ explain the symmetry.

Try D = 61. A modest radicand can have a very large first solution. Explain its index using the period, then reproduce it in Sage Lab 5.

1. ProportionsTotal cattle = 50,389,082t.
2. SquareWhite + black bulls force t = 4,456,749r².
3. TriangleX² − 4,729,494Z² = 1, with Z = 9,314r.

The least positive unit ε = u + v√4,729,494 generates every positive Pell solution: ε^j = X_j + Z_j√4,729,494. The cattle condition is 9,314 divides Z_j. Follow the coefficients modulo 9,314 to find the first admissible exponent.

Minimality needs both steps. The residue recurrence checks every exponent from 1 through 2,329. Its first zero second coefficient occurs at j = 2,329. Since all positive Pell solutions are powers of ε and their second coefficients increase, this gives the smallest herd: 206,545 decimal digits, approximately 7.760271406 × 10²⁰⁶⁵⁴⁴.

Follow the full reduction in the book → · Ilan Vardi, Archimedes’ Cattle Problem (preprint).