Concept

Quadratic irrational

A number of the form (a + b√D)/c with whole numbers a, b, c and D not a perfect square. These are the irrational roots of quadratic equations with whole coefficients, and they are exactly the numbers whose continued fractions eventually repeat.

Named by 7 essays across 3 fields — each of them below, with the objects they name alongside it.

Whole-number points on x² − 2y² = 1. The branch of the hyperbola x² − 2y² = 1 in the first quadrant, with the whole-number points on it marked and labelled, and the lattice drawn faintly behind.

One solution that makes all the others

The equation x² − 2y² = 1 has infinitely many whole-number solutions, and every one of them is a power of the smallest. The multiplication that produces them is what multiplying two numbers of the form a + b√2 comes to when the √2 terms are collected — so an equation about a hyperbola turns out to carry a group.

number · Pell
The states the continued fraction of √61 can be in. A grid of whole-number pairs with the band of reduced states shaded, the pairs that qualify marked, and the cycle of states visited by the expansion of the square root numbered in order.

Why the expansion has to repeat

The continued fraction of √61 runs 7; 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14 and then starts again. It must: each step's state is a pair of whole numbers trapped in a small band, and only 14 pairs fit. The expansion of √61 visits 11 of them in a cycle, the other 3 form a cycle of their own, and the period reads the same backwards before its last term, which is twice the first.

number · Pell
How many digits Pell's first solution has, for D up to 1000. A scatter of the digit count of the smallest solution of Pell's equation against D, with the record-setting values ringed and labelled.

Sixty needs two digits and sixty-one needs ten

The smallest solution of x² − 60y² = 1 is x = 31. For x² − 61y² = 1 it is x = 1,766,319,049. The jump is not an accident of 61: the solution is exactly the product of one period's complete quotients, so its size is set by how long the continued fraction takes to come home — and for the cattle problem Archimedes is said to have posed, that product has 103,273 digits.

number · Pell
The first 600 terms of the continued fraction of ∛2. Partial quotients of ∛2 computed exactly: 1, 3, 1, 5, 1, 1, 4, 1, 1, 8, 1, 14, 1, 10, 2, 1, 4, 12, 2, 3 …; the largest in the first 600 is 7451.

A cube root that looks like chance

The continued fraction of √61 repeats after eleven terms, and every square root's must. The cube root of 2 obeys no such law. Computed exactly, by a method of Lagrange's that never leaves the whole numbers, its terms run 1, 3, 1, 5, 1, 1, 4 … and throw up a 534 at the thirty-fifth place and a 7,451 at the 571st — and in every statistic anyone has measured, they look like the terms of a number picked at random.

number · Pell
The tree of solutions of x² + y² + z² = 3xyz. Markov triples from (1, 2, 5) to depth 4; 231 Markov numbers below 10^15; first 1, 2, 5, 13, 29, 34, 89, 169, 194, 233, 433, 610, 985, 1325.

The numbers that come after the golden ratio

The golden ratio is the number worst approximated by fractions: no constant larger than √5 works in Hurwitz's theorem. Set it aside and √2 is worst, with √8. Set that aside and the next is a number whose continued fraction repeats 2, 2, 1, 1, with √221/5. Markov proved in 1879 that the list continues for ever below 3, one entry for each solution of x² + y² + z² = 3xyz.

geometry · Golden ratio
Three cycles of forms for the prime 229. Reduced forms of discriminant 229 in 3 cycles of lengths 6, 6, 2; checked: h(5) = h(13) = 1, h(229) = h(257) = 3, h(401) = 5, h(577) = 7, h(1129) = 9.

Real fields that factorise uniquely

Among the fields Q(√−d) only nine have unique factorisation, and Gauss guessed as much. Among the real fields Q(√p) he guessed the opposite — infinitely many — and the guess is still unproved. Counting cycles of reduced quadratic forms for the 16,900 primes p ≡ 1 mod 4 below 400,000 finds 79 per cent with class number one, drifting slowly down towards the 75.45 per cent that Cohen and Lenstra's heuristic predicts — and finds the reason the real fields behave so differently: their units are enormous.

number · Pell
A square cut into rectangles of ratio φ squared. 1/r@(0.0000,0.0000) 0.1273×0.3333; r@(0.1273,0.0000) 0.8727×0.3333; 1/r@(0.0000,0.3333) 0.1273×0.3333; r@(0.1273,0.3333) 0.8727×0.3333; 1/r@(0.0000,0.6667) 0.1273×0.3333; r@(0.1273,0.6667) 0.8727×0.3333

The rectangles a square can be cut into

Cut a square into rectangles that all have the same shape — the same ratio of long side to short, in any sizes, lying either way. Two to one is easy. The golden ratio is impossible, however many pieces are used and however small. The ratio φ squared works with six. What decides it is not the ratio but its partner: the number obtained by changing the sign of the square root, which must be positive. Changing that sign is a map that keeps every sum and product and scrambles every size, and areas that must add up to one cannot all turn negative.

computation · Scissors congruence

Named alongside it

The objects these essays reach for when they reach for this one.

Continued fractionsPell equationFundamental solutionConjectureGolden ratioPeriodicityQuadratic formAlgebraic numberAlgorithmBinary treesConjugateConvergent

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