Matpic
mathematics, in pictures
Series

Pell — the series

8 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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.

    part 1 · number
  2. Bhāskara's cyclic method on x² − 61y² = 1, step by step. A table of the cyclic method's rows: the helper m chosen at each step and the near miss a² − Db² = k it produces, ending at k = 1 with the fundamental solution.

    A method that is allowed to miss

    Bhāskara's cyclic method solves x² − Dy² = 1 by aiming at the wrong target. It keeps a pair a, b with a² − Db² = k for some small k, combines it with a helper chosen so that k can be divided out, and repeats until k is 1. For D = 61 it reaches the ten-digit fundamental solution in 13 steps, where walking the convergents of √61 takes 22 — and for every D up to 100 it is faster.

    part 2 · number
  3. 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.

    part 3 · number
  4. The solutions of x² − 2y² = N, class by class. Rows for several right-hand sides N, each marking the solutions of x² − 2y² = N at the logarithm of x + y√2, coloured by class, over alternately shaded windows one unit-step wide.

    Two families of solutions, and a box that holds both

    Replace the 1 in Pell's equation by 7 and x² − 2y² = 7 still has infinitely many solutions — but they fall into exactly two families, each one an orbit of the same multiplication, and every family has a member inside a box whose size is fixed in advance. How many families there are is then a count of factors, and 3 has none.

    part 4 · number
  5. 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.

    part 5 · number
  6. Which D up to 400 make x² − Dy² = −1 solvable. D from 2 to 400: 69 solvable, 302 blocked by 4 or a prime 4k + 3, 9 unsolvable with no such obstruction, first at 34.

    When minus one can be reached

    x² − Dy² = 1 always has a solution. Put −1 on the right and it sometimes does and sometimes does not. A remainder rules out most D at a glance, every prime of the form 4k + 1 is guaranteed one, and between the two lies a set of D decided by arithmetic several layers deep — of which, a million numbers in, 78% are solvable, on the way to a limit of 58% that was proved only in 2022.

    part 6 · number
  7. 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.

    part 7 · number
  8. 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.

    part 8 · number

All series