Pell equation
Named by 8 essays across one field — each of them below, with the objects they name alongside it.
The square that cannot shrink
The usual proof that the square root of two is irrational is about even and odd numbers. There is a proof about squares instead, in which a supposed solution is folded into a smaller one — and the folding is a drawing.
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.
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.
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.
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.
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.
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.
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.
Named alongside it
The objects these essays reach for when they reach for this one.
Continued fractionsFundamental solutionQuadratic irrationalConvergentHyperbolaLogarithmPeriodicityQuadratic formUnitAlgorithmConjectureConjugate