Chinese remainder theorem
Named by 3 essays across 3 fields — each of them below, with the objects they name alongside it.
Two dials at once
Watch one number on two clocks with different faces. If the faces share no factor, every pair of readings occurs exactly once — so two remainders name a number, and a hard calculation can be split into two easy ones.
A collision that finds a factor
A walk through the remainders modulo a number must eventually repeat, and it repeats modulo each hidden prime factor long before it repeats modulo the number. Pollard saw that the earlier repeat can be detected without knowing the prime — and that its timing is the birthday problem, so the cost is the square root of the factor.
A number that is its own square
Square 625 and the answer, 390625, ends in 625. Square 9376 and it ends in 9376. Keep extending them and two endless strings of digits appear, running leftward, each equal to its own square — and their product is nought, though neither is. Base ten does this because ten is two times five: a number in base ten is secretly a pair of numbers, one in base two and one in base five, and these two strings are the pair (one, nought) and (nought, one). In a prime base no such thing can happen.
Named alongside it
The objects these essays reach for when they reach for this one.
Modular arithmeticGreatest common divisorBijectionBirthday problemCollisionCounting two waysCyclic groupFactoringHensel lemmaIterationLatticeLifting