Hamiltonian cycle
Named by 2 essays across 2 fields — each of them below, with the objects they name alongside it.
Every pair side by side, once
Seat an odd number of guests at round tables for as many nights as it takes, the same table sizes every night, so that every two guests sit side by side on exactly one night. For a single table a zigzag turned a notch each night does it for any number of guests. For other table plans the answer is almost always yes — and for six guests at two tables of three, nine at tables of four and five, and eleven at three, three and five, an exhaustive search proves it is no.
Seven points and a knot they cannot avoid
Put seven points anywhere in space and join every pair with a straight segment. There are 360 closed paths that visit all seven points once each, and at least one of them is knotted — however the points are placed. The reason is a parity, as it was for six points and a linked pair: a number read off each path's knot adds up, over all 360, to something odd. Six points are not enough; seven always are.
Named alongside it
The objects these essays reach for when they reach for this one.
Complete graphAlexander polynomialBipartite graphCounterexampleExhaustive searchGraph decompositionIntrinsic knottingKirkmanKnotParitySeatingStick number