Conjecture
Named by 18 essays across 6 fields — each of them below, with the objects they name alongside it.
The primes on a spiral, and a pattern nobody ordered
Wind the whole numbers outward in a square spiral, mark the primes, and they line up on diagonals. The observation is a hundred years old and there is still no proof it means anything.
Every fifth one divides
p(4) is 5, p(9) is 30, p(14) is 135, and every partition count at a number leaving four on division by five is divisible by five. Ramanujan read it off a table; the explanation is a way of splitting those partitions into five equal heaps.
The triangle nobody can settle
Does every triangular billiard table have a path that closes on itself? Acute triangles do, right triangles do, triangles with rational angles do — and for the rest the question has been open since it was asked.
How many ways to sort it
An order says some things come before others and leaves the rest open. Counting the orderings consistent with it measures how much is still unknown — and the counting is as hard as any counting problem gets.
Which infinitudes are proved
The primes never stop, and neither — apparently — do the twin pairs, the primes one more than a square, or the Mersenne primes. Three of those four statements are theorems and one is not, and counting the members of each family tells nobody which.
The sum of the parts, taken again
Replace a number by the sum of its proper divisors and do it again. Most numbers fall to 1, a few land on a perfect number or a cycle, and some climb for hundreds of steps. Below a thousand there are twelve whose fate nobody knows — and the thing that keeps them climbing is a perfect number hiding in their factorisation.
The fewest ordinary lines a polygon allows
Take the corners of a regular polygon and add the points at infinity where its parallel chords meet. Every chord then carries three points, the line at infinity carries all the new ones, and the only lines left with exactly two points are the tangents at the corners — half as many as there are points. Dirac guessed in 1951 that nothing does better, and Green and Tao proved it in 2013.
Rows of three, planted on a cubic
Nine trees can be planted in ten rows of three, and the arrangement that does it is not a grid or a star but nine points on a cubic curve. On the curve three points are in line exactly when their angles add up to a right angle, so choosing the points as a cyclic group turns collinearity into addition — and the count of rows it produces is the number Green and Tao proved is the most any planting can reach.
Envy that any single item would cure
Envy-free up to one item lets a person's envy be excused if removing the envied bundle's best item would cure it. The stronger standard asks that removing any item would — even the one that person values least. Every allocation of three people's items can be searched, and an allocation meeting the stronger standard was there every time; for two people cut and choose finds one, for three it took until 2020 to prove, and for four nobody knows.
Every power of x that draws a hyperoval
In a plane of order 2^h, the graph of x^k plus two points at infinity is sometimes a hyperoval — as many points as a plane allows with no three in line. Searching every exponent in every plane from order 4 to 4096 finds hundreds that work, and once six symmetries of the problem are applied they fall into exactly the families already known: the conic, the translation curves, Segre's x⁶ and Glynn's two. Whether that list is complete in every order is open.
The crossings a graph cannot avoid
Five points all joined need one crossing, six need three, seven need nine. Euler's formula gives a lower bound that grows like the square of the number of points and is badly wrong; a sampling trick turns the same formula into a bound that grows like the fourth power and is right to within a constant. And for drawings with straight edges the count turns out to be something else entirely: the number of quadrilaterals the points make.
How often two generates every remainder
For some primes p, the powers of 2 run through every non-zero remainder before repeating; for others they get stuck in a smaller cycle. Among the primes up to two million the first kind make up 37.43% — and Artin predicted 37.40% in 1927 by treating each way of getting stuck as an independent accident. For the base 5 the same prediction is wrong, because two of the accidents are secretly the same one, and nobody can yet prove that 2 works for infinitely many primes at all.
Fifteen numbers decide every number
Lagrange proved that x² + y² + z² + w² takes every whole value. Ramanujan asked which other sums of four weighted squares do, and in 1917 listed fifty-five. One of them is wrong: x² + 2y² + 5z² + 5w² misses 15. The mistake points at a theorem found eighty years later — to know whether such a form reaches every number, it is enough to check that it reaches 1, 2, 3, 5, 6, 7, 10, 14 and 15. Nine checks decide infinitely many cases, and the nine numbers come out of an escalation any one can run.
A labelling every tree seems to have
Number the points of a tree 0 to n − 1 and write on each edge the difference of the numbers at its ends. The labelling is graceful if the edges then carry 1 to n − 1, each exactly once. Every tree anyone has ever checked — every one of the 551 trees on twelve points, and every tree up to about thirty-five — has such a labelling, and no one knows why. A graceful tree also tiles a complete graph by rotation, which is why the question was asked.
Every edge on exactly two cycles
Draw a graph on paper without crossings and its faces are cycles, each edge on exactly two of them. George Szekeres and Paul Seymour asked in the 1970s whether every graph without a bridge has such a set of cycles, drawn on paper or not. The Petersen graph needs five, the flower snark on twenty points five as well, every random graph tried has one, and nobody has proved it — though a minimal counterexample would have to be a snark.
Three unit fractions for every four over n
Two neighbouring fractions in the tree always differ by a unit fraction, so stepping down the tree writes any fraction as a sum of them — four over n in at most four steps. Erdős and Straus asked in 1948 whether three always suffice. Three identities settle every n except those leaving remainder 1 on division by 24, a search settles every one anyone has tried, and nobody has a proof.
A third kind of member
Two sides always have a stable matching, and the proof is a procedure. Add a third kind of member — capitals who rank numbers, numbers who rank letters, letters who rank capitals — and every market small enough to check still has a stable grouping. Ten million markets of three have at least ten each. Nobody can prove it continues, because the structure the two-sided proof stands on is gone.
What a junction can save
Join a set of towns by roads that may meet only at towns, and the shortest such network is easy to find. Allow junctions anywhere and the network gets shorter — but never, it is conjectured, by more than 13.4%, the saving the Fermat point makes on an equilateral triangle. Computed exactly on 1,304 random sets of up to seven towns, the average saving is about 3%. An adversarial search drives it down to the bound and no further, and a hexagon with its centre attains it exactly. The proof of the bound that was accepted for twenty years has a gap.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchCounting argumentIncidenceOpen problemClassificationComplete graphCounterexampleDensityExtremal configurationOrdinary linePlanar graphPoint set