Concept

Conjecture

A statement believed true and not proved, usually on the strength of computation or of a pattern holding in every case anybody has checked. Its value is in what it organises, and the proof of a good one generally needs machinery invented for the purpose.

Named by 29 essays across 8 fields — each of them below, with the objects they name alongside it.

Ulam's spiral to 900. The integers up to 900 laid out in a square spiral, with the primes marked; they crowd onto diagonal lines.

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.

discrete · Prime distribution
Every fifth partition count divides, and the rank that says why. A row of partition counts with the ones in a congruence class marked, and a histogram of partitions sorted by rank.

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.

number · Partitions
A path of 4 bounces that closes, in a triangle of 100°, 40°, 40°. A triangular billiard table with a periodic path found by an exhaustive sweep of starting positions and directions.

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.

dynamics · Billiards
16 ways to sort it. A small order with its 16 linear extensions counted, and for each incomparable pair the fraction of extensions putting one before the other.

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.

discrete · Posets
Five families of primes, counted below 100,000. Five counting curves on logarithmic axes: every prime, the primes one more than a multiple of four, the twin pairs, the primes one more than a square, and the Mersenne primes. Two of the five families are known to be infinite and three are open questions.

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.

number · Infinitude of primes
Where the aliquot sequence of every number up to 1000 goes. A grid of the starting values 2 to 1000 coloured by the fate of each aliquot sequence: 964 reach 1, 19 reach a perfect number, 3 enter a cycle, and 13 pass 10²² undecided.

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.

number · Perfect numbers
Böröczky's 12 points and their 6 ordinary lines. A disc standing for the projective plane: the 6 corners of a regular polygon inside, and 6 points at infinity marked in pairs on the rim. All 22 connecting lines are drawn, the 6 ordinary ones solid.

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.

geometry · Ordinary lines
9 trees on a cubic, 10 rows of three. The curve y = 1/(1 + x²) with 8 marked points on it and a 9th at infinity. Every line through three of them is drawn: 10 in all.

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.

geometry · Ordinary lines
Envy-free up to one item, and up to any: three people, five items. A value matrix for three people, five items with two allocations beneath it: one envy-free up to one item but not up to any, and one envy-free; the counts over all 243 allocations beside it.

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.

applied · Fair division
Every power of x that draws a hyperoval, in the planes of order 4 to 4096. q = 4: 1 exponents in 1 classes (conic); q = 8: 3 exponents in 1 classes (conic); q = 16: 3 exponents in 1 classes (conic); q = 32: 11 exponents in 3 classes (conic, translation/Glynn I/Glynn II, Segre); q = 64: 3 exponents in 1 classes (conic); q = 128: 23 exponents in 5 classes (conic, translation, Segre/Glynn II, translation, Glynn I); q = 256: 9 exponents in 2 classes (conic, translation); q = 512: 27 exponents in 5 classes (conic, translation, Segre, translation, Glynn I/Glynn II); q = 1024: 9 exponents in 2 classes (conic, translation); q = 2048: 45 exponents in 8 classes (conic, translation, Segre, translation, translation, Glynn II, translation, Glynn I); q = 4096: 9 exponents in 2 classes (conic, translation).

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.

computation · Finite geometry
The complete graph on 7 points, drawn straight with 9 crossings. A straight-line drawing of the complete graph on 7 vertices with 9 crossings, the minimum for straight drawings; each crossing comes from four vertices in convex position.

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.

discrete · Planarity
How often 2 is a primitive root, measured against the prediction. base 2: 37.429% of primes up to 2000000, predicted 37.396%.

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.

number · Fermats little theorem
The fifty-four sums of four weighted squares that reach every number. The 54 universal diagonal quaternary forms, as coefficient lists: 1,1,1,1; 1,1,1,2; 1,1,1,3; 1,1,1,4; 1,1,1,5; 1,1,1,6; 1,1,1,7; 1,1,2,2; 1,1,2,3; 1,1,2,4; 1,1,2,5; 1,1,2,6; 1,1,2,7; 1,1,2,8; 1,1,2,9; 1,1,2,10; 1,1,2,11; 1,1,2,12; 1,1,2,13; 1,1,2,14; 1,1,3,3; 1,1,3,4; 1,1,3,5; 1,1,3,6; 1,2,2,2; 1,2,2,3; 1,2,2,4; 1,2,2,5; 1,2,2,6; 1,2,2,7; 1,2,3,3; 1,2,3,4; 1,2,3,5; 1,2,3,6; 1,2,3,7; 1,2,3,8; 1,2,3,9; 1,2,3,10; 1,2,4,4; 1,2,4,5; 1,2,4,6; 1,2,4,7; 1,2,4,8; 1,2,4,9; 1,2,4,10; 1,2,4,11; 1,2,4,12; 1,2,4,13; 1,2,4,14; 1,2,5,6; 1,2,5,7; 1,2,5,8; 1,2,5,9; 1,2,5,10.

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.

geometry · Figurate numbers
Four trees on seven points, gracefully labelled. a path: 0 6 1 5 2 4 3; a star: 0 1 2 3 4 5 6; a caterpillar: 0 5 1 6 2 4 3; a spider: 0 1 4 5 3 6 2.

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.

discrete · Labelled trees
Five cycles of the Petersen graph that cover every edge twice. Petersen graph: 57 cycles; minimum cycle double cover has 5 cycles of lengths 5, 8, 6, 6, 5; 1654 search steps.

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.

discrete · Eulerian paths
The tree writes 4/17 as four unit fractions, and a search finds three. Left-neighbour chain 4/17, 3/13, 2/9, 1/5, 0/1, gaps 1/221, 1/117, 1/45, 1/5; three-term decomposition 1/5 + 1/30 + 1/510.

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.

number · Stern brocot
Three kinds of member, and a triple that breaks a grouping. Two copies of one market with three members of each of three kinds and cyclic preferences: on the left a grouping blocked by the triple B, 1, p; on the right a stable grouping.

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.

applied · Stable matching
Seven towns joined with junctions and without. Seven random points: spanning tree 1.65407, Steiner tree 1.54997 with 2 junctions, ratio 0.93707.

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.

geometry · Triangle centres
The tree of solutions of x² + y² + z² = 3xyz. Markov triples from (1, 2, 5) to depth 4; 231 Markov numbers below 10^15; first 1, 2, 5, 13, 29, 34, 89, 169, 194, 233, 433, 610, 985, 1325.

The numbers that come after the golden ratio

The golden ratio is the number worst approximated by fractions: no constant larger than √5 works in Hurwitz's theorem. Set it aside and √2 is worst, with √8. Set that aside and the next is a number whose continued fraction repeats 2, 2, 1, 1, with √221/5. Markov proved in 1879 that the list continues for ever below 3, one entry for each solution of x² + y² + z² = 3xyz.

geometry · Golden ratio
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.

number · Pell
A walk on Gaussian primes stopped by a moat. Step ≤ 2 from 1 + i: 720 primes reached, farthest 17 + 42i at 45.310.

A walk on Gaussian primes stopped by a moat

Start at 1 + i and step from Gaussian prime to Gaussian prime, never more than 2 at a time: the walk reaches 720 primes and stops at 17 + 42i, surrounded by a moat. Allow steps of √10 and it reaches a quarter of a million primes before stopping at distance 1,024; steps of 4 carry it to 4,313. Whether some fixed step lets a walker reach infinity is a question Basil Gordon asked in 1962, and it is open.

number · Sums of two squares
Where 3003 appears eight times. 3003 = C(3003,1) = C(78,2) = C(15,5) = C(14,6) and their mirror images; N(3003) = 8.

How often a number can appear in Pascal's triangle

Every number bigger than one appears in Pascal's triangle at least twice, next to the 1 at each end of its own row. A few appear more often: 120 appears six times, and 3003 eight — as C(3003, 1), C(78, 2), C(15, 5) and C(14, 6), each twice. Counting every entry up to a million million finds seven numbers that appear more than four times and none that appear five, seven or more than eight. Whether any number appears ten times, or whether there is any limit at all, is unknown.

discrete · Pascals triangle
Ten thousand steps of disorder, then a highway. Langton's ant from an empty grid: highway from step 9976, moving (-2, -2) every 104 steps; position after 12000 steps (-56, -32).

An ant that builds a highway

Two rules: on a white cell turn right, on a black cell turn left, and flip the colour as you leave. From an empty grid the ant wanders for 9,975 steps without visible order, then falls into a cycle of 104 steps that carries it diagonally for ever. That it always escapes to infinity is proved; that it always ends on a highway is not.

dynamics · Cellular automata
Right triangles of area five, six and seven. area 5: 3/2, 20/3, 41/6; area 6: 3, 4, 5; area 7: 24/5, 35/12, 337/60.

The areas a right triangle can have

A right triangle with sides 3, 4 and 5 has area 6; one with sides 3/2, 20/3 and 41/6 has area 5. No right triangle with rational sides has area 1, 2 or 3. Which whole numbers are areas is a thousand-year-old question, and its best answer — two counts of solutions to quadratic equations — rests on one of the great unproved conjectures.

geometry · Pythagoras
Forty-two squares in a row. p = 366791: partial sums of (n/p) for n ≤ 200: 0, 10, 20, 30, 40, 46, 54, 60, 64, 72, 78, 78, 88, 92, 98, 102, 108, 116, 122, 126, 132; least non-residue 43.

The first number that is not a square

Modulo a prime, half the numbers are perfect squares. Count up from 1 and the first non-square is usually 2 or 3 — but modulo 366,791 the numbers 1 to 42 are all squares. How far the run of squares can last is one of the oldest open questions about primes: the truth is tiny, the proofs are huge, and the Riemann hypothesis would close most of the gap.

number · Quadratic reciprocity
The numbers below a hundred as sums of three cubes. Smallest solution size per n < 100: 1:9 2:1 3:1 6:2 7:2 8:16 9:2 10:2 11:3 12:11 15:2 16:2 17:2 18:3 19:3 20:3 21:16 24:2 25:3 26:3 27:6 28:3 29:3 30:10d 33:16d 34:3 35:3 36:3 37:4 38:4 39:6d 42:17d 43:3 44:8 45:4 46:3 47:8 48:4 51:796 52:11d 53:3 54:3 55:3 56:4 57:4 60:5 61:5 62:3 63:4 64:6 65:4 66:4 69:5 70:21 71:4 72:4 73:4 74:15d 75:9d 78:55 79:35 80:4 81:3 82:14 83:4 84:8d 87:4271 88:5 89:7 90:4 91:4 92:4 93:7 96:22 97:5 98:5 99:4.

Thirty-three as three cubes

Which numbers are a sum of three cubes, allowing negatives? Nothing like the rule for two squares is known — only that numbers leaving 4 or 5 on division by 9 are impossible. Every other number below a hundred has a sum with cubes under 5,000, except eight; 33 needed sixteen-digit cubes and was found in 2019.

number · Sums of two squares
The chance a random formula can be satisfied. n=20: 3:1.000 3.25:0.988 3.5:0.988 3.75:0.938 4:0.863 4.25:0.762 4.5:0.563 4.75:0.400 5:0.200 5.25:0.275 5.5:0.100 5.75:0.050 6:0.013; n=40: 3:1.000 3.25:1.000 3.5:1.000 3.75:0.963 4:0.863 4.25:0.575 4.5:0.425 4.75:0.138 5:0.050 5.25:0.025 5.5:0.037 5.75:0.000 6:0.000; n=80: 3:1.000 3.25:1.000 3.5:1.000 3.75:1.000 4:0.912 4.25:0.613 4.5:0.275 4.75:0.075 5:0.050 5.25:0.000 5.5:0.000 5.75:0.000 6:0.000.

A threshold nobody can place

Random three-literal clauses on n variables: below about 4.27 clauses a variable a formula can almost always be satisfied, above it almost never. The change is proved to be sharp and nobody has proved where it is. Counting solutions overshoots to 5.19, counting pairs of them fails outright, and every algorithm with a proof stops well short.

probability · Random graphs
Five squares on one octagon. Octagon [[1.0756,0.5516],[0.6082,1.2964],[-0.466,1.2294],[-0.8758,0.3153],[-0.797,-0.3348],[-0.3706,-0.7683],[0.2718,-0.6558],[1.1693,-0.4453]]; squares with sides 1.4975, 1.4588, 1.4407, 1.3929, 1.3921.

An odd number of squares on every polygon

Does every closed curve that does not cross itself pass through the four corners of some square? Toeplitz asked in 1911, and for polygons and smooth curves the answer is yes, because the squares can be counted and the count is odd. For rectangles of a fixed shape the count is even, which is why they needed a different proof — and for an arbitrary continuous curve the square is still not proved to exist.

topology · Jordan curve
What junctions save in space. tetrahedron SMT/MST 0.81305; 12-point helix 0.79472.

A chain of tetrahedra saves more

In the plane, junctions never seem to shorten a network of roads by more than the equilateral triangle's 13.4%. In space the regular tetrahedron allows 18.7%, and the obvious guess — that it is the worst case, as the triangle is in the plane — is wrong. Points spaced along a chain of tetrahedra glued face to face do better with every point added: 20.5% at twelve, 21.4% at sixty, and about 21.6% in the limit. Random points come nowhere near, and a search that only goes downhill does not find the chain.

geometry · Triangle centres

Named alongside it

The objects these essays reach for when they reach for this one.

Exhaustive searchCounting argumentPrimesDensityElliptic curveHeuristicIncidenceOpen problemQuadratic formClassificationComplete graphContinued fractions

All concepts