Heuristic
Named by 12 essays across 5 fields — each of them below, with the objects they name alongside it.
The question nobody can answer
Halve it if it is even, triple it and add one if it is odd. Every number anyone has tried comes down to one. Nobody can prove they all do, and the reason is not that the problem is hard to state.
The heuristic that cannot be a proof
There is a two-line argument that the Collatz conjecture is true, it is convincing, and everybody who works on the problem believes it. It also cannot be turned into a proof, and understanding exactly where it fails is more instructive than the argument itself.
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.
Two primes where Fermat holds twice
Fermat's theorem says p divides 2^(p−1) − 1. Usually p² does not. It does at 1093 and at 3511 and at no other prime anyone has found, in searches reaching past 10^19. The leftover, (2^(p−1) − 1)/p taken mod p, behaves like a random number, so a prime has about a one-in-p chance of the extra divisibility — and a random count with that chance grows so slowly that two by now is unremarkable, while nobody can prove there are any more, or that there are infinitely many primes where it fails.
The boards a knight can tour
A knight can visit every square of a chessboard once and land a move from where it began. On a 5 × 5 board it cannot, on a 4 × 100 board it cannot, and on a 3 × 8 board it cannot, though on 3 × 10 it can in sixteen ways. Allen Schwenk found in 1991 the complete list of rectangles without a closed tour, and the reasons on it are of two kinds — a colouring that counts squares, and a second colouring that a tour would have to make agree with the first.
Twenty-seven million cubics, sorted by their symmetry
Write down every cubic x³ + ax² + bx + c with coefficients up to 150 and sort each by its Galois group. 98.7% have all six symmetries of three roots. The rest are reducible at a rate of fifteen in every H², or cyclic at five in every H to the three-halves — and the cyclic ones turn up four times as often as chance would allow, because discriminants are not random numbers.
A sum of factorials that converges at every prime
1 + 1 + 2 + 6 + 24 + 120 + … grows faster than any geometric series and has no sum in the real numbers. Measure size by divisibility instead and the same series converges at every prime at once, to a different number each time. Whether any of those numbers ends in zero is a question Đuro Kurepa asked in 1971, and a search through every prime below 100,000 says no while a coin-toss model says it should have said yes about twice.
When Euclid's number is itself prime
Multiply the primes up to p and add one. Euclid needed only that the result has a prime factor beyond p; it is itself prime for p = 2, 3, 5, 7 and 11, and then not again until 31, and then not until 379. Up to 1,100 there are nine such primes of each sign, against thirteen that a careful estimate expects. The estimate predicts infinitely many and grows by about 1.8 every time x is multiplied by e; nobody can prove there is even one more.
The numbers that fool every base
561 = 3 · 11 · 17 passes Fermat's test for every base it shares no factor with, and so do 1,105, 1,729 and 102 more numbers below ten million. They were proved infinite only in 1994. Counted to 10²¹ there are about twenty million, growing like x to a power that has crept from 0.21 to 0.35 — just past the bound the proofs guarantee and nowhere near the power of one that Erdős's argument predicts.
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.
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.
Nine cubes for two numbers
Every whole number is a sum of at most nine positive cubes, and only two numbers, 23 and 239, need all nine. Fifteen need eight, the last of them 454; 121 need seven, the last 8,042; six are needed for the last time at 1,290,740. The count every number needs is nine, fixed by two small accidents. The count every large number needs is somewhere between four and seven, and the computation says four — a question about how the cubes spread out, not about the cubes themselves.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchModular arithmeticOpen problemConjecturePrimesCollatzFactorialIterationLogarithmParityPrimeSum of cubes