Quadratic form
Named by 20 essays across 5 fields — each of them below, with the objects they name alongside it.
![The level curve, and the axes the matrix chooses. The curve xᵀAx = 1 for the matrix [2, 0.8, 0.8, 1.4], drawn by solving for the radius at each angle, with the two eigen-directions marked; they cross at a right angle and are the axes of the curve.](/f/the-level-curve-and-the-axes-the-matrix-chooses-81b169af.svg)
Symmetry forces a right angle
A matrix equal to its own reflection across the diagonal always has real stretches and always has perpendicular directions to stretch along. Neither is true of matrices in general, and both follow from one line of algebra.
One sign decides which curve
The general quadratic in two variables has six coefficients and draws a conic. Which of the four it draws is settled by a single combination of three of them, and the other three cannot change the answer however they are chosen.
Solutions that come in multiples of p
Count the solutions of x² + y² + z² = 0 in the field with five elements and there are 25; with seven, there are 49. Whenever a system of equations has more unknowns than its total degree, its number of solutions is a multiple of the characteristic — which forces a solution besides zero, and the reason is a sum over the field that vanishes because its non-zero elements form one cycle.
Why the expansion has to repeat
The continued fraction of √61 runs 7; 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14 and then starts again. It must: each step's state is a pair of whole numbers trapped in a small band, and only 14 pairs fit. The expansion of √61 visits 11 of them in a cycle, the other 3 form a cycle of their own, and the period reads the same backwards before its last term, which is twice the first.
Three triangular numbers, and no fewer
On 10 July 1796 Gauss wrote in his diary: ΕΥΡΗΚΑ — num = Δ + Δ + Δ. Every whole number is a sum of three triangular numbers. Two are not enough, and not by a little: the numbers that are sums of two thin out to a share of nought. Both facts are statements about squares in disguise, and one picture translates them.
The highest point on the sphere is an eigenvalue
For a symmetric matrix, walk a unit arrow over every direction and record the value of xᵀAx. The highest value reached is the largest eigenvalue, the lowest is the smallest, and every eigenvalue in between is a saddle height, a minimum of maxima. From that one description comes a theorem no formula for the roots could give: delete a row and its column, and every eigenvalue of what is left sits between two of the original's.
How many changes undo a knot
Cut the string at a crossing, pass it through the other strand and join it up again, and any knot can be undone by doing that often enough. The fewest changes needed is the unknotting number, and proving that fewer will not do needs a number that each change can move only a little. The signature moves by at most two per change — enough to settle thirteen of the fourteen knots up to seven crossings, and not the fourteenth.
A fraction on the circle forces a whole point
If a number is a sum of two squares of fractions, it is a sum of two squares of whole numbers. Draw the circle, mark the rational point, join it to the nearest lattice point and follow the line to where it meets the circle again: the new point is rational too, with a smaller denominator. Repeat, and the denominators fall until they reach one. The argument needs no primes at all — only the fact that every point of the plane is within distance one of the lattice.
Counting the classes that break factorisation
The class number measures how badly unique factorisation fails in a field, and defined through ideals it looks impossible to compute. Gauss computed it by hand, for every field he wanted, by counting quadratic forms — and every form can be squeezed, by changes of variable that keep its values, into exactly one small standard shape.
The orders a plane cannot have
Every counting condition allows a projective plane of order six, and there is none. The proof that rules it out looks at one matrix identity — each point on seven lines, each two points on one — and turns it, by way of Lagrange's four squares, into the statement that six would have to be a sum of two squares. Run on the planes that do exist, the same argument hands back their orders as sums of two squares; run on six, it asks for something no arithmetic can supply.
Six numbers that need five pentagons
In 1638 Fermat wrote that every whole number is a sum of three triangular numbers, four squares, five pentagonal numbers, six hexagonal numbers, and so on for every polygon — and that he had a proof he would not write down. The claim is true; Cauchy proved it in 1813. What the claim hides is how unequal the cases are. Triangles and squares need their full count infinitely often. For pentagons, only six numbers ever need all five — 9, 21, 31, 43, 55 and 89 — and from hexagons on, two apiece.
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.
Euler's sixty-five convenient numbers
For some n, whether a prime can be written as x² + ny² is settled by its remainder on division by 4n alone, the way a prime's remainder on division by four settles whether it is a sum of two squares. Euler found sixty-five such n, from 1 to 1,848, called them convenient, and used the largest to prove that 18,518,809 is prime. Every one of them is a number whose class group has no element of order more than two — and whether the list is complete is still not known.
Class groups drawn at random
The class number of an imaginary quadratic field measures how badly unique factorisation fails there, and it looks arbitrary: 1, 2, 4, 3, 5, 12, 1. Count how often 3 divides it and the answer is not one time in three but noticeably more — heading, by a conjecture of Cohen and Lenstra, for 44 per cent. The reason is that class groups behave like finite groups chosen at random with each group weighted by one over its number of symmetries, so groups with few symmetries, like the cyclic group of order 3, turn up more often than the naive count suggests.
The signs no completion can change
A quadratic in several variables can be completed square by square in many orders, and each order hands back different coefficients. What no order changes is how many come out positive and how many negative — and that count, read off a completion of A − tI, says how many eigenvalues lie below t without finding a single one.
Regression is a square completed halfway
A bell in two variables, sliced at a fixed x, is a bell in y — and completing the square in y alone says where it is centred: at ρx, on a line flatter than the ellipse's own axis. That gap is regression to the mean, and the same completion says it runs backwards in time just as well as forwards.
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.
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.
Holding one still turns the other over
A quantity can rise with an outcome and still predict it downwards once a second quantity is held fixed. Completing the square in two predictors says exactly when: the pair of correlations has to fall in a wedge whose angle is arcsin of the predictors' own correlation. Averaged over every possible correlation matrix the chance is exactly 1/4 − 1/π², about one coefficient in seven; two coefficients never reverse together, and with two dozen predictors almost two in five do.
The class number is a degree
The integers of ℚ(√−23) fall into three classes of ideals, and that 3 can be found by counting forms. It is also the degree of a single number: the modular function j, evaluated at the three forms' points, gives three values that are not whole but are the roots of one cubic with whole-number coefficients. When there is only one class, the value is a whole number outright — and that is why e^(π√163) misses being whole by less than a trillionth.
Named alongside it
The objects these essays reach for when they reach for this one.
Class numberEigenvalueExhaustive searchSums of two squaresClass groupCompleting the squareConjectureDiscriminantUnique factorisationConditional probabilityContinued fractionsCorrelation