The circle of radius √25 on the integer lattice
lattice-circle is one function. Everything below came out of it during this
build, at parameters taken from the essays rather than invented for this page — so a figure
here is the same figure a reader meets in an essay, and if the generator changes, this page
changes with it.
With nothing chosen
Rational points on the unit circle
From a rational point on x² + y² = 17 to a whole one
Denominators falling to one
Every point within distance one of the lattice
A walk on Gaussian primes stopped by a moat
What it checks while it draws
Collected by running the family and recording what it asserted, not written here. The count is how many separate times the claim was put to the test while these drawings were made.
- the count on the circle of squared radius 1 is 4(d₁ − d₃) ×64
- 3, 4, 5 is a Pythagorean triple ×10
- the points on the circle of radius √25 are 4(d₁ − d₃) ×8
- 5 = 2² + 1² ×7
- the tetrahedron of height 1 has no lattice point inside it ×7
- the area is 20 interior points plus half of 7 on the edge, less one ×5
- the count at dilation 1 is the Ehrhart polynomial's value ×4
- the count at k = 1 matches the cubic fitted from the first value ×4
- n is a whole number between 2 and 70 ×2
- the number of columns in the grid is a whole number between 4 and 8 ×2
- the range is a whole number between 10000 and 2000000 ×2
- 1 has solutions both on and off its family ×1
- a group with entries only up to two fails to identify two of the classes ×1
- a nine-degree sector holds its share to within ten per cent at every bound ×1
- a polygon's boundary count and its dual's add to twelve ×1
- a prime is a sum of two squares exactly when it is one more than a multiple of four ×1
- a square root of −1 exists modulo a prime one more than a multiple of four ×1
- a² + b² lies at most 2b past n ×1
- and a self-dual one has six boundary points, which is half of twelve ×1
- and dividing by the radius itself tames it further still ×1
- and every basis along the way spans a cell of the same area ×1
- and exactly four on it, which are its own corners ×1
- and from entries up to three onward the count stops moving ×1
- and is one more than a multiple of four ×1
- and it is the first one below it — the row above is still above ×1
- and it really is in the lattice ×1
- and Scott's inequality caps the boundary count at nine ×1
- and so within 2√2 · n^(1/4) + 1 ×1
- and some attain the bound ×1
- and stays inside three times the two-thirds power throughout ×1
- and stays inside twice the cube root over the range drawn ×1
- and stopping one step earlier would not have given two squares ×1
- and the answer is one of the points the circle of that radius actually passes through ×1
- and the candidate really does square to −1 ×1
- and the density is falling, which is the statement the constant qualifies ×1
- and the disc of radius two catches thirteen ×1
- and the flat ones are excluded rather than ignored ×1
- and the point it came from is on the unit circle ×1
- and the rest pair off ×1
- and they have between three and six corners ×1
- and twelve has six ×1
- at every bound the primes' angles are more even than independent draws ×1
- at least two heights are drawn ×1
- at width X^(−0.6) there are far more empty sectors than chance allows ×1
- between one and five powers are drawn ×1
- both polynomial families add up for every t ×1
- cubes leave only 0, 1 or 8 on division by 9 ×1
- D is a whole number between 2 and 40 and not a perfect square ×1
- D is between 2 and 40 ×1
- D is not a perfect square, or the equation has only the trivial solution ×1
- each of these walks stops inside its disc ×1
- each recorded representation adds up exactly ×1
- each reflection strictly lowers the denominator ×1
- every cosine averages below one per cent at two million ×1
- every edge lies at lattice distance one from the interior point ×1
- every lattice triangle in the grid satisfies the identity ×1
- every marked point really is on the circle ×1
- every one of the 2300 triples of grid points was reached ×1
- every one-degree band holds its share to within four per cent ×1
- every power is again a solution ×1
- every recorded triple adds up ×1
- every solution for 2 in the box lies on its family ×1
- every solution up to the largest drawn is a power of the smallest ×1
- every step length up to 112 is stopped by a gap below two million ×1
- for some height the linear coefficient is negative, so it counts nothing ×1
- for x² + 3y² the centre sits at exactly 1 ×1
- four consecutive sums of two squares never happen ×1
- four of the sixteen are their own dual ×1
- in five dimensions the typical error is a fraction of one shell's population ×1
- Jacobi's four-square formula ×1
- most empty sectors lie beside slopes whose two terms are both odd ×1
- n names one circle and is read only by the default view; the other views take their own parameters ×1
- no gap exceeds the elementary bound ×1
- no number 3 more than a multiple of 4 is a sum of two squares ×1
- no number leaving 4 or 5 on division by 9 is reached ×1
- no polygon with an interior point has more boundary points than Scott allows ×1
- no two slopes give the same triple ×1
- one extra for the hole repairs it, which is the Euler characteristic in disguise ×1
- one has one divisor ×1
- one of them attains Scott's bound ×1
- one to three rational turns ×1
- p names the prime the two squares are produced for and is read only by the cornacchia and reduce views ×1
- Pick's formula with one interior point makes twice the area the boundary count ×1
- Pick's theorem holds for every polygon swept ×1
- primes a² + 1 found two ways ×1
- some circles miss the lattice entirely, which is the whole question ×1
- sums of three squares miss exactly the numbers 4ᵃ(8b + 7) ×1
- the answer satisfies the congruence the chain started from ×1
- the average count grows by between one and two per tenfold enlargement ×1
- the base point is on the circle ×1
- the bound is attained only at one interior point ×1
- the bound the count runs to is a whole number between 2000 and 200000 ×1
- the box misses exactly the eight numbers the large searches were needed for ×1
- the box of side 5,000 reaches between 85 and 92 per cent ×1
- the box the polygons are drawn from is a whole number between 3 and 4 ×1
- the centre of a square cell is at squared distance 1/2 from the corners ×1
- the chain ends on a lattice point ×1
- the circle of squared radius one carries four points ×1
- the count times √(log x) over x is near Landau's constant at the top of the range ×1
- the counts dip beside the slope 1/3 ×1
- the disc of radius one catches five points ×1
- the divisors of everything up to six add to fourteen ×1
- the dual of one of the sixteen is another of the sixteen ×1
- the error averages out to nought ×1
- the error divided by the cube root is the tamer ratio ×1
- the error divided by the two-thirds power is the tamer ratio ×1
- the error is not bounded by twice the square root over this range ×1
- the factorisation test and the circle agree about every number up to two hundred ×1
- the first remainder below the square root, with its partner, squares and adds to the prime ×1
- the identity as stated gives the wrong area for a region with a hole ×1
- the largest bound counted is a whole number between 200 and 4000 ×1
- the largest gap shrinks from the smallest bound ×1
- the largest prime to test is a whole number between 10 and 200 ×1
- the largest radius counted is a whole number between 40 and 400 ×1
- the largest squared radius drawn is a whole number between 12 and 64 ×1
- the last number is a whole number between 59 and 99999 ×1
- the lattice half-width is a whole number between 1 and 16 ×1
- the leading coefficient is the volume for every height drawn ×1
- the measured growth of the largest error matches the known orders ×1
- the new point is on the same circle ×1
- the number is a whole number between 1 and 200 ×1
- the number is prime ×1
- the number of cells is a whole number between 60 and 400 ×1
- the polygon has between 3 and 16 whole-numbered corners and some area ×1
- the polygon is one of blob, triangle, comb, thin ×1
- the prime is one more than a multiple of four ×1
- the prime test and a direct search agree ×1
- the prime the lattice is built from is a whole number between 5 and 200 ×1
- the prime the two squares are found for is a whole number between 5 and 4001 ×1
- the remainder after the largest square is at most 2a ×1
- the second differences are twice the area ×1
- the share of Gaussian integers near radius r that are prime is about 4/(π ln r²) ×1
- the shortest vector of the lattice has length squared exactly the prime ×1
- the slices add up to the count from sums of three squares ×1
- the slopes are proper positive fractions ×1
- the solution found satisfies the equation ×1
- the starting point is on the circle ×1
- the sweep found a substantial family ×1
- the sweep grid is between 3 and 6 points across ×1
- the three-square count equals Gauss's class-number formula ×1
- the view is one the family draws ×1
- the volumes differ while the counts do not ×1
- there are sixteen classes to pair up ×1
- there are sixteen lattice polygons with a single interior point ×1
- three cubes reach every remainder but 4 and 5 ×1
- twice the area is a whole number, which is why halves are the only fractions in the formula ×1
- upTo names a range and is read only by the gaussian, count and density views ×1
- while a polygon with no interior point can exceed the bound, which is why it needs one ×1
- wide sectors fluctuate far less than chance and thin ones nearly as much ×1
- with steps of √10 the farthest prime reached is at distance 1,024.4 ×1
- with steps of √18 the walk reaches the edge of a disc of radius 4,500 ×1
- with steps of length at most 2 the walk from 1 + i is trapped, its farthest prime 17 + 42i ×1
Where it is called
Every figure on this list is drawn by the same rule, so a change to the rule changes all of them at once. That is why the list is published.
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.
NumberA tree that holds every triple
Three fixed matrices, applied to 3-4-5 over and over, produce every primitive Pythagorean triple there is — each of them once, none of them twice, and with no test for common factors anywhere in the procedure.
NumberA 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.
NumberAlmost no number is one
Sums of two squares look common — a sixth of all numbers up to a million are one. The fraction is falling to nothing, at a rate so slow that no computation will ever make it obvious, and the constant in front of it has been computed to fifty places and identified with nothing.
DiscreteArea by counting dots
Draw a polygon with every corner on a grid of dots. Count the dots strictly inside, add half the dots on the edge, subtract one — and the answer is the area, exactly, with no measuring anywhere.
NumberEvery triple, on one circle
Draw a line of rational slope through a single point of a circle. Wherever it comes out is a rational point, and clearing the denominators turns it into a Pythagorean triple — so every triple there is comes from one line through one point.
AlgebraOne point in every big enough shape
A determinant measures a lattice, not the basis that happened to describe it — and that measurement is an exchange rate. Any symmetric convex region with more than four times that area has to swallow a lattice point.
NumberOne solution that makes all the others
The equation x² − 2y² = 1 has infinitely many whole-number solutions, and every one of them is a power of the smallest. The multiplication that produces them is what multiplying two numbers of the form a + b√2 comes to when the √2 terms are collected — so an equation about a hyperbola turns out to carry a group.
NumberOne way to factor, and no other
Every number breaks into primes in exactly one way. That is so familiar it is hard to see as a claim at all — until it is put beside an arithmetic where it is false, and where six has two different factorisations that cannot be reconciled.
NumberSectors that shrink with the primes
Hecke proved that the angles of the primes a² + b² are evenly spread, so every fixed sector eventually gets its share. Let the sector shrink as the primes grow, to width X^(−α), and the question is open beyond small α. Measured on the 581,517 such primes between twenty and forty million, wide shrinking sectors fluctuate far less than chance and thin ones as much as chance — and the thin sectors that go empty are not scattered: they gather beside the directions of small rational slope, most of all slopes like 1/3 and 1/1 whose two terms are both odd.
DiscreteSixteen polygons with one dot inside
Fix one of Pick's two counts at one and ask what is left. The answer is a finite list, the list has exactly sixteen entries, each one is its own kind of object with a dual that is another entry, and the whole classification is a search a page can carry out.
DiscreteThe dots a ball catches
Count the lattice points inside a ball and compare with its volume. In five dimensions the error is about the size of one spherical shell, and that is a theorem; in four a logarithm sneaks in; in three and two the true size of the error is unknown. The difficulty runs backwards because the arithmetic of sums of squares is most erratic when there are fewest squares to add.
DiscreteThe dots a circle catches
Pick's theorem gives a lattice polygon's area exactly, with no error term anywhere. Ask a circle the same question and the exactness is gone: the count is the area plus something, the something has been measured for two centuries, and nobody knows how big it is.
AlgebraThe identity that multiplies sums of squares
A sum of two squares times a sum of two squares is a sum of two squares, and the same is true of four and of eight. It is true of three of nothing: 3 and 5 are each a sum of three squares, 15 is not, and one small pair of numbers rules the case out forever.
AlgebraThe integers among the quaternions
The obvious integer quaternions are the ones with whole coordinates, and they are the wrong ones. Sixteen more, with every coordinate a half, have to be let in — and with them comes a division algorithm, twenty-four units instead of eight, and a regular solid that exists only in four dimensions.
DiscreteThe theorem that has no version in space
A lattice polygon's area is decided completely by two counts of dots. The obvious guess is that a lattice solid's volume is decided by the same two counts in three dimensions, and there is a family of tetrahedra with identical counts and every volume that says otherwise.
NumberThe two squares actually produced
Three proofs say a prime one more than a multiple of four is a sum of two squares, and not one of them hands over the squares. Running the Euclidean algorithm half-way does — and where to stop is the whole of the correctness argument.
NumberThe wait for the next sum of two squares
Sums of two squares thin out to a share of nought, and yet the gaps between them stay short. Take the largest square below any number; what is left over is small, and a small number is always close to a square. Two squarings in a row say the next sum of two squares is never more than about 2√2 times the fourth root of n away — a bound proved in 1947 that nobody has improved, sitting far above every gap anyone has found.
NumberThirty-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.
NumberTwo squares, and a lattice
Whether a prime is the sum of two squares is decided entirely by its remainder on division by four. A fact about circles is settled by a fact about remainders, and neither statement contains any hint of the other.
AlgebraUnits that form a lattice
In the whole numbers only 1 and −1 have whole-number reciprocals. In the integers of a bigger field there can be infinitely many such units, and they are not scattered: take logarithms of their sizes under each way of placing the field in the real or complex numbers, and the units land exactly on a lattice. How many dimensions that lattice has is a count of those placements, and the area of its cell is a number no formula gives.
NumberWhich primes a form takes
A prime is the sum of two squares exactly when it is 1 modulo 4. Change the form slightly, to x² + 27y², and no congruence on p decides it at all — which is where the elementary subject ends and its successor begins.
NumberWhich way a prime's two squares point
A prime one more than a multiple of four is a sum of two squares in exactly one way, and the two squares make a point on a circle. Draw that point for every such prime and ask which way it faces. It faces every way equally — the angles spread evenly over the sector, which Hecke proved by giving each angle a remainder and copying Dirichlet. Measured over seventy-four thousand primes, they are even more evenly spread than random angles would be.