Sums of two squares — the series

1. 

   Two 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.

   part 1 · number
2. 

   The 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.

   part 2 · number
3. 

   Almost 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.

   part 3 · number
4. 

   The 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.

   part 4 · number
5. 

   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.

   part 5 · number
6. 

   Which 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.

   part 6 · number
7. 

   Sectors 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.

   part 7 · number
8. 

   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.

   part 8 · number
9. 

   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.

   part 9 · number
10. 

    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.

    part 10 · number
11. 

    Nineteen fourth powers, and then sixteen

    Edward Waring claimed in 1770 that every number is a sum of at most nineteen fourth powers. It is true, and it is needed: 79 takes nineteen, because its remainder on division by sixteen forces fifteen ones. But only seven numbers need nineteen, the last of them 559; the last to need eighteen is 1,248, and the last to need seventeen is 13,792. Counting every number to ten million shows the stragglers thinning out and the one obstruction that never goes away — the remainder modulo sixteen, which makes 31, 496, 7,936 and every 31·16ᵐ need sixteen for ever.

    part 11 · number

All series