The dots a circle catches

Pick’s theorem is exact. A lattice polygon’s area is its interior count plus half its boundary count minus one, with no error term, no approximation and no hypothesis beyond the corners being on the lattice.

That exactness comes entirely from the corners. A polygon’s boundary is a union of segments between lattice points, and every one of the theorem’s ingredients is a count of things the boundary passes through in a controlled way. Replace the polygon by a circle and every ingredient fails at once: the boundary passes through the lattice at no points at all, or at a few, in a way that depends on arithmetic rather than on geometry.

What survives is an approximation. The number of lattice points inside a circle of radius $r$ is about $\pi r^2$, which is obvious, and the interest is entirely in the word about.

What is known, and the gap between the two halves #

Write $N(r)$ for the count and $E(r)$ for the difference $N(r) - \pi r^2$. Two things are known and they do not meet.

Gauss, 1837: the error is at most a constant times $r$. The argument is one line. Give each lattice point a unit square centred on it; the squares tile the plane; every square meeting the disc of radius $r$ lies inside the disc of radius $r + \sqrt{2}/2$ and every square inside the disc of radius $r - \sqrt{2}/2$ meets it. So the count is between the areas of two discs whose radii differ by a constant, and the error is at most the area of the annulus between them, which is proportional to $r$.

That bound has been improved many times — to $r^{2/3}$ by Sierpiński in 1906, and by increments since to about $r^{0.6297}$ — and each improvement is a substantial piece of analysis.

And the error is at least $r^{1/2}$, infinitely often. That is Hardy’s, from 1915, and it is a lower bound: the error genuinely oscillates at that size, so no proof will ever push the exponent below a half.

Nobody knows which end is the truth. The conjecture is that a half is right, up to an arbitrarily small amount, and it is one of the oldest open questions in the subject.

The figure measures both claims on its own range. Dividing the largest error by $r^{1/2}$ leaves a ratio reaching 5.31; dividing by $r^{2/3}$ leaves 2.60. Neither ratio settles anything, and their behaviour over a longer range is the whole question.

One radius, worked #

The abstraction is easier to hold after one concrete count.

At radius 160 the disc contains 80381 lattice points and its area is 80424.8, so the error is about $-43.8$ — the count falls short of the area by about forty-four points out of eighty thousand, which is a relative error of one part in two thousand. Over the whole range drawn the error never exceeds 53.8 in absolute value.

Those numbers are worth holding beside the two candidate exponents. The square root of 160 is about 12.6, and the two-thirds power about 29.6; the largest error observed is a small multiple of either. Both candidates are entirely plausible at this scale, and that is the point — the range a figure can carry is the range where the question looks settled and is not.

The relative error is what makes the main term useful and the absolute error is what makes the problem hard. Anybody wanting to know roughly how many lattice points a large disc holds has known the answer since antiquity. The question here is a question about the second digit of a quantity nobody needs, and it is asked because the answer would say something about cancellation in arithmetic sums that a great many other problems also need.

Why the lower bound is the harder half #

The two sides of the question are not symmetric, and the asymmetry is instructive.

Proving the error is small means proving that a sum of oscillating terms cancels — a positive statement about conspiracy failing to happen, established by estimating each piece and adding up. That is difficult and it is the kind of difficulty analysis is built for, which is why the exponent has come down steadily from 1 to 2/3 to 0.6297 over a century.

Proving the error is large, infinitely often, means exhibiting radii where the conspiracy does happen. Hardy’s argument does this without producing a single such radius: it shows that if the error were always smaller than $r^{1/2}$ then a certain average would be too small, and the average can be computed. An existence proof with no example, of the same kind as a colouring nobody has ever seen.

That asymmetry explains why the conjectured answer is the lower bound rather than something in the middle. The lower bound is a fact about the oscillation’s true size; every upper bound is a limit on how well anybody can currently show the oscillation cancels. The truth is at the bottom of the range and the work is all at the top, and the gap measures ignorance rather than uncertainty about the phenomenon.

Why the circle is hard and the polygon is not #

The comparison with Pick’s theorem is worth making carefully, because it locates the difficulty precisely.

A lattice polygon’s boundary meets the lattice in a controlled way: each edge from one lattice point to another passes through exactly as many further lattice points as the greatest common divisor of its two coordinate differences, and that is arithmetic anybody can do. The boundary’s interaction with the lattice is finite and computable, so the count is exact.

A circle’s boundary passes through a lattice point exactly when its squared radius is a sum of two squares, and how many such points there are is decided by the prime factorisation of that number.

So the count inside a disc is a sum of an erratic arithmetic function, and the error term measures how much that function’s fluctuations fail to cancel. The circle problem is not a geometry problem at all; it is a question about cancellation in a sum of arithmetic terms, and the exponents in it are the exponents that appear whenever such a sum is estimated.

The same question under a hyperbola #

There is a second problem with the same shape, the same two exponents, and the same gap, and the two have moved together for a century.

Count the lattice points under the hyperbola $xy = n$ in the positive quadrant. A point $(a, b)$ with $ab \le n$ corresponds to a divisor, so the count is the total number of divisors of every number up to $n$ — and Dirichlet showed in 1849 that it is about $n \log n + (2\gamma - 1)n$, with an error he bounded by the square root.

The two problems’ exponents differ by exactly a factor of two — a half against a quarter, two-thirds against a third — because the hyperbola’s count is a sum over a region of area $n$ where the circle’s is over a region of area $r^2$. Every method that has improved one exponent has improved the other, in the same year and by the same person, and the current records are 0.6297 for the circle and 0.3149 for the divisors, which is the same number halved.

Two problems that have never been separated by any technique are, for practical purposes, one problem, and the fact that neither has yielded is the strongest evidence available that the obstruction is real rather than a failure of ingenuity.

What the sum is really doing #

There is a formula for the error, and it explains both the half and the two-thirds.

The count inside a disc, written as a sum over the lattice, has an expansion — Hardy’s, from the same 1915 paper — as a sum of Bessel functions indexed by the ways of writing whole numbers as sums of two squares. Each term oscillates; each has size about $r^{1/2}$ divided by a power of its index; and the error is what those oscillations add to.

If the terms were independent the sum of a great many oscillations of size $r^{1/2}$ would be about $r^{1/2}$, which is the conjecture. They are not independent, they conspire occasionally, and every known proof of an upper bound is a way of bounding the conspiracy. The gap between $1/2$ and $0.6297$ is the amount of conspiracy nobody can rule out.

That framing is worth having because it says what a solution would look like. It would not be a better estimate of anything geometric. It would be a statement that a particular arithmetic sum does not conspire — the same kind of statement as the prime number theorem’s error term, and hard for the same reason.

What a better bound would buy #

It is fair to ask why anybody wants the exponent, given that the main term has been good enough for two hundred years, and the answer is that the circle problem is a test case rather than an application.

The sum being estimated is a sum of an arithmetic function over a range, with a smooth main term subtracted. That description fits a great many problems: the distribution of the primes, the number of squarefree integers below a bound, the number of lattice points under any smooth curve, the size of class numbers. In every one of them the difficulty is the same difficulty and the exponents that appear are the same exponents.

The circle problem is the cleanest member of the family — the geometry is trivial, the arithmetic function is completely understood, and the only thing left is the cancellation — so it functions as the place where a new method is tried first. The exponent record is therefore a scoreboard for a technique rather than a fact anybody needs, and the several improvements since Sierpiński each introduced a method that then went elsewhere.

That is what makes an old open problem worth keeping open. It is not that the answer matters; it is that the problem is stripped of everything except the obstruction, so a method that moves it has moved the obstruction rather than worked around it.

What can be said about the shape of the error #

The figure’s oscillation looks structureless and is not, and two of its features are worth naming.

It is not symmetric about zero in the short run. Over the range drawn, the error spends more time negative than positive, which is a real effect: the count at radius $r$ includes the points on the circle itself, and the average of that count over $r$ contributes a term the main term does not include.

And the extremes are attained at special radii. The largest excursions happen where the circle passes through many lattice points at once — squared radii with many divisors that are 1 mod 4 — which is again the arithmetic showing through. A picture of the error is a picture of the arithmetic function’s partial sums, and the geometry contributes only the smooth part that has been subtracted off.

There is a third feature and it is a warning. The error’s size grows so slowly that on a linear plot over a short range it looks like noise about zero with no trend at all — and it is not noise; every value is determined exactly by arithmetic, and the same radius always gives the same answer. A deterministic sequence that passes for random is the hardest kind of object to reason about, and it is why every argument here is about averages rather than about individual radii.

Both observations are stable over the range measured and neither is a theorem. That is the ordinary condition of this problem: everything anybody has noticed about the error is either proved and weak, or observed and unproved.

The three-dimensional surprise #

One more comparison, because it is the one that makes the difficulty look strange rather than merely hard.

Ask the same question of a sphere: how many lattice points inside a ball of radius $r$, against its volume. In dimensions four and above the answer is essentially settled — the error is of the order of $r^{d-2}$, and the proofs are not especially difficult, because the number of ways of writing a number as a sum of four or more squares is regular enough to control.

In three dimensions it is open, and in two it is open, and the reason in both cases is the same: the number of representations as a sum of two or three squares is erratic in a way that four squares is not.

So the problem is hard in low dimensions and easy in high ones, which is the reverse of the usual arrangement, and it says clearly where the difficulty lives. It is not in the geometry, which is simplest in the plane. It is in the arithmetic of sums of squares, which is most erratic when there are fewest squares to sum.

That is the same reason Pick’s theorem fails in space while the plane case is exact — the plane is where the lattice and the region interact most tightly — and the two failures point in opposite directions from the same fact.

What the pictures cannot show #

The measurements run to radius 160 and to 1200 for the divisors, which are small ranges for a question about exponents. An exponent of a half and an exponent of 0.6297 differ by a factor of about two at radius 160, which is well within the noise of a curve that oscillates. No range a figure can carry distinguishes the two candidates, and the essay says so rather than inviting a reader to squint.

The upper bounds quoted — Sierpiński’s two-thirds and the current record — are stated with no argument. Both are genuine pieces of analysis using exponential sum estimates, and neither has a picture.

And the Bessel expansion is described and not computed. Drawing its first few terms against the measured error would be a good figure and a long one, and it is not here.

Where the ladder goes next #

This rung closes the ladder: counting dots exactly for a polygon, the failure in space, the classification when one count is fixed, and now what happens when the boundary stops being made of lattice segments.

Named here as debts. The Bessel expansion, drawn against the measured error, which would show the oscillation being assembled. And the higher-dimensional versions, where the sphere’s error term is settled in four dimensions and above and open in three — the reverse of the usual pattern, and worth a rung for that reason alone.

Sideways, the exact theorem this abandons is Pick’s, the arithmetic driving the error is which numbers are sums of two squares, the classification at fixed interior count is the rung below, and the analogous error term for the primes is the one Riemann’s function estimates.

What is worth carrying away #

An exact formula and a good approximation are different kinds of object, and which one a problem admits is decided by how its boundary meets the structure being counted.

Pick’s theorem is exact because a lattice polygon’s edges meet the lattice in a way arithmetic controls completely. The circle problem has an error term because a circle’s boundary meets the lattice according to a factorisation, and no geometric argument reaches a factorisation.

The habit worth taking is to ask what the boundary is made of. A count over a region is the region’s area plus a boundary effect, and every difficulty in these problems is in the second term — so the useful first question is whether the boundary is a finite combinatorial object or a curve that has to be approximated.

The corollary is about reading a measurement of an error term. Two centuries of computation have not distinguished the candidate exponents here, and no amount more will: the difference between $r^{1/2}$ and $r^{0.63}$ is a factor of four at a radius of a million, against oscillations of the same order. A quantity whose exponent is in doubt cannot be settled by computing it, which is worth knowing before starting.