The Dots in a Circle, and the Wobble No One Can Pin Down
How many ways can a number be a sum of two squares? Its divisors decide. Add those counts across a whole disc and you get πR², plus an error that has stayed open for nearly two centuries.
> The last drop asked whether a number is a sum of two squares. This one asks how many ways, counting signs and order, and the answer turns out to be fixed entirely by the number's divisors. Add those counts up over a whole disc and you are counting lattice points inside a circle, a quantity Gauss showed is πR² to a very good approximation. How good, exactly, is a question no one has managed to answer in almost two centuries.
01 · the count from the divisors
Count the ways n is a sum of two squares with signs and order, r₂(n). It is not random: it is four times the surplus of n’s divisors that are ≡ 1 mod 4 over those that are ≡ 3.
The count of representations r₂(n), taken with all four sign choices and both orders, is not random: it is four times the number of divisors of n that are one-more-than-a-multiple-of-four, minus those that are one-less. That is Jacobi's two-square theorem (1834). Divide each divisor by four, tally the ≡1 class against the ≡3 class, multiply the surplus by four, and you have counted the lattice points on the circle without hunting for a single one. When the ≡3 divisors exactly balance the ≡1 divisors the count is zero and the circle threads no integer point at all.
02 · fill the disc
Stop asking about one circle. Count every lattice point inside a disc of radius R, and race the count against the disc’s area, πR².
| R | points N(R) | area πR² | gap E(R) |
|---|---|---|---|
| 10 | 317 | 314.2 | +2.8 |
| 100 | 31,417 | 31,415.9 | +1.1 |
| 1,000 | 3,141,549 | 3,141,592.7 | −43.7 |
Now stop asking about one circle and fill a whole disc. Count every lattice point (x, y) with x² + y² ≤ R². Gauss's insight: each point owns a unit square, the squares tile the disc, so the count is very nearly the disc's area, πR². The running total is exactly 1 (for the origin) plus the sum of r₂(n) over every n up to R², so the divisor counting from console 01 is doing the work under the hood. Dial R and watch the point count chase πR², always close, never landing.
03 · the wobble no one can pin down
The count is always near πR², never exactly on it. How fast the gap E(R) can grow is a question open since Gauss, trapped between two exponents no one has managed to bring together.
Each curve is the largest error seen out to radius R, divided by a power of R, measured live to R = 2,000. Dividing by √R the curve climbs (2.9 → 7.0): the error genuinely outgrows √R, exactly as Hardy and Landau proved. Dividing by R it collapses (0.76 → 0.16): far under Gauss’s trivial ceiling. And dividing by R^0.6298 it sits nearly flat (2.1 → 2.6), hugging the proven ceiling. The true exponent is somewhere in between, and no finite plot like this one can ever name it. That is the open problem.
| who | exponent | note |
|---|---|---|
| Gauss, 1837 | ≤ 1 | the rim has length ~2πR |
| Sierpiński, 1906 | ≤ 2/3 | ≈ 0.6667 |
| Huxley, 2003 | ≤ 131/208 | ≈ 0.6298 — the record |
| Hardy–Landau, 1915 | ≥ 1/2 | the error must exceed √R |
| Hardy (conjecture) | = 1/2 | believed, still open |
Write the error as E(R) = N(R) − πR², the gap between the honest point count and the smooth area. How fast can it grow? Gauss saw at once it is no bigger than the boundary is long, roughly R, because only the points straddling the rim are in doubt. That was 1837. Everyone since has been pushing the exponent down: Sierpiński reached 2/3 in 1906; the record, Huxley in 2003, is 131/208 ≈ 0.6298. From below, Hardy and Landau proved in 1915 that the error genuinely exceeds √R infinitely often, so the exponent cannot drop under 1/2. Hardy conjectured 1/2 is the whole truth. The gap between 0.5 and 0.6298 is the Gauss circle problem, open since it was posed. The three curves above are the same live measurement seen through three lenses, and the honest lesson is that the one that comes closest to holding the error level sits near the proven ceiling, yet no plot at any radius you can compute will ever pin the exponent down. That is why it is still open.
The last question, counted
A previous drop asked a yes-or-no question: is a given number a sum of two squares? Thirteen is (it is 4 + 9); seven is not, and no search will ever turn one up. Fermat settled which numbers say yes with a single glance at the remainder mod 4.
This drop asks the follow-up that the yes-or-no framing hides: how many ways? Count them honestly, with signs and with order, so that 5 = 1² + 2² is not one representation but eight, the point (1, 2) together with (2, 1), (−1, 2), (2, −1), and the rest of the eight ways to plant a signed, ordered pair on the circle of radius √5. Write that count r₂(n). It looks like it should be as unruly as the primes. It is not. It is fixed, completely, by the number's divisors.
The divisors already know the answer
Here is the surprise, and it is exact. Look at the divisors of n and sort them by their remainder mod 4. Let d₁ be how many are ≡ 1 and d₃ how many are ≡ 3 (the even divisors are ignored). Then
r₂(n) = 4 · (d₁ − d₃).
That is Jacobi's two-square theorem (1834). Take n = 25: its divisors are 1, 5, 25, all ≡ 1 mod 4, so d₁ = 3, d₃ = 0, and r₂(25) = 12 — the four axis points (±5, 0), (0, ±5) plus the eight ways to arrange (3, 4). Take n = 45 = 9 · 5: divisors 1, 3, 5, 9, 15, 45, of which 1, 5, 9, 45 are ≡ 1 and 3, 15 are ≡ 3, so 4·(4 − 2) = 8. And take any number all of whose "odd part" is built from primes ≡ 3 mod 4 to odd powers — say 21 = 3 · 7 — and the two classes cancel exactly, d₁ − d₃ = 0, the count is zero, and the circle threads no lattice point. Fermat's yes-or-no rule is just the question of whether that difference is zero.
Console 01 lets you dial n and watch the divisors fall into their two columns while the circle of radius √n lights up every signed, ordered point it passes through. The two numbers always agree, because they are the same fact told twice.
Then fill the whole disc
Jacobi counts the points on one circle. Gauss asked the bigger question: how many lattice points lie inside a disc of radius R — every (x, y) with x² + y² ≤ R²?
His argument is one of those ideas that feels like cheating. Give every lattice point the unit square just below-and-left of it. Those squares tile the plane without gap or overlap, and the ones belonging to points inside the disc very nearly cover the disc itself. So the number of points is very nearly the disc's area:
N(R) ≈ πR².
And the two views are one: the count inside the disc is exactly 1 (the origin) plus the sum of r₂(n) over every whole number n from 1 up to R². The divisor bookkeeping from console 01, summed, is the area of a circle. Console 02 fills the disc for a dialable R and reads out the point count beside πR². They stay stubbornly close — a disc of radius 100 holds 31,417 points against an area of 31,415.9 — and yet the count never quite lands on the area. It wobbles.
The wobble no one can pin down
Write the wobble down as
E(R) = N(R) − πR²,
the gap between the honest integer count and the smooth area. The whole of the Gauss circle problem is the single question: how fast can E(R) grow?
Gauss himself gave the easy bound in 1837. Only the points straddling the rim are ever in doubt, and the rim has length about 2πR, so the error is at worst about R. Everyone since has been shaving the exponent down. Sierpiński got it to 2/3 in 1906. Van der Corput, then a long relay of analysts, chipped away through the century; the current record, Huxley in 2003, is 131/208 ≈ 0.6298. From the other side, Hardy and Landau proved in 1915 that the error genuinely does exceed √R infinitely often — so the exponent can never drop below 1/2. Hardy conjectured that 1/2 is the whole truth, that E(R) is O(R^(1/2+ε)) and no better.
That conjecture is still open. The true exponent is trapped somewhere in
[ 0.5 , 0.6298… ]
and nobody has closed the gap in more than a hundred years. Console 03 measures the peak error live — the largest |E(k)| out to each radius R — and then divides it by three powers of R to see which one keeps it level. Divide by √R and the curve climbs: the error really does outgrow √R, exactly as Hardy and Landau proved. Divide by R and it collapses toward zero: the error is far under Gauss's trivial ceiling. Divide by R^0.6298 and it sits nearly flat, hugging Huxley's proven bound. The exponent that would hold the error perfectly level is somewhere between the diverging ½ and the flat 0.6298 — and here is the honest part: no finite plot can ever name it. At any radius you can actually compute, the constants and slow logarithmic factors blur the true rate, which is precisely why a question this simple to state has resisted a definitive answer for so long.
Why it is hard, in one sentence
Counting points is arithmetic; the error is the arithmetic leaking into the geometry, the same divisor sums from console 01 refusing to average out as smoothly as an area should. The best modern bounds come from bounding exponential sums that encode exactly how lattice points cluster near a smooth curve — the same machinery behind estimates for the Riemann zeta function on the critical line. The circle problem and the zeros of zeta are cousins. Which is a long way of saying: a question a child can state — how many dots fit in a circle? — sits on top of some of the deepest unfinished business in the subject.
A sequel to the two-squares drop, built around Jacobi's counting formula and the Gauss circle problem. Everything is integer-exact and recomputed in your browser on load, no stored figures. Verified offline before shipping: Jacobi's r₂(n) = 4(d₁ − d₃) matches a direct signed-and-ordered lattice count for every n from 1 to 20,000 with zero mismatches; the disc count N(R) equals 1 + Σ r₂(n) exactly (checked at R = 10, 25, 50, 100). Console 03 avoids a fragile log-log slope fit, which is pre-asymptotic and range-unstable, and instead normalises the running-max error M(R) by three powers of R: measured live to R = 2,000, M(R)/√R rises from 2.9 to 7.0 (the error outgrows √R, per Hardy–Landau), M(R)/R falls from 0.76 to 0.16 (far under Gauss's trivial bound), and M(R)/R^(131/208) stays near 2.1–2.8 (hugging Huxley's proven ceiling). The exact exponent in [½, 0.6298] is the open problem, and no finite computation can decide it.