A fraction on the circle forces a whole point

The number 2 is a sum of two squares: $1^2 + 1^2$. It is also a sum of two squares of fractions in infinitely many ways — $(1/5)^2 + (7/5)^2 = 50/25$, $(7/13)^2 + (17/13)^2 = 338/169$, and so on. Is the converse true? If some number $n$ can be written as $(a/c)^2 + (b/c)^2$, with fractions, can it always be written with whole numbers?

It can, and there are two ways to see it. One goes through primes: clear the denominators and compare what the rule about primes says about $n$ and about $nc^2$. The other goes through a picture, and it is one of the most economical arguments in elementary number theory. It was found by Léon Aubry in 1912, rediscovered by Harold Davenport and J. W. S. Cassels, and it uses nothing about primes at all: only that every point of the plane lies within distance one of some point of the integer lattice.

Rational points everywhere, lattice points rarely #

A sum of two squares of fractions is a rational point on a circle: $x^2 + y^2 = n$ with $x$ and $y$ rational. For the unit circle these are plentiful.

The picture is the construction of every Pythagorean triple: a line of rational slope through one rational point of a conic meets it again at another rational point, because the second intersection solves a quadratic whose other root is already rational. On the unit circle this fills the circle densely with rational points, and among them the whole-number points are a mere four.

So a circle with one rational point has infinitely many, and the question is not whether the rational points are rare — they are not — but whether the presence of any rational point forces the presence of a lattice point. For $n = 3$ there is no rational point at all, and no lattice point; for $n = 2$ there are infinitely many of the first and four of the second. The claim is that the two situations are the only ones: a circle $x^2 + y^2 = n$ has either no rational points, or some lattice points.

The step that lowers the denominator #

Start with a rational point $P$ on the circle $x^2 + y^2 = n$, written with a common denominator: $P = (x/d,\ y/d)$ in lowest terms, with $d > 1$. Let $Q$ be the lattice point nearest to $P$. Draw the line through $P$ and $Q$, and let $P'$ be its second intersection with the circle.

Three things are true of $P'$, and the figure checks each of them in exact rational arithmetic.

$P'$ is rational. The line through $P$ and $Q$ has rational slope, since both points are rational, and it meets the circle at $P$; the other intersection is the second root of a quadratic with rational coefficients whose first root is rational, so it is rational too. This is the same fact that drew the unit circle’s rational points.

$P'$ is on the same circle, by construction.

Its denominator is smaller. Write $v = Q - P$ for the short step to the lattice point. Working out the second intersection gives an explicit formula, and from it the new denominator divides $d \cdot |v|^2$ — which is a whole number, because

$$d\,|v|^2 = d\,|Q|^2 - 2\,Q \cdot (x, y) + \frac{x^2 + y^2}{d} = d\,|Q|^2 - 2\,Q\cdot(x,y) + n\,d,$$

using $x^2 + y^2 = n d^2$. And since $Q$ is the nearest lattice point, $|v|^2 < 1$: the new denominator is a whole number strictly smaller than $d$.

So each step lowers the denominator, and a strictly decreasing sequence of positive whole numbers must stop. It stops only at denominator 1 — a lattice point on the circle. That is the whole proof. In the figure the denominators fall $3757 \to 205 \to 25 \to 5 \to 1$, the squared distances to the nearest lattice points being about 0.055, 0.12, 0.2 and 0.2.

Why the nearest lattice point, and not any #

The choice of $Q$ matters, and it is the only choice the argument makes. The formula for the new denominator holds for any lattice point $Q$: the second intersection of the line $PQ$ with the circle is always rational, and its denominator always divides $d\,|P - Q|^2$. What changes with $Q$ is only the size of $|P - Q|^2$, and the argument needs it below one. Choose a lattice point farther away and the new denominator can be larger than the old; the construction still produces a rational point, but it climbs instead of descending.

So the argument is really a statement about the lattice’s covering radius — the largest distance from any point of the plane to the nearest lattice point. For the square lattice it is $\sqrt 2/2 \approx 0.71$, comfortably below one. That number is a geometric invariant of the lattice, of the same family as the determinant that decides when a convex region must contain a lattice point: Minkowski’s theorem asks how large a region must be to be sure of catching a lattice point, and the covering radius asks how large a disc round every point must be to be sure of the same thing.

A chain on the smallest circle #

The same thing happens on the circle of radius $\sqrt 2$, where the lattice points are the four corners $(\pm 1, \pm 1)$.

The starting point here was manufactured: $(1, 1)$ turned through the angles of the 3–4–5 and 5–12–13 triangles, which multiplies the denominator by 5 and then by 13. The descent does not undo those turns. It goes $65 \to 17 \to 5 \to 1$ and arrives at $(1, -1)$ rather than back at $(1, 1)$ — the chain is determined by the geometry of the nearest lattice points, not by the history of the point.

The small circle also shows the steps at their most cramped. The circle of radius $\sqrt 2$ passes through four lattice points and encloses only the origin, so every rational point on it is near one of the four corners or near the origin, and its nearest lattice point is always one of those five. The reflections therefore zigzag across a very small region, and the drawing has to show several chords almost on top of one another. On larger circles, where many lattice points lie close to the curve, the steps spread out and each can be followed separately — which is why the circle of radius $\sqrt{17}$ was chosen for the first picture.

Two features of the argument are worth separating. The first is a descent: a quantity that is a positive whole number and strictly decreases, which is Fermat’s method for proving that things cannot go on for ever, run here to prove that something does exist. The second is a covering: the proof needs a lattice point within distance less than one of every rational point, and it uses nothing else about the lattice.

Six chains, and what they have in common #

The descent can be run from any rational point on any such circle.

Every row falls, and falls fast. The step multiplies the denominator by the squared distance to the nearest lattice point, which is at most $1/2$ for the square lattice — so each step at least halves it, and a denominator $d$ needs at most $\log_2 d$ steps. The chain from $3757$ on the circle of seventeen took four steps; the one from $3757$ on the circle of sixty-five took two. In that last row the chain started from a point manufactured out of $4^2 + 7^2$ and landed on $8^2 + 1^2$: sixty-five is a sum of two squares in two essentially different ways, and the descent is free to find either. It keeps no memory of where the point came from.

That freedom is a reminder of what the argument does and does not produce. It produces a lattice point, not a particular one. If the goal is every representation, the right tool is still the Gaussian integers and the factorisation of $n$; the descent answers only the question of existence, and answers it without factoring anything.

The one hypothesis, and where it fails #

The argument used a single property of the lattice and the circle together: for every point of the plane, some lattice point is at squared distance strictly less than one, measured in the same quantity $x^2 + y^2$ that defines the circle.

For $x^2 + y^2$ the worst point is the centre of a square, at squared distance $1/2$. For $x^2 + y^2 + z^2$ it is the centre of a cube, at $3/4$ — still less than one, so the argument works for sums of three squares too. For $x^2 + 2y^2$ the worst point is at $1/4 + 2/4 = 3/4$, and the argument works. For $x^2 + 3y^2$ the worst is $1/4 + 3/4 = 1$ exactly, and the descent can stall: a step that multiplies the denominator by one does not lower it. For four squares the worst point is at $4/4 = 1$ too.

The general statement is the Davenport–Cassels lemma: if a positive definite quadratic form with integer coefficients has the property that every rational point lies at form-distance less than one from some lattice point, then any whole number it represents with rationals it represents with integers. The condition is sufficient, not necessary. Four squares represent every whole number anyway, by Lagrange’s theorem, so for them the conclusion holds even though the lemma does not apply.

The proof it replaces #

The prime-by-prime proof is worth setting beside the picture, because the contrast is the point.

Suppose $n c^2 = a^2 + b^2$ with whole numbers. The right side is a sum of two squares, so by the rule about primes every prime of the form $4k + 3$ divides $nc^2$ to an even power. The factor $c^2$ contributes even powers of everything. So every such prime divides $n$ to an even power, and by the same rule $n$ is a sum of two squares of whole numbers. Three lines — but each of them leans on a substantial theorem: Fermat’s characterisation of the primes that are sums of two squares, unique factorisation, and the multiplicativity of sums of two squares.

The descent uses none of that. It never mentions a prime, never factors anything, and works verbatim in settings where no rule about primes is available. That is why it matters beyond this one fact: it is how rational representations are turned into integer ones in the proof of Gauss’s three-square theorem, the statement that a whole number is a sum of three squares unless it has the form $4^a(8b + 7)$ — the theorem behind the three triangular numbers. The hard part of that proof is showing that such $n$ are sums of three squares of rationals, which follows from a theorem about quadratic forms over the rationals and over every $p$-adic field; the step from rationals to integers is then Aubry’s descent in three dimensions, where the worst point of the cubic lattice is at $3/4 < 1$. Jean-Pierre Serre’s A Course in Arithmetic proves the three-square theorem exactly this way, and the lattice-covering step takes half a page.

What the descent says about the unit circle #

The unit circle is a pleasant special case. Every rational point on it — the whole dense family of the first figure — can be walked down, by nearest-lattice-point reflections, to one of just four points: $(\pm 1, 0)$ or $(0, \pm 1)$. The Pythagorean triples, which correspond to those rational points, are thereby all connected to the four trivial ones by chains of reflections through lattice points.

It also says something about where the Pythagorean triples come from that the parametrisation does not. The parametrisation builds every rational point from one of them by a line of rational slope, which is a construction outward from a single point. The descent works inward from any point, and it arrives at a lattice point by a route that depends on the point — so the rational points are organised into basins, one for each of the four lattice points, according to where the chain from them ends.

There is a classical cousin of this. Every primitive Pythagorean triple can be reached from $(3, 4, 5)$ by repeatedly applying three fixed matrices — the tree of Pythagorean triples found by Berggren in 1934 and rediscovered several times since — and running the tree backwards is a descent that ends at $(3, 4, 5)$, or in degenerate form at $(1, 0, 1)$. The two descents are not identical, but they share the principle: a rational point is carried to a simpler one by a map with integer coefficients, and simplicity is measured by a denominator that cannot fall for ever.

What the figures leave unproved #

The starting points were built, not found. Each chain starts from a lattice point turned through rational angles, so its conclusion — that the chain ends on the lattice — was guaranteed from the start. The figures demonstrate the mechanism on those points; the theorem is that it works from every rational point, which is the inequality $|P - Q|^2 < 1$ and not something a finite list of chains can show.

The squared distance is not always the new denominator. The new denominator divides $d\,|P - Q|^2$ and can be smaller still after reducing to lowest terms. The figures report the reduced denominators, which is why the ratios between consecutive entries in a row are not exactly the squared distances drawn.

The discs in the covering picture are drawn, not checked. That every point of the plane is within $\sqrt 2/2$ of the lattice is a one-line computation — the farthest point from the four corners of a unit square is its centre — and the figure checks that number rather than sampling the plane. The picture is an illustration of the covering; the proof of it is the computation.

The right-hand covering picture shows a borderline, not a counterexample. For $x^2 + 3y^2$ the argument as stated stalls on points at exactly distance one, and the figure makes no claim that the conclusion fails for that form; whether it does is a separate question with a separate answer.

Still open: which forms turn fractions into whole numbers #

The property the descent proves — represented by rationals implies represented by integers — makes sense for any positive definite form, and for most forms it is false. The obstruction is that a number can be represented rationally, and locally at every prime, by several inequivalent integer forms at once; they make up a genus, and a rational representation says only that some form in the genus represents the number, not which. When a genus contains a single class of forms, rational representability and the local conditions pin the answer down.

Which genera contain a single class is a finite question in each dimension beyond the smallest, and the lists have been computed — in three variables they run to hundreds of genera — but they are lists, found by computer, and no structural explanation predicts them. Aubry’s covering condition picks out a small, clean subset of the forms on those lists, the ones where the lattice is dense enough in its own geometry. Why the property holds for the others, form by form, is known only case by case.

For two variables the story is cleaner and older. The forms $x^2 + ny^2$ for which the primes represented are described by congruence conditions alone are exactly those whose genus has one class, and Euler found sixty-five such $n$ — his idoneal numbers — the largest being 1848. Which primes such a form takes is then a matter of remainders, just as it is for $x^2 + y^2$. Whether Euler’s list is complete is not known: it is complete apart from possibly one more value, and ruling that last value out would need a case of the generalised Riemann hypothesis.

A lattice close enough to every point #

The proof has no arithmetic in it beyond one inequality. A rational point on the circle is never far from the lattice, and being close is enough: the line from the point to its nearest lattice neighbour cuts the circle again at a simpler point, and simplicity cannot decrease for ever. The primes, which govern which circles have lattice points at all, are never consulted — the descent reaches the same conclusion from a fact about how tightly unit discs pack round the lattice, which is a fact about geometry rather than about numbers.

The two proofs of the same fact sit at opposite ends of the subject. One is arithmetic through and through — primes, their remainders, the identity that multiplies sums of squares — and needs the whole theory of which primes split. The other is a covering inequality and a descent, and needs nothing but the lattice. That the second exists at all, and that it generalises to three squares where the first does not easily go, is the reason it is remembered. A number theorist reaching for it is usually in exactly that position: holding a rational solution, produced by some local-to-global principle, and needing a whole one.