The areas a right triangle can have

Every Pythagorean triple sits on one circle: draw a line of rational slope through a point of the unit circle, and where it comes out is a rational point, which cleared of denominators is a triple. That essay ended with a theorem of Fermat’s — no right triangle with whole-number sides has an area that is a perfect square — and a question it opened. Which whole numbers are the area of some right triangle whose sides are rational?

Six is, because of the triangle with sides 3, 4 and 5. Five is too, but the smallest triangle that proves it has sides $\tfrac32$, $\tfrac{20}3$ and $\tfrac{41}6$. Seven needs $\tfrac{24}5$, $\tfrac{35}{12}$ and $\tfrac{337}{60}$. A whole number that is such an area is called congruent, a name that goes back to Fibonacci’s congruum of 1225, though Arab mathematicians had studied the question in another form in the tenth century. One, two and three are not congruent: no rational right triangle at all, however thin or strangely shaped, has area 1, 2 or 3.

The question of which numbers are congruent sounds like a puzzle about triangles. It turns out to be a question about rational points on a curve, and its best answer — a theorem of Jerrold Tunnell from 1983 — decides it by counting solutions of three quadratic equations, completely and correctly in one direction, and in the other direction only if one of the hardest open conjectures in mathematics is true.

Three squares in a row #

The name comes from a different but equivalent puzzle. In 1225, at the court of the emperor Frederick II, Johannes of Palermo set Leonardo of Pisa — Fibonacci — the problem of finding a rational square that stays a square when 5 is added to it and when 5 is subtracted. Fibonacci found one: $\left(\tfrac{41}{12}\right)^2$, since

$$\left(\tfrac{41}{12}\right)^2 - 5 = \left(\tfrac{31}{12}\right)^2, \qquad \left(\tfrac{41}{12}\right)^2 + 5 = \left(\tfrac{49}{12}\right)^2.$$

Three squares in arithmetic progression with common difference 5 — and he called such a difference a congruum. The connection to triangles is one identity: if $(a, b, c)$ is a right triangle of area $n$, then $(c/2)^2 \pm n = \left((a \pm b)/2\right)^2$, because $(a \pm b)^2 = c^2 \pm 4n$ once $a^2 + b^2 = c^2$ and $2ab = 4n$ are used. The triangle $(\tfrac32, \tfrac{20}3, \tfrac{41}6)$ of area 5 gives exactly Fibonacci’s three squares, and the $(3, 4, 5)$ triangle gives $(\tfrac12)^2$, $(\tfrac52)^2$ and $(\tfrac72)^2$, three squares a distance 6 apart. Conversely three rational squares with common difference $n$ give back a triangle of area $n$, so congruent numbers are exactly the possible common differences of three rational squares in arithmetic progression.

Fibonacci also stated, without a complete proof, that no square number is a congruum — the case Fermat later settled by descent. A congruent number can always be scaled by a square, so the question is really about square-free numbers, and every figure here uses only those.

From triangles to triples #

Every rational right triangle is a scaled copy of a primitive Pythagorean triple — whole numbers $(m^2 - k^2, 2mk, m^2 + k^2)$ for coprime $m > k$ of opposite parity, all of which a single tree of three matrices produces. The triple’s area is $mk(m - k)(m + k)$, and scaling the triangle by a rational factor $t$ scales its area by $t^2$. So $n$ is congruent exactly when some triple’s area is $n$ times a square: then shrinking the triple by the square root of that square gives a rational triangle of area $n$.

That turns the question into a search. For each triple, factor the area, strip out the squares, and what is left — the area’s square-free part — is a congruent number. The four factors $m$, $k$, $m - k$ and $m + k$ are pairwise coprime, so the square-free part of their product is the product of their square-free parts, which makes the search fast. The triple $(3, 4, 5)$ has $m = 2$, $k = 1$ and area 6, which is square-free, so 6 is congruent. The triple with $m = 5$, $k = 4$ is $(9, 40, 41)$, of area $180 = 5 \times 36$, so 5 is congruent — and dividing every side by 6 gives the triangle in the figure.

The areas 1, 2 and 3 never appear, and never will. Fermat’s proof that 1 is not congruent was the first proof by infinite descent: from a right triangle of square area he built a smaller one, and since whole numbers cannot shrink for ever, there is none. The same kind of argument disposes of 2 and 3. Among the congruent numbers that do appear, most are proved by only one, two or three of the eighteen thousand triples in this range, and several — marked with a dot — appear in none of them. A congruent number’s triangle can be very large.

Why one is not an area #

Fermat’s argument for 1 is worth seeing in the form the squares in arithmetic progression give it. If 1 were congruent, there would be three rational squares a distance 1 apart; clearing denominators, three whole-number squares $u^2 - w^2$, $u^2$, $u^2 + w^2$ that are all squares, and multiplying the outer two, a solution of $u^4 - w^4 = z^2$ in whole numbers. Fermat showed that any such solution can be turned into a strictly smaller one — the same move as folding a square that cannot shrink in the proof that $\sqrt2$ is irrational — and whole numbers cannot decrease for ever. So there is no solution, and 1 is not congruent.

The equation $u^4 - w^4 = z^2$ has a famous corollary: it has no solutions with $z$ a square, so $x^4 + y^4 = z^4$ has none either, which is the exponent-four case of Fermat’s last theorem — proved by this descent three and a half centuries before the general case. The essay that met every triple on one circle told that story from the side of the triples; here it is the first entry in the list of non-congruent numbers, and the descent that produced it does not extend to settle the general question. For 2 and 3 similar descents work. For most numbers neither a short descent nor a small triangle settles the matter, and the answer has to come from the curve.

How large the smallest triangle can be #

Extending the search to every $m$ up to 5,000 makes the point sharper.

For half of the congruent numbers below 400 the search finds a triple, sometimes tiny — 6 needs $m = 2$, 5 needs $m = 5$ — and sometimes in the thousands. For the other half it finds nothing, and those numbers are congruent all the same, as the next section’s criterion shows. The most famous of them is 157: the simplest rational triangle of area 157, found by Don Zagier, has sides that are fractions with more than twenty digits above and below the line, and its generating triple is far beyond any search.

That is why searching can never settle the question on its own. A search that finds a triangle proves a number congruent. A search that finds nothing proves nothing, because no bound is known on how large the smallest triangle can be in terms of $n$. Deciding that a number is not congruent needs a different kind of argument.

A curve where triangles are points #

The different kind of argument comes from geometry of a different sort. A rational right triangle $(a, b, c)$ of area $n$ gives a rational point

$$x = \left(\frac c2\right)^2, \qquad y = \frac{c\,(b^2 - a^2)}{8}$$

on the curve $y^2 = x^3 - n^2 x$, and every rational point on that curve with $y \ne 0$ gives back a triangle: $a = |(x^2 - n^2)/y|$, $b = |2nx/y|$, $c = |(x^2 + n^2)/y|$. Congruent numbers are the $n$ for which this elliptic curve has rational points other than the three obvious ones on the axis.

Elliptic curves carry an extraordinary structure: a line through two rational points of the curve meets it in a third rational point, and a tangent at one rational point meets it in another. The (3, 4, 5) triangle is the point $(12, 36)$ on the curve for $n = 6$. Its tangent there has slope $\tfrac{11}2$ — found by differentiating the curve’s equation as it stands — and meets the curve again at $(\tfrac{25}4, \tfrac{35}8)$. Reflecting that point in the horizontal axis gives $(\tfrac{25}4, -\tfrac{35}8)$, and the formula turns it into a second triangle of area 6:

$$\frac{7}{10}, \qquad \frac{120}{7}, \qquad \frac{1201}{70}.$$

Doubling again gives a third, much larger, and so on for ever: a congruent number has infinitely many triangles, and the tangent-and-chord rule produces them. The same rule makes the points of any cubic curve into a group, which is how nine trees can be planted in ten rows of three: collinear triples of points on a cubic are triples that add to zero.

The triangles grow fast as the doubling is repeated. On the curve for 6, the hypotenuse of the triangle from $2P$ has a numerator of four digits, from $4P$ thirteen, from $8P$ fifty, from $16P$ a hundred and ninety-eight, and from $32P$ seven hundred and ninety-one — the number of digits multiplies by about four at each doubling. That is the height of a rational point at work: doubling a point raises the size of its coordinates’ numerators and denominators to roughly the fourth power, so their lengths in digits roughly quadruple. It is also why the smallest triangle for a number like 157 can be enormous: if the curve’s simplest point is already large, every triangle built from it is larger still, and there is nothing smaller to find.

Two counts that decide #

The curve’s rational points are governed by a deep theory, and in 1983 Tunnell extracted from it a criterion that needs no curve at all. For an odd square-free $n$, count the integer solutions of two equations:

$$A = \#\{x^2 + 2y^2 + 8z^2 = n\}, \qquad B = \#\{x^2 + 2y^2 + 32z^2 = n\}.$$

If $n$ is congruent then $A = 2B$. For even $n$ the same holds with $4y^2$ in place of $2y^2$ and $n/2$ in place of $n$.

The first direction is a theorem: a number with $A \ne 2B$ is certainly not congruent, and the counts are finite and quick. That settles 1, 2 and 3 in a line each, and every dot off the line in the figure. The converse — that $A = 2B$ implies congruence — Tunnell proved only on the assumption that the Birch and Swinnerton-Dyer conjecture is true. That conjecture, one of the Clay Institute’s seven Millennium Prize problems, says that the number of independent rational points on an elliptic curve can be read off from the behaviour of a function built from counting the curve’s points modulo every prime. For congruent numbers it would mean that the counts $A$ and $B$, which measure that function’s value at one point, decide everything.

The counts are close relatives of the representation numbers that count lattice points in a ball and of Ramanujan’s question about which numbers a sum of weighted squares takes. Here they decide a question about triangles: the numbers of points on two ellipsoids with integer coordinates, compared, say whether a curve has infinitely many rational points.

The remainder mod 8 decides most of it #

Applying Tunnell’s criterion to every square-free number up to 1,000 finds 361 congruent numbers, and their distribution has a striking shape.

Every one of the 308 square-free numbers of the form $8k + 5$, $8k + 6$ or $8k + 7$ passes the test. Among those of the form $8k + 1$, $8k + 2$ and $8k + 3$, only 22, 19 and 12 do. The pattern is predicted by the Birch and Swinnerton-Dyer conjecture through a parity argument: for $n$ in the first three classes the curve’s root number forces an odd number of independent points, so at least one, so a triangle. For primes the predictions are theorems in several classes — Kurt Heegner and later Paul Monsky proved, for instance, that every prime of the form $8k + 5$ or $8k + 7$ is congruent — and Tian Ye proved in 2014 that each of the three classes contains infinitely many congruent numbers. That every square-free number of the form $8k + 5$, 6 or 7 is congruent is still not proved.

In the other three classes congruent numbers are expected to become rarer and rarer. Alexander Smith proved in 2017 that among square-free numbers of the forms $8k + 1$, $8k + 2$ and $8k + 3$ the congruent ones make up a proportion that tends to nought, by a delicate analysis of how the curves’ rational points behave on average; the 22, 19 and 12 congruent numbers in those columns below 1,000 are, in that sense, the start of a long thinning out. The other three classes hold the hard cases. A number like 41 or 34, in the classes where congruence is rare, is congruent for reasons the residue does not reveal, and Tunnell’s counts are the only general way to find out.

What a search and a count cannot settle #

Every congruent number in the figures is certified in one of two ways, and the two are of different strength. A triangle found by the search is a proof, checked by arithmetic. A number passing Tunnell’s test, with no triangle found, is congruent if the Birch and Swinnerton-Dyer conjecture holds; for many such numbers, including 157, a triangle has since been found by other means, so the conditional is not needed for them, but for the criterion as a whole it is.

The search figure also shows only what the search reached. That 74 congruent numbers below 400 have no triangle with $m \le 5{,}000$ says nothing about how large their triangles are, only that they are larger. The curves’ rational points can be searched far more efficiently than the triples, by computing heights and using the group structure, and the largest known triangles were found by a different route altogether: Zagier obtained 157’s triangle by constructing a point on its curve from the theory of modular forms, the method of Heegner points, which produces a rational point directly when the theory guarantees one exists. The simple search here is meant to show the difficulty, not to measure it.

And the counts are exact for the numbers drawn and say nothing about larger ones. The pattern by residue mod 8 is visible in every range anyone has checked, and that is evidence, not proof.

Still open: the Birch and Swinnerton-Dyer conjecture #

Is every square-free $n$ with $A = 2B$ congruent? Equivalently, for these curves, is the Birch and Swinnerton-Dyer conjecture true? The conjecture is known in special cases — when the relevant function vanishes to order zero or one, by the work of Benedict Gross, Don Zagier and Victor Kolyvagin in the 1980s — and those cases are enough to settle the congruent number problem for many individual $n$ and for some infinite families. The general statement is open, and a proof would settle a question about triangles that the Arab mathematicians of the tenth century already asked.

In the meantime, the problem has a curious status. It is decidable in practice for any particular number anyone has cared to test: compute $A$ and $B$, and if they agree, search the curve for a point. Every number with $A = 2B$ that has been examined closely has yielded a triangle in the end. What is missing is a guarantee that the search always ends.

Areas decided by counting #

A whole number is the area of a rational right triangle exactly when the curve $y^2 = x^3 - n^2x$ has a rational point off its axis; six, five and seven are, and one, two and three are not. A triangle found by search proves congruence, and a tangent to the curve turns it into infinitely many — (3, 4, 5) becomes (7/10, 120/7, 1201/70). Tunnell’s two counts of solutions to quadratic equations prove non-congruence outright and congruence on the strength of the Birch and Swinnerton-Dyer conjecture; applied to every square-free number up to 1,000 they find 361 congruent numbers, including every one of the form $8k + 5$, 6 or 7 — and 157, whose smallest triangle has fractions of more than twenty digits.