The Congruent Number Problem

Definition 6.5 (Congruent Number)   We call a nonzero rational number $n$ a congruent number if $\pm n$ is the area of a right triangle with rational side lengths. Equivalently, $n$ is a congruent number if the system of two equations

$$a^2+b^2 = c^2$$

$$\frac{1}{2}ab = n$$

has a solution with $a,b,c\in\mathbb{Q}$ .

For example, $6$ is the area of the right triangle with side lengths $3$ , $4$ , and $5$ , so $6$ is a congruent number. Less obvious is that $5$ is also a congruent number; it is the area of the right triangle with side lengths $3/2$ , $20/3$ , and $41/6$ . It is nontrivial to prove that $1$ , $2$ , $3$ , and $4$ are not congruent numbers. Here is a list of the integer congruent numbers up to $50$ :

$$5, 6, 7, 13, 14, 15, 20, 21, 22, 23, 24, 28, 29, 30, 31, 34, 37, 38, 39, 41, 45, 46, 47.$$

Every congruence class modulo $8$ except $3$ is represented in this list, which incorrectly suggests that if $n\equiv 3\pmod{8}$ then $n$ is not a congruent number. Though no $n\leq 218$ with $n\equiv 3\pmod{8}$ is a congruent number, $n=219$ is a congruent number congruent and $219\equiv 3\pmod{8}$ .

Deciding whether an integer $n$ is a congruent number can be subtle since the simplest triangle with area $n$ can be very complicated. For example, as Zagier pointed out, the number $157$ is a congruent number, and the ``simplest'' rational right triangle with area $157$ has side lengths

$$a = \frac{6803298487826435051217540}{411340519227716149383203}$$   and$$b = \frac{411340519227716149383203}{21666555693714761309610}.$$

This solution would be difficult to find by a brute force search.

We call congruent numbers ``congruent'' because of the following proposition, which asserts that any congruent number is the common ``congruence'' between three perfect squares.

Proposition 6.5   Suppose $n$ is the area of a right triangle with rational side lengths $a, b, c$ , with $a\leq b<c$ . Let $A=(c/2)^2$ . Then

$$A-n, \quad A,$$    and $$A+n$$

are all perfect squares of rational numbers.

Proof. We have

$$a^2+b^2 = c^2$$

$$\frac{1}{2}ab = n$$

Add or subtract $4$ times the second equation to the first to get

$$a^2\pm2ab +b^2 = c^2\pm 4n$$

$$(a\pm b)^2 = c^2\pm 4n$$

$$\left(\frac{a\pm b}{2}\right)^2 = \left( \frac{c}{2}\right)^2 \pm n$$

$$= A \pm n$$

$\qedsymbol$

The main motivating open problem related to congruent numbers, is to give a systematic way to recognize them.

Open Problem 6.5   Give an algorithm which, given $n$ , outputs whether or not $n$ is a congruent number.

Fortunately, the vast theory developed about elliptic curves has something to say about the above problem. In order to understand this connection, we begin with an elementary algebraic proposition that establishes a link between elliptic curves and the congruent number problem.

Proposition 6.5 (Congruent numbers and elliptic curves)   Let $n$ be a rational number. There is a bijection between

$$A = \left\{(a,b,c) \in \mathbb{Q}^3 : \frac{ab}{2} = n, a^2 + b^2 = c^2\right\}$$

and

$$B = \left\{(x,y) \in \mathbb{Q}^2 : y^2 = x^3 - n^2 x, \text{\rm with } y \neq 0\right\}$$

given explicitly by the maps

$$f(a,b,c) = \left(-\frac{nb}{a+c}, \frac{2n^2}{a+c}\right)$$

and

$$g(x,y) = \left(\frac{n^2-x^2}{y}, -\frac{2xn}{y}, \frac{n^2+x^2}{y}\right).$$

The proof of this proposition is not deep, but involves substantial (elementary) algebra and we will not prove it in this book.

For $n\neq 0$ , let $E_n$ be the elliptic curve $y^2 = x^3 - n^2 x$ .

Proposition 6.5 (Congruent number criterion)   The rational number $n$ is a congruent number if and only if there is a point $P=(x,y)\in E_n(\mathbb{Q})$ with $y\neq 0$ .

Proof. The number $n$ is a congruent number if and only if the set $A$ from Proposition 6.5.7 is nonempty. By the proposition $A$ is nonempty if and only if $B$ is nonempty. $\qedsymbol$

Example 6.5   Let $n=5$ . Then $E_n$ is $y^2=x^3-25x$ , and we notice that $(-4,-6)\in E_n(\mathbb{Q})$ . We next use the bijection of Proposition 6.5.7 to find the corresponding right traingle:

$$g(-4,-6) = \left(\frac{25-16}{-6},-\frac{-40}{-6}, \frac{25+16}{-6}\right) = \left(-\frac{3}{2}, -\frac{20}{3}, -\frac{41}{6}\right).$$

Multiplying through by $-1$ yields the side lengths of a rational right triangle with area $5$ . Are there any others?

Observe that we can apply $g$ to any point in $E_n(\mathbb{Q})$ with $y\neq 0$ . Using the group law we find that $2(-4,-6) = (1681/144, 62279/1728)$ , and

$$g(2(-4,-6)) = \left( -\frac{1519}{492}, -\frac{4920}{1519}, \frac{3344161}{747348} \right).$$

Example 6.5   Let $n=1$ , so $E_1$ is defined by $y^2=x^3-x$ . Since $1$ is not a congruent number, the elliptic curve $E_1$ has no point with $y\neq 0$ . See Exercise 6.11.

Example 6.5.9 foreshadows the following theorem.

Theorem 6.5 (Infinitely Many Triangles)   If $n$ is a congruent number, then there are infinitely many distinct right triangles with rational side lengths and area $n$ .

We will not prove this theorem, except to note that one proves it by showing that $E_n(\mathbb{Q})_{\tor } = \{\O,(0,0),(n,0),(-n,0)\}$ , so the elements of the set $B$ in Proposition 6.5.7 all have infinite order, hence $B$ is infinite so $A$ is infinite.

Tunnell has proved that the Birch and Swinnerton-Dyer conjecture (alluded to above), implies the existence of an elementary way to decide whether or not an integer $n$ is a congruent number. We state Tunnell's elementary way in the form of a conjecture.

Conjecture 6.5   Let $a, b, c$ denote integers. If $n$ is an even square-free integer then $n$ is a congruent number if and only if

$$\char93 \left\{(a,b,c)\in\mathbb{Z}^3 : 4a^2 + b^2 + 8c^2 = \frac{n}{2} : c\text{\rm {} is even}\right\}$$

$$\qquad\qquad\qquad\qquad\qquad = \char93 \left\{(a,b,c) : 4a^2 + b^2 + 8c^2 = \frac{n}{2}: c\text{\rm {} is odd}\right\}.$$

If $n$ is odd and square free then $n$ is a congruent number if and only if

$$\char93 \left\{(a,b,c) : 2a^2 + b^2 + 8c^2 = n : c\text{\rm {} is even}\right\}$$

$$\qquad\qquad\qquad\qquad\qquad = \char93 \left\{(a,b,c) : 2a^2 + b^2 + 8c^2 = n: c\text{\rm {} is odd}\right\}.$$

Enough of the Birch and Swinnerton-Dyer conjecture is known to prove one direction of Conjecture 6.5.12. In particular, it is a very deep theorem that if we do not have equality of the displayed cardinalities, then $n$ is not a congruent number. For example, when $n=1$ ,

The even more difficult (and still open!) part of Conjecture 6.5.12 is the converse: If one has equality of the displayed cardinalities, prove that $n$ is a congruent number. The difficulty in this direction, which appears to be very deep, is that we must somehow construct (or prove the existence of) elements of $E_n(\mathbb{Q})$ . This has been accomplished in some cases do to groundbreaking work of Gross and Zagier ([#!gross-zagier!#]) but much work remains to be done.

The excellent book [#!koblitz:cong!#] is about congruent numbers and Conjecture 6.5.12, and we encourage the reader to consult it. The Birch and Swinnerton-Dyer conjecture is a Clay Math Institute million dollar millennium prize problem (see [#!cmi!#,#!wiles:cmi!#]).