Motivation: The Congruent Number Problem

Definition 4.1 (Congruent Number)   A rational number $n$ is called a congruent number if $\pm n$ is the area of a right triangle with rational side lengths. Equivalently, $n$ is congruent if the system of two equations

$$n=ab/2$$   and$$\qquad a^2 + b^2 = c^2$$

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 congruent numbers up to $50$:

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

Every congruence class modulo $8$ except $3$ is represented in this list, which suggests that if $n\equiv 3\pmod{8}$ then $n$ is not a congruent number. This is true for $n\leq 218$, but $n=219$ is a congruent number congruent to $3$ mod $8$. Something very subtle is going on.

This is another example which hints at the subtlety of congruent numbers. The number $157$ is a congruent number, and Don Zagier showed that the simplest rational right triangle with area $157$ has side lengths

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

This solution would take a long time to find without understanding more about congruent numbers.

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

The following proposition establishes a link between elliptic curves and the congruent number problem. This link connects the congruent number problem with the Birch and Swinnerton-Dyer conjecture, which some consider to be the most important open problem in the theory of elliptic curves.

Proposition 4.2   Let $n$ be a rational number. There is a bijection between

$$A = \left\{(x,y,z) \in {\mathbb{Q}}^3 \,:\, \frac{xy}{2} = n,\, x^2 + y^2 = z^2\right\}$$

and

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

given explicitly by the maps

$$f(x,y,z) = \left(-\frac{ny}{x+z},\,\, {2n^2}{x+z}\right)$$

and

$$g(r,s) = \left(\frac{n^2-r^2}{s},\,\, -\frac{2rn}{s},\,\, \frac{n^2+r^2}{s}\right).$$

Corollary 4.3   The rational number $n$ is a congruent number if and only if the elliptic curve $E_n$ defined by $y^2 = x^3 - n^2 x$ has a solution with $y\neq 0$.

Proof. The number $n$ is a congruent number if and only if the set $A$ from Proposition 4.2 is nonempty. By the proposition $A$ is nonempty if and only if $B$ is nonempty, which proves the corollary. $\qedsymbol$

Example 4.4   Let $n=5$. Then $E_n$ is defined by $y^2=x^3-25x$, and we find by a brute force search the solution $(-4,-6)$. Then

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

Multiplying through by $-1$ yields the side lengths of a rational right triangle with area $5$.

Theorem 4.5   Let $n$ be even and squarefree, and let $E$ be the elliptic curve

$$y^2=x^3-n^2 x.$$

Then $L(E,1)=0$ if and only if

$$\char93 \Bigl\{(a,b,c) : 4a^2 + b^2 + 8c^2 = \frac{n}{2} : c \Bigr\}$$

$$= \char93 \Bigl\{(a,b,c) : 4a^2 + b^2 + 8c^2 = \frac{n}{2}: c \Bigr\}.$$

So far I have told you nothing about the meaning of ``$L(E,1)=0$''. Suffice for now to know that (a consequence of) the Birch and Swinnerton-Dyer conjecture is the assertion that the set of rational solutions to $y^2 = x^3 - n^2 x$ is infinite if and only if ``$L(E,1)=0$''. Also, it is easy to prove that this set of solutions is infinite if and only if $n$ is a congruent number.

When $n=6$, we get

$$\char93 \emptyset = 2\cdot \char93 \emptyset.$$

When $n=2$, we get

$$\char93 \{(0,1,0)\} \neq 2\cdot \char93 \{(0,1,0)\},$$

so the BSD conjecture predicts that $y^2=x^3-4x$ has no interesting solutions and $2$ is not a triangle number.

In fact, this is true. The implication $L(E,1)\neq 0$ implies $y^2 = x^3 - n^2 x$ has no interesting solutions was proved by Coates and Wiles (this is the same Wiles who proved Fermat's Last Theorem).

The other implication:

$$L(E,1) = 0 \Longrightarrow y^2 = x^3 - n^2 x$$    has lots of solutions 

is a fascinating open problem.