A nonsingular plane algebraic curve is the set of solutions to a (nonsingular) polynomial:
$$F(X,Y) = 0$$
A rational point is $(x,y) \in \mathbf{Q}\times\mathbf{Q}$ such that $F(x,y) = 0$.
Definition: An integer $n$ is a congruent number if $n$ is the area of a right triangle with rational side lengths.
Open Problem: Give an algorithm to decide whether or not an integer $n$ is a congruent number.
This is a 1000-year old open problem, perhaps the oldest open problem in mathematics.
{{{id=39| T = line([(0,0), (3,0), (3,4), (0,0)],rgbcolor='black',thickness=2) lbl = text("3",(1.5,-.5),fontsize=28) + text("4",(3.2,1.5),fontsize=28) lbl += text("5",(1.5,2.5),fontsize=28) lbl += text("Area $n = 6$", (2.1,1.2), fontsize=28, rgbcolor='red') show(T+lbl, axes=False) /// }}} {{{id=40| /// }}} {{{id=3| /// }}}Theorem: A proof of the Birch and Swinnerton-Dyer Conjecture would also solve the congruent number problem.
Proof: Suppose $n$ is a positive integer. Consider the cubic curve $y^2 = x^3 - n^2 x$. Using algebra (see next slide), one sees that this cubic curve has infinitely many rational points if and only if there are rationals $a,b,c$ such that $n=ab/2$ and $a^2 + b^2 = c^2$. The Birch and Swinnerton-Dyer conjecture gives an algorithm to decide whether or not any cubic curve has infinitely many solutions.
{{{id=56| /// }}} {{{id=55| /// }}} {{{id=13| /// }}}
In fact, there is a bijection between
$$
A = \left\{(a,b,c) \in \mathbf{Q}^3 \,:\, \frac{ab}{2} = n,\, a^2 + b^2 = c^2\right\}
$$
and
$$
B = \left\{(x,y) \in \mathbf{Q}^2 \,:\, y^2 = x^3 - n^2 x, \,\,\text{with } y \neq 0\right\}
$$
given explicitly by the maps
$$
f(a,b,c) = \left(-\frac{nb}{a+c},\,\, \frac{2n^2}{a+c}\right)
$$
and
$$
g(x,y) = \left(\frac{n^2-x^2}{y},\,\,-\frac{2xn}{y},\,\, \frac{n^2+x^2}{y}\right).
$$
Let $C$ be a cubic curve (+ a technical condition I'm not mentioning). For each prime number $p$, let $N_p$ be the number of solutions to the cubic modulo $p$.
Definition: For any cubic curve $C$, let $a_p = p-N_p$.
Theorem (Hasse): $|a_p| < 2\sqrt{p}$.
Theorem (Wiles et al.): The function $$L(C,s) = \prod_p\left(\frac{1}{1-a_p p^{-s} + p^{1-2s}}\right)$ extends to an entire complex-analytic function on $\mathbf{C}$.
{{{id=49| L = EllipticCurve([-2009^2,0])._pari_().elllseries show(line([(i,L(i)) for i in [0,0.03,..,2]]), figsize=[7,1.5], ymax=10) /// }}} {{{id=27| /// }}}Heuristic Observation: If $C$ has infinitely many rational points, then the numbers $N_p$ will tend to be "large". Since $L(C,1) "=" \prod_p \frac{p}{N_p}$, the number $L(C,1)$ will tend to be small.
Theorem (Mordell): There is a finite set $P_1, \ldots, P_r$ of rational points on $C$ so that all (non-torsion) rational points can be generated from these using a simple geometric process (chords and tangents).
We call the smallest $r$ in Mordell's theorem the rank of $C$.
Conjecture (Birch and Swinnerton-Dyer): $$\text{ord}_{s=1} L(C,s) = \text{rank}(C)$$
This problem, exactly as stated, is the Clay Math Institute Million Dollar prize problem in number theory. We proved above that its solution would also resolve the 1000-year old congruent number problem.
{{{id=34| /// }}} {{{id=33| /// }}}Theorem: If $\text{ord}_{s=1} L(C,s) \leq 1$ then the Birch and Swinnerton-Dyer conjecture is true for $C$.
The proof involves Heegner points, modular curves, Euler systems and Galois cohomology.
