Fun facts from elliptic curves

(Zhao Yu here.) It's been a long time since I've touched olympiads and I have a really bad memory, so today I'll try something different and talk about something unrelated to olympiads. I will talk about 3 interesting facts that follow from the theory of elliptic curves. I will not be precise, and I will black-box many things. The point is just to give you a taste of elliptic curves and hopefully this will motivate you to learn the actual theory in the future!

First, let me tell you what (complex) elliptic curves are. As with any nice object, there are several ways to view it. Here are the two especially useful ones.

1. Weierstrass form. Points of an elliptic curve are the solutions $(x,y)\in \mathbb C^2$ to the equation $y^2 = x^3+ax+b$, plus a point at infinity. The graph below only shows the real $(x,y)$ points (there are complex points that are not graphed here!).

2. Complex torus. An elliptic curve is a parallelogram piece of the complex plane with vertices $0,1,\tau,1+\tau$. Also, the parallel sides of the parallelogram are glued together, forming a torus as shown. 

The connection between the two is not easy to see, and it requires the use of some complex analysis. But essentially one can do a change of variables from one form to another. The Weierstrass form can be seen as an algebraic interpretation of an elliptic curve (it is defined by an algebraic equation), while the complex torus is an analytic interpretation (it is part of the complex plane). We will see how interchanging both interpretations can be useful in the examples below.

Another good thing to keep in mind is the difference between real dimension and complex dimension. Complex elliptic curves, as the name suggests, have $1$ complex dimension and thus $2$ real dimensions. We see this in both interpretations above. Note that the plotted graph of the Weierstrass equation is only a $1$ real dimensional slice of the full complex curve. 

Our first example is a parody of fruit equations memes. Those memes are usually easy, but this one is actually impossible!

(1) Meme

Let $a$ stand for apple, $b$ stand for banana and $c$ for pineapple, we get a more readable version: $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} = 4.$$

You might be able to find the solutions $a=-1, b=4, c=11$, but we want positive integer solutions. Surprisingly, the smallest positive integer solutions are 80-digits long! $$a=154476802108746166441951315019919837485664325669565431700026634898253202035277999, \\b=36875131794129999827197811565225474825492979968971970996283137471637224634055579, \\c=4373612677928697257861252602371390152816537558161613618621437993378423467772036.$$ Substituting this back into the original equation gives exactly $4$! This is impossible to solve by brute force, even with the help of a computer. The only way to find this solution is using the theory of elliptic curves, which we explain now. By multiplying the denominators, we get a degree $3$ equation in $a,b,c$: $$a(a+c)(a+b)+b(b+c)(a+b)+c(a+c)(b+c)=4(b+c)(a+c)(a+b).$$As this is a homogeneous equation, we can substitute $x=a/c,y=b/c$ (while allowing possible "points at infinity" when $c=0$) to get $$x(x+1)(x+y)+y(y+1)(x+y)+(x+1)(y+1)=4(x+y)(x+1)(y+1)$$which is now a degree $3$ equation in $2$ variables, so this should define a $1$-dimensional complex curve. One could show that all such degree $3$ equations cut out an elliptic curve, by finding a change of variables into the Weierstrass form, which we will not explicitly describe.

One reason why elliptic curves are nice is because they have an additive structure, that is, you can add and subtract points on the curve. This is clear when you view an elliptic curve as a complex torus. Here, points are just complex numbers, and you can just add them as complex numbers! In the Weierstrass form, to add points $P,Q$, you draw the line $PQ$ which will intersect the elliptic curve again at another point $R$. Then, draw a line parallel to the $y$-axis passing through $R$, which will intersect the elliptic curve again at the point $P+Q$. One thing to note is that adding two rational points gives another rational point. Here, rational means that $(x,y)\in \mathbb Q$ in the Weierstrass form, which then corresponds to some integer solution $(a,b,c)\in \mathbb Z$.

Back to the problem. Since our equation defines an elliptic curve, we can then make a change of variables into either the Weierstrass form or the complex torus form. If we write out this change of variables, we realise that only part of the curve corresponds to the positive solutions (where $a,b,c>0$). We already know a rational point $P$ on the curve corresponding to $(a,b,c)=(-1,0,1)$, but this does not lie on the positive part of the curve. We can repeatedly add $P$ to itself, and $nP$ will also be a rational point for each $n\in \Z$. Eventually, we get a rational point $9P$ in the desired positive area. Translating back to the original equation, we get the positive integer solution (positive because it lies in the positive part of the curve, integer because it is a rational point)! 

Here is an illustration of the points $P,2P,\ldots, 9P$ on the complex torus form and the Weierstrass form. The purple part corresponds to the part of the curve that corresponds to positive solutions.

Our next example is an application of the theory of elliptic curves to a physics problem.

(2) A physics problem

Problem: Consider a particle in a potential $V(x)$, where $V(x)$ is a real quartic polynomial with two local minima. For $E$ in a suitable range, there are two possible distinct periodic oscillations with energy $E$, around each local minima. Show that the period of the two oscilations are equal. (Source: Riemann Surfaces, Donaldson, Chapter 6).

For those who are not as familiar with kinematics, just imagine a slope in the shape of $V(x)$. You can release a ball at height $E$ and it rolls around the valley (this is essentially correct up to some minor corrections). There are two possible oscillations as indicated in the digram above, and we want to show they have the same period. 

Among the three examples, this is my favourite -- even though the oscillations of the particles don't overlap, there is some "spooky action from a distance" that makes the oscillation periods the same. It is also a cool example of how mathematics and physics are intertwined, and at first glance you wouldn't expect that this has anything to do with elliptic curves.

Let's tackle this problem. We start by translating the physical statement into a mathematical problem. By conservation of energy, we have the equation $E = \frac 1 2 m v^2 + V(x)$. By scaling, we obtain $E = \left(\frac{dx}{dt}\right)^2 + V(x)$, and after rearranging we get $\frac{dx}{\sqrt{f(x)}} = dt$, where $f(x)\coloneqq E-V(x)$ is a quartic polynomial. Translate $x$ such that $0$ is a root of $f(x)$, and then substitute $u=\frac 1 x$. We now get the equation $$\frac{du}{\sqrt{g(u)}} = dt$$ where $g(u)$ is a cubic. Again, one can translate and scale $u$ such that $g(u)=u^3+au+b$. We essentially reduced this to the form of an elliptic integral.

Now, consider the elliptic curve given by the Weierstrass equation $y^2=u^3+au+b$, with the differential $du/\sqrt{g(u)}$ on the complex curve. It turns out that when we view this elliptic curve as a complex torus with complex coordinate $z$, this differential simply becomes $dz$!. (This is not a coincidence, both differentials satisfy the property of an "invariant differential", which means that they are invariant under translation.) Thus, our equation simplifies to $dz=dt$, and thus we have $z=t+C$, where the constant $C$ is different for each oscillation. 

Here is an illustration of the paths that the two particles take on the Weierstrass form and the complex torus form. We see that in the complex torus form, the particles move in the horizontal direction as $z=t+C$. As the complex torus is a parallelogram, both paths have period exactly $1$ (after scaling from the various change of variables), and are the same. 

Lastly, we have our final example.

(3) $e^{\pi\sqrt{163}}$ is almost an integer

Plugging this into a calculator, we get $$e^{\pi\sqrt{163}}= 262537412640768743.99999999999925\ldots .$$ This is sometimes known as Ramanujan's constant. Given how close it is to an integer, you would be right to suspect that there is a deeper reason for this. 

We introduce the $j$-invariant of an elliptic curve $E$, which we denote as $j(E)\in \mathbb C$. This is an important invariant -- two elliptic curves are isomorphic (same up to change of variables) if and only if their $j$-invariant is the same. If we let $E$ be in the complex torus form, we can compute the $j$-invariant by the following power series formula: $$j(E) = \frac{1}{q}+744+196884q+\cdots ,$$ where $q=e^{2\pi i \tau}$.

If you dabble a little in algebraic number theory, you may recognize the number $163$. Indeed, the ring $\mathbb Z[\frac{1+\sqrt{-163}}{2}]=\{a+b\left(\frac{1+\sqrt{-163}}{2}\right)\colon a,b\in \mathbb Z\}$ is a unique factorisation domain (UFD)! This means that each element of $R$ can be uniquely written as a product of prime elements (up to $\pm$ sign). This is quite a special property -- the only negative integers which give UFDs in a similar way are $\{-1,-2,-3,-7,-11,-19,-43,-67,-163\}$.

The fact that these two results both involve the number $163$ is not a coincidence. By the theory of complex multiplication, the fact that $\mathbb Z[\frac{1+\sqrt{-163}}{2}]$ is a UFD implies that $j(E)$ is an integer (this is the main step and unfortunately requires a lot of theory, so it is black-boxed). Now, we use our power series formula for $j(E)$ where $q=e^{2\pi i \tau} = -e^{-\pi\sqrt{163}}$, to get: $$e^{\pi\sqrt{163}}=-\frac{1}{q} = -j(E)+744+196884q+\cdots .$$ Note that $q$ is tiny, and it happens that the coefficients of the power series don't grow too quickly, so the terms $196884q+ \cdots$ is tiny as well. Hence, $$e^{\pi\sqrt{163}}\approx -j(E)+744 \in \Z$$ is almost an integer. As an additional side note, there is another formula expressing $j(E)$ as a cube, so $e^{\pi\sqrt{163}}-744\approx 640320^3$ is approximately a cube.

These three facts are just random and fun applications of elliptic curves. There are many real and important uses for elliptic curves, such as in cryptography. It also plays a huge role in the proof of Fermat's Last Theorem and is a big part of Number Theory. I hope you enjoyed reading this post!