# Monday, March 28, 2016

### Fermat's Last Theorem and the 2016 Abel Prize

**by** **Jonathan** **Kujawa**

On March 15th it was announced that Andrew Wiles won the 2016 Abel Prize. Established in 2002, the Abel Prize has become arguably the most prestigious prize in mathematics. In contrast to the Fields medal, which is awarded to those under 40, the Abel prize set itself as the prize which recognizes long term contributions to mathematics.

In keeping with tradition (see here: 2015, 2014) we're taking the opportunity to check out the math of behind the Abel Prize. This is the rare instance when the prizewinner's work appeared in the New York Times and may well need no introduction. Wiles won for

...for his stunning proof of Fermat’s Last Theorem by way of the modularity conjecture for semistable elliptic curves, opening a new era in number theory.

-- from the Abel Prize Announcement

Fermat's Last Theorem (FLT) is the claim that, for any n greater than or equal to three, there are no integer solutions to the equation

That is, you can't find numbers a, b, and c from among 0, 1, -1, 2, -2,... which can be plugged into

and have the same number on both sides of the equal sign. The same goes if the six is replaced with a 3, or 2016, or 187,201, or any other number greater than or equal to three.

If you haven't heard of FLT before, it's hard to see why anyone should give a rat's rear end about whether nor not there are integer solutions to this equation. On the other hand, Fermat conjectured FLT in 1637 and here we are in 2016 giving Wiles the Abel Prize for proving that yes, indeed, there are no such solutions. Something interesting must be going on.

First off, FLT is a mathematical romance. Fermat was a French mathematician who squandered his days (and paid his bills) by being a lawyer. We ran into him last year when talking about Pascal's Wondertorium. Like any respectable mathematician of the day, he had a copy of Diophantus's *Arithmetica*. This was a book from the 3rd century which contained a series of interesting algebra problems which would be familiar to anyone who has wrestled with the quadratic formula [1]. One of the problems was to show that each square can be written as the sum of two squares (and so is closely related to the FLT when the n is two). Next to this problem, Fermat wrote:

It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second, into two like powers. I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

To this day we have no idea if Fermat actually had a proof [2]. As is often the case in math, if a problem is easy to state and it's not trivial, then it is devilishly hard. Some progress was made in the intervening 350 years, but it was pretty much still a wide open unsolved problem when a 10 year old Andrew Wiles came across FLT at his local library. Wiles was one of those rare kids who actually achieved his childhood dream [3].

Part of the problem is that FLT ramps up from trivial to impossible on a nearly vertical trajectory.

If n equals one and you give me any integers a and b, then I can simply take their sum to find the corresponding c. Every pair (a,b) has its c. If n equals two, then things get a bit harder. If I take a and b to both be one, then I get 1^{2}+1^{2}=c^{2}. That is, we need to find an integer which squares to 2. A quick check shows that nearly all integers are too big and the few that are small enough just don't do the job [4]. A little experimentation shows that 3, 4, and 5 make a perfectly good solution to the equation. And so does 6, 8, and 10, or 9, 12, and 15. In fact there are infinitely many solutions when n equals two. Once we've found one (like 3, 4, 5) we can multiply all three by any number we like and a little algebra shows we get another solution. When n equals two the solutions are called Pythagorean triples. We'll revisit them in a moment.

But when n equals three we hit the wall. We go from solutions being a dime a dozen when n equals one, to harder to find but still infinitely many when n equals two, to conjecturally zero when n equals three. It's hard to get any traction or see a pattern. What to do when you're stuck on a problem? A mathematician's favorite ruse is to change the question to one we *can* answer. It's a bit like setting out to discovery a drug which treats heart disease and ending up with one which gives four hour erections instead. But, hey, a win's a win, right?

One direction we *can* go is to dig in on those not-trivial, non-impossible solutions when n equals two. These are called Pythagorean triples for the simple reason that the equation

is nothing but the formula from the Pythagorean Theorem. This is the equation which relates the three sides of a right triangle (and is flubbed in the Wizard of Oz). It has many, many proofs, including one hit upon by President James Garfield while discussing mathematics with members of Congress [6]. There are numerous "proofs without words" which go by geometry alone. Here on the side is one from a 1594 Arabic edition of Euclid's elements.

The study of Pythagorean triples goes back millennia. There is a Babylonian clay tablet from 1800 BC called the Plimpton 322 which contains a list of them. It's no surprise, then, that we know quite a lot about Pythagorean triples. From above we saw that if you have a Pythagorean triple, then any multiple will give you another triple. Let's call a triple primitive if it *can't* be gotten from another triple in this way.

If you pick any positive integers m and n, with m larger than n, the Euclid's formula:

gives you a Pythagorean triple. As long as m and n have no common factors and one is an even number and one is an odd number, then you get a primitive triple. Even better, every primitive triple can be made in this way. That is, every Pythagorean triple is nothing but the multiple of one obtained from Euler's formula!

Once you have the a and b of a Pythagorean triple, algebra gives you the c. We may as well ignore the c for a moment. If we plot the pairs (a,b) for the first few thousand Pythagorean triples, we see intriguing patterns:

Indeed, exactly one of a and b is divisible by 3 and exactly one of them is divisible by 4. Exactly one of a,b,c is divisible by 5 and the largest number to divide the product abc is 60. The number c is always odd; in fact, it must be of the form 4k+1 for some integer k. Their strange properties go on and on. Pythagorean triples are far from random. Like the numbers in Pascal's Triangle they hold all sorts of surprises.

Nearly one hundred years ago, the famous Indian mathematician Ramanujan had the clever idea to study "near misses" of FLT. What if you found an a,b,c which *almost* gave a solution to the FLT when n equals three? Say you were off by only one. For example, 135^{3}+138^{3}=172^{3}-1. Close, but no cigar! You can see this example and others in a page from Ramanujan's notebook. Ramanujan clearly had FLT on his mind.

Ramanujan had no luck in proving the FLT. In his notebook it looks like he was thinking about finding a solution when n equals three so as to show that FLT fails. In any case, he came up with a machine which generates infinitely many of these near misses. So while FLT itself has no solutions when n equals three, if you give yourself just the slightest flexibility you find you can make infinitely many "near misses". Much like how there is infinitely many Pythagorean triples when n equals two. Just last year Ken Ono and Sarah Trebat-Leder showed that you can use Ramanujan's century old machine to generate infinitely many solutions to very modern elliptic curves.

With elliptic curves we come full circle. We ran across them here at 3QD them nearly two years ago. They are certain curves which play a fundamental role in modern mathematics. While they were invented while studying such seemingly useless questions like FLT, it turns out that they can be used to provide exotic number systems at the heart of modern cryptography (sometimes the things you find along the way turn out to be the billion dollar idea). They are also key to Wiles's proof of FLT. To each equation like those in FLT you can pair it with a certain elliptic curve. In this case the FLT equation is paired with the elliptic curve given by:

If you could find a solution a,b,c to FLT, then those numbers plugged into this equation would give you an elliptic curve with peculiar properties.

But there is another way to make elliptic curves. There are certain special functions on the upper half of the complex plane (called modular functions) and to each of these you can also attach an elliptic curve. Because of how nice modular functions are, these elliptic curves have to be equally nice. In the 1980s Frey, Serre, and Ribet proved that an elliptic curve coming from a solution to FLT would be so strange as to definitely not be nice enough to be one also coming from a modular function. But, on the other hand, the Taniyama-Shimura conjecture was that *every* elliptic curve of the kind coming from FLT must also come from a modular function! If Taniyama-Shimura was right, then every such elliptic curve is nice, which means no weird ones coming from solutions to FLT, which means no solutions to FLT, which means Fermat was right!

This connection brought FLT right into Wiles's wheelhouse. He is an expert on these sorts of questions. Wiles worked in secret [7] for seven years to prove the Taniyama-Shimura conjecture (or at least enough of it to cover the elliptic curves involving FLT). Excruciatingly, a gap in the proof was found after Wiles announced his result. Fortunately Wiles and Taylor were able to circumvent the gap and the complete proof was published in 1995.

While proving FLT is the shiny star on top, Wiles real contribution was to lead the way to a proof of the Taniyama-Shimura conjecture. The deep connections between number theory, geometry, and representation theory implied by this work go far, far beyond proving FLT. This is Wiles's real contribution and the reason for the Abel Prize. As always, deepening our understanding is the real goal. Wiles put it perfectly [8]:

Perhaps I can best describe my experience of doing mathematics in terms of a journey through a dark unexplored mansion. You enter the first room of the mansion and it's completely dark. You stumble around bumping into the furniture, but gradually you learn where each piece of furniture is. Finally, after six months or so, you find the light switch, you turn it on, and suddenly it's all illuminated. You can see exactly where you were. Then you move into the next room and spend another six months in the dark. So each of these breakthroughs, while sometimes they're momentary, sometimes over a period of a day or two, they are the culmination of—and couldn't exist without—the many months of stumbling around in the dark that proceed them.

[1] According to the modern library of Alexandria, Wikipedia, Arithmetica originally came in thirteen volumes, with only six surviving to the present. Amazingly, another four volumes were found in a shrine in Iran in 1968! It had been wrongly shelved at some point in the past. It makes you wonder how many ancient texts that we only know by reputation which are still out there somewhere.

[2] Almost certainly he didn't. Using the math Fermat knew, there are near proofs which deceptively look like they could do the trick but ultimately fail. The technology used by Wiles is literally centuries beyond Fermat's grasp. It is as if he said he knew how to design a fusion reactor, but the margin was too narrow to fit the schematics.

[3] I'm still holding out hope on being the world's first cowboy/astronaut/paleontologist.

[4] In fact, we need c=√2, and the square root of two is definitely not an integer. In fact it is famously irrational (which means that it's not even the ratio of two integers). The discovery that the diagonal of a one by one square wasn't a fraction was a major shock to the Pythagoreans. Certainly this is not the sort of solution we, Diophantus, or Fermat would like to see.

[5] From the amazing History of Science collection at the University of Oklahoma. See here.

[6] How far we have fallen!

[7] FLT is one of those problems so famous (and famously hard) that it is embarrassing to admit you've spent time on it. I spent a few sunny summer days as an undergraduate thinking about 3D geometric arguments along the lines of the geometric proofs of the Pythagorean theorem. I only had a tan to show for my efforts.

Posted by Jon Kujawa at 12:40 AM | Permalink