Fermat's One Proof

In his life, Pierre de Fermat only left one proof in relation to number theory. He used his method of infinite descent to show that the area of a right triangle cannot be a square within the domain of whole numbers.

The proof itself was found after his death in his notes on Diophantus's Arithmetica.

In what follows, I will go through each sentence of Fermat's proof and provide details to make his step clear. My goal is to shed light on Fermat's thinking in his one existent proof.

I will show how this proof can also be used to prove the case of Fermat's Last Theorem where n = 4. In my next blog, I go over a simpler proof for n=4.

My analysis is based on the work by Harold M. Edwards called Fermat's Last Theorem. The translation is taken from T. L. Heath's translation.

Fermat writes:

(1) Fermat: "If the area of a right-angled triangle were a square, there would exist two biquadrates the difference of which would be a square number."

A biquadrate is a value to the fourth-power. So, the biquadrate of 2 is 2⁴ = 16.

This corresponds to the following equation:
p⁴ - q⁴ = z².

When I first read this, it was surprising to me that Fermat assumed that this equation was obvious from the problem of showing that right triangle's area cannot be equal to a square.

This equation proves n = 4 since if x⁴ + y⁴ = z⁴, then:
x⁴ = (x²)² = z⁴ - y⁴

So, proving there is no right triangle that has an area equal to a square will also prove Fermat's Last Theorem for n=4.

The steps to this equation can be traced as follows:

(a) A right triangle is characterized by the Pythagorean Theorem:
a² + b² = c².

(b) The area of a rectangle is base x height.

(c) A right triangle of a given base and height is created by dividing the same rectangle across the diagonal.

(d) Therefore, the area of a right triangle is base * height / 2.

(e) Or, using a,b,c from above: area = ab/2

(f) From the solution to Pythagorean Triples, we know that:
a = (2pq)d
b = (p² - q²)d
where we know that p,q are relatively prime.

(g)So, saying the area of a right triangle is equal to a square comes down to proving that there are no solutions for:
z² = ((2pq)d[p² - q²]d)/2 = (pq)(d²)[p² - q²]

(h) From a previous result, we know that d will divide z so, we are left with showing no solutions for:
(z/d)² = (pq)[p² - q²]

(i) Since p,q are relatively prime, it follows that (pq) and [p² - q²] are relatively prime [see here for details]

(j) From this we conclude, that pq is a square and p² - q² are squares.

(k) And since p,q are relatively prime, p,q are themselves squares.

(l) So, there exist P,Q such that p = P² and q = Q²

(m) Since pq is a square, there exists a value k such that k² = pq.

(n) And from (m), k divides (z/d), so we get:

(z/dk)² = p² - q² = P⁴ - Q⁴

(2) Fermat: "Consequently there would exist two square numbers the sum and difference of which would both be squares."

This one is a bit easier to derive
z² = P⁴ - Q⁴ = (P² + Q²)(P² - Q²)

Now, all, we need to show is that (P² + Q²) is relatively prime to (P² - Q²) which means that p + q is realtively prime to p - q. [ The trick is to remember that p,q are relatively prime and they are of different parity (see 13(b) here). See here for details]

In other words, there exist two square numbers (P²,Q²) the sum (P² + Q²) and difference (P² - Q²) are both squares.

(3) Fermat: "Therefore we should have a square number which would be equal to the sum of a square and the double of another square..."

We know that P² + Q² is equal to a square. Let's say S².
We also know that P² - Q² is equal to a square. Let's say T²
So that P² = Q² + T²
Which combined with the other equation gives us:
Q² + T² + Q² = T² + 2Q² = S²

We can assume that S,P, Q are relatively prime and also that P,T,Q are relatively prime with S,P,T odd and Q even. [See here for details]

We can also assume that Q,S,T are relatively prime. [See here for details]

(4) Fermat: "...while the squares of which this sum is made up would themselves have a square number for their sum."

And as stated before: P² = Q² + T².

(5) Fermat: "But if a square is made up of a square and the double of another square, its side, as I can very easily prove, is also similarly made up of a square and the double of another square."

So, he is saying this:
T² + 2Q² = S² --> S = t² + 2q².

This means that

2Q² = S² - T² = (S - T)(S + T)

And:

Q² = (1/2)(S-T)(S+T)

Now Q,S,T are relatively prime. We know that Q, S-T, S+T are even. [see above for details]
Let S-T = 2u, S + T = 2v
So that Q² = (1/2)(2u)(2v) = 2uv
Now, u,v are relatively prime [see here for details]
And, either u or v is even, let's assume u
So that, u = 2w and Q² = 2(2w)v
And w,v are relatively prime so w,v are squares.
Let w = W², v = V²
And S + T + S - T = 2S = 2u + 2v = 2(2W² + V²)
So that S = 2W² + V²

(6) Fermat: "From this we conclude that the said side is the sum of the sides about the right angle in a right-angled triangle and that the simple square contained in the sum is the base and the double of the other square is the perpendicular."

So, he is saying this:
S = 2W² + V² -> (2W² )² + ( V²)² is a square.

(2W² )² + ( V²)² = (2w)² + ( v)² = (u)² + ( v)² =
[(1/2)(S-T)]² + [(1/2)(S+T)]² =
[S² - 2ST + T²]/4 + [S² + 2ST + T²]/4 =
= [2S² + 2T²]/4 = (S² + T²)/2.

Now, since S² = 2Q² + T²
We have:
(2Q² + T² + T²)/2 = Q² + T²

Which equals P².

(7) Fermat: "This right-angled triangle will thus be formed from two squares, the sum and differences of which will be squares."

We can once again apply the Solution to the Pythagorean Triples so that:

2W²= 2mn
V² = m² - n²
P= m² + n²

So we've proven that the difference is a square. Now, we need to prove that the sum is a square.

Since Q² = 2(2w)v = 4W²V².

Q²= 2(2mn)(m² - n²)=4mn(m² - n²).

And

(Q/2)² = mn(m² - n²)

Since m,n and m² - n² are relatively prime, then m,n,m²-n² are all squares.

Letting m = M² and n = N²

We get:
V² = M⁴ - N⁴ = (M² - N²)(M² + N²)

Since M² - N², M² + N² are relatively prime, both the sum and differences are squares.

(8) Fermat: "But both these squares can be shown to be smaller than the squares originally assumed to be such that both their sum and differences are squares."

So M² + N² ≤ P² + Q².

Since M² + N² ≤ m + n is less than P which is less than P² + Q².

(9) Fermat: "Thus if there exist two squares such that their sum and differences are both squares, there will also exist two other integer squares which have the same property but a smaller sum"

We have proven that:
if P² + Q² is a square and P² - Q² is a square, then there exists M,N such that:
(a) M²+N² is less than P² + Q²
(b) M² + N² is a square
(c) M² - N² is a square.

(10) Fermat: "By the same reasoning we find a sum still smaller than the last found, and we can go on ad infinitum finding integer square numbers smaller and smaller which have the same property."

This leads to an infinite number of smaller solutions.

(11) Fermat: "This is, however, impossible because there cannot be an infinite series of numbers smaller than any given integer we please."

So that we have a proof by Infinite Descent.

(12) Fermat: "The margin is too small to enable me to give the proof completely and with all detail."

At least, this time we were able to reconstruct the proof. :-)

-Larry

Infinite Descent

Pierre de Fermat was very proud of his technique known as Infinite Descent. He wrote that this method "will lead to marvelous advancement in the theory of numbers" (quoted from: http://fermat.workjoke.com/flt2.htm)

As he often did, Fermat left very few details on how to apply the method. He did provide one example of this method in his proof that the area of a right triangle cannot be equal to a square number. An elegant application of this proof is found in the case of FLT: n=4 where the proof rests on the method of infinite descent and the solution to Pythagorean Triples.

The basic method is very straight forward. One proves that if an assumption is true, it means that the assumption must also be true for an element that is smaller. In other words, an assumption leads to the proposition of an infinite number of cases where it is true.

This technique is especially useful in the domain of positive integers. In this case, infinite descent is impossible since it contradicts the Well Ordering Principle. In other words, there must always be a smallest element in any set of positive integers. But if there is a smallest element, then, there cannot be an infinite descent. In this way, the method can be used as way to prove by contradiction certain negative assumptions. It can also be used to prove a positive conclusion as I will show below.

Fermat is known to have used this technique to prove that there is no square equal to a right triangle. He left a proof using this technique for the case of n = 4 which I will go over in a future blog. He also wrote about its use in the proof for the case of n = 3 in a letter to Christian Huygens.

If Fermat had really found a proof for his theorem, it was without doubt based on this method.

To show the technique in action, I will use it to prove the following theorem:

Thereom: Relatively Prime Divisors of an n-power are themselves n-powers.

This theorem says that if gcd(v,w) = 1 and vw = zⁿ
Then, there exists x,y such that v = xⁿ, w = yⁿ

(1) So, we start with gcd(v,w) = 1, vw = zⁿ
(2) Assume that v is not equal to any number xⁿ
(3) v ≠ 1 since 1 is an xⁿ power
(4) Now, v is divisible by a prime number p. [Fundamental Theorem of Arithmetic]
(5) So, there exists k such that v = pk
(6) p divides z since zⁿ = vw = pkw [By applying Euclid's Lemma]
(7) So, there exists m such that z=pm
(8) So, zⁿ = vw = pkw = (pm)ⁿ = pⁿmⁿ
(9) Dividing p from both sides gives us:
kw = p^(n-1)mⁿ
(10) From Euclid's Lemma, p divides k or w.
(11) It can't divide w since it already divides v and gcd(v,w)=1. Therefore, it divides k
(12) We can apply this same argument for each p in p^(n-1)
(13) So, we can conclude that p^(n-1) divides k.
(14) So, there exists V such that k = p^(n-1)*V
(15) So, kw = p^(n-1)mⁿ = p^(n-1)*V*w
(16) Dividing p^(n-1) from both sides gives us:
Vw = mⁿ
(17) Now, gcd(V,w)=1 since V is a divisor of v and gcd(v,w) = 1
(18) Likewise, V cannot be an n-power. If it were, then v = pⁿV would make v an n-power which goes against our assumption.
(19) Finally, V is less than v since p^(n-1) > 1.
(20) Thus, we have a contradiction by infinite descent.

QED

We proved that assuming that a relatively prime divisor of an n-power is not itself an n-power means that there must necessarily be a smaller relatively prime divisor that is also not an n-power and so on and so on.

Here is what Fermat wrote about Infinite Descent in a letter to another mathematician:

"As ordinary methods, such as are found in books, are inadequate
to proving such difficult propositions, I discovered at last
a most singular method...which I called the infinite descent.
At first I used it to prove only negative assertions, such as
'There is no right angled triangle in numbers whose area is a
square'... To apply it to affirmative questions is much harder,
so when I had to prove 'Every prime of the form 4n+1 is a sum
of two squares" I found myself in a sorry plight (en belle
peine). But at last such questions proved amenable to my methods."
-Quoted from Andre Weil's Number Theory

Pythagorean Triples: Solution

OK, here's the solution for Diophantus's problem of determining the solution for:
x² + y² = z²

(1) We know that we can assume that x,y,z are coprime. [See my previous blog for details]

(2) The second important insight is that z has to be odd.

(a) Assume the opposite that z is even.

(b) Then, there exists another value Z such that z = 2 * Z

(c) Also, z² is then divisible by 4 since:

z² = (2 * Z)² = 4 * (Z²)

(d) We know that x,y must both be odd because of (1).

(e) Since they are odd, there must also exist values X,Y such that:

x = 2 * X + 1
y = 2 * Y + 1

(f) But x² + y² cannot be divisible by 4 since:

x² + y² = (2 * X + 1)² + (2 * Y + 1)² =
= 4X² + 4X + 1 + 4Y² + 4Y + 1 =
= 4[ X² + X + Y² + Y ] + 2

(g) So, we have a contradiction and we reject our assumption.

(3) Since z is odd, either x or y must be even since an odd number is always the sum of an odd and an even number.

(4) Let's assume x is even. The same argument will also work if y is even.

(5) Now, we know that:

x² = z² - y² = (z - y)(z + y)

(6) And, z - y and z + y must be even since z,y are odd.

(7) So, we know that there must exist u,v,w such that:
x = (2u)
z + y = (2v)
z - y = (2w)

(8) Which means that:
(2u)² = (2v)(2w) [From (5) and (7)]

(9) Dividing both sides by 4 gives us:
u² = v * w

(10) We need 1 more insight before the solution. Here it is: v,w are coprime

(a) Assume that v,w are not coprime.

(b) Then, there exists d such that d > 1 and d divides both v,w

(c) Then d divides both v + w and v - w

(d) But:

z + y + z - y =
2v + 2w
So 2z = 2v + 2w which means that z = v + w
So d divides z

(e) And:

z + y - (z - y) = 2v - 2w
So 2y = 2v - 2w which means that y = v - w
So d divides y

(f) Which is a contradiction since z,y are coprime [by (1)].

(g) So, we reject our assumption.

(11) By the properties of coprimes, we know from (9),(10) that v,w are themselves squares (see here for proof). [For those who need a review of coprimes, here is a link.]

(12) So, there exists p,q such that:

v = p²
w = q²

(13) And, we have our solution since:

z = v + w = p² + q²
y = v - w = p² - q²
x = 2u = 2pq [Since u² = vw means u = pq]

We also know that:

(a) p,q are relatively prime. [Otherwise, z,x,y would not be relatively prime]

(b) p,q are opposite parity (that is, one is odd and one is even) [Since z is odd]

(14) For sure enough:

(p² + q²)² = (2pq)² + (p² - q²)²

(15) Now, to generate our answer, we can pick any p,q we want so long as they are integers.

For example, if p = 2 and q = 1, we get:

z = (2)² + (1)² = 5.
y = (2)² - (1)² = 3.
x = 2pq = 2*2*1 = 4

(16) We can do even better than this because we know that for each x,y,z, if they have common factors, the relation still holds.

z = d[p² + q²]
y = d[p² - q²]
x = d[2pq]

For example, if p = 2 and q = 1 and d = 2, we get:
z = (2)(5) = 10
y = (2)(3) = 6
x = (2)(4) = 8

And sure enough, 6² + 8² = 36 + 64 = 100.

QED

What's also nice about this result is that it is not too difficult to apply Fermat's method of infinite descent and prove Fermat's Last Theorem for n=4.

Coprime Numbers

In the last blog, I spoke about Diophantus's problem: to divide a square into the sum of two smaller squares.

In other words, to find solutions for x,y,z where:
x² + y² = z².

The first step in solving this problem is to realize that we can assume that x,y,z are coprime (or another way to say it, relatively prime). That is, no two of these values are divisible by the same prime. So, if p is a prime that is a factor of x, then we know that it is not a factor of y and not a factor of z.

When we have a situation where the three numbers are not coprime (for example, 6,8,10), we will be able to divide out common factors and end up with three numbers that are.

In the case of 6,8,10, the three numbers share the prime 2. If we divide out 2, then we are left with 3,4,5 which are coprime.

This assumption is important because it greatly simplifies the task of analyzing the conditions for when a solution exists. In my next blog, I will show how this assumption gives us the solution to Diophantus's problem.

Interestingly, we can apply this same assumption to Fermat's Last Theorem. From this point on, we will only need to consider the case where x,y,z are relatively prime.

One of my goals in this project is to provide complete proofs each of the conclusions presented. This blog relies on one lemma. A lemma is an intermediate statement that requires proof and is used in a larger theorem.

Lemma: All solutions to xⁿ + yⁿ = zⁿ can be reduced to a form where x,y,z are coprime. [Here is the proof.]

Pythagorean Triples

When Fermat wrote his note in the margin, he was making a comment on the problem of determining Pythagorean Triples.

In Book II, Problem 8 of the Arithmetica, Diophantus poses the problem of how to divide a given square number into the sum of two smaller squares.

In other words, solve the problem:
x² + y² = z².

Any three numbers that satisfy this equation are called Pythagorean Triples. They are called Pythagorean Triples since this is the same equation as the Pythagorean Theorem.

The Pythagorean Theorem is so well known that I refer people to this link if you would like to see a proof for it.

An example of a Pythagorean Triple is 3, 4, and 5 since 3² + 4² = 5².

I encourage everyone who has not already seen the solution to Diophantus's problem to try and solve it. This is without doubt what Fermat did and in solving this problem, he stumbled upon his famous generalization.

If you solve the problem, you should be able to prove there an infinite number of Pythagorean Triples and find a method for listing them out.

You can find the solution here.

This blog is based on the following sources:

• Timeline of Fermat's Last Theorem

Diophantus of Alexandria

When Fermat made his famous note in the margin, he was making a comment on a problem from Diophantus of Alexandria. While Fermat is today considered the father of number theory, he would probably have given this title to Diophantus.

Not much is known about Diophantus's life. He was a Hellenized Babylonian who lived in Alexandria, Egypt. He was born around 200 AD and he died around 284 AD.

His classic work is the Arithmetica which consists of 130 problems in 13 books. Today, only 6 of the books survive. It was written around 240 AD.

He focused on problems that have a positive rational number as a solution (a rational number is any fraction made from two whole numbers). He considered negative numbers and irrational numbers to be "useless," "meaningless," and "absurd."

When he spoke about quadratic equations, he only offered one solution. It is not clear if he knew that all quadratic equations have two solutions.

Diophantus is considered to be the father of algebra. Diophantine equations are named after him.

Much of what we know about him comes from the words written on his tombstone, which is itself an algebra problem for determining his age:

This tomb hold Diophantus. Ah, what a marvel! And the tomb tells scientifically the measure of his life. God vouchsafed that he should be a boy for the sixth part of his life; when a twelfth was added, his cheeks acquired a beard; He kindled for him the light of marriage after a seventh, and in the fifth year after his marriage He granted him a son. Alas! late-begotten and miserable child, when he had reached the measure of half his father's life, the chill grave took him. After consoling his grief by this science of numbers forfour years, he reached the end of his life.

This blog is based on the following sources:

• The Biographies of Mathematicians Web Site
• J.R. Newman, The World of Mathematics
• Wikipedia

Fermat's Achievements

Pierre de Fermat published very little in his lifetime, and yet by most accounts, he is considered to be one of the greatest mathematicians of all time.

I was very surprised when I learned that Fermat's Last Theorem was published after Fermat's death. It is pretty amazing that the mathematical world took a note in the margin so seriously.

Fermat was very precise about his mathematical pronouncements. When he was not certain about a finding, he labeled it a conjecture. Indeed, shortly after Fermat's death, all but one of his theorems had a proof just as Fermat had claimed.

Fermat was born in Toulouse, France on August 17, 1601. He was born into a wealthy family. He studied law at the university in Orleans and became a councillor at the parliament in 1631. He quickly moved up the ranks and in 1652, he became the chief magistrate of the criminal court.

In 17th century France, magistrates were required to spend large amounts of time in isolation. It was during this time, that he worked on mathematics.

His achievements in mathematics are incredible.

Sir Isaac Newton said that his invention of calculus was based a large part on Fermat's method of tangents.

In 1654, Blaise Pascal wrote a letter asking about Fermat's views on probability. Their series of correspondences became the foundation of probability theory.

Rene Descartes is perhaps most famous for his invention of Cartesian coordinates and his classic work La Geometrie. Fermat independently came up with a three-dimensional geometry. While Fermat's version of analytic geometry was more complicated and advanced, Descarte's classic work became more popular because its notation was more convenient. Today, both men are seen as the fathers of analytic geometry.

He made advances in optics. He came up with Fermat's Principle, the idea that light always traverses the path that takes the least time.

With Wiles' amazing proof, we are still left with a question. Did Fermat really have a proof?

Wiles' proof rests on twentieth century mathematics including the theories of elliptic curves, modular forms, and Galois Representations.

Wile's proof goes way beyond the mathematics that Fermat helped to create. It is very possible that Fermat's proposed proof made an incorrect assumption about cyclotomic factorization. This was a mistake that got Gabriel Lame into falsely believing that he had found a proof of Fermat's famous problem (I will be talking more about this in a later blog).

It is also possible that Fermat, in his lifetime, figured out that he had made a mistake. If he had truly found such a marvelous proof, why did he never tell anyone about it while he was alive? Perhaps, he had figured that he had made a mistake but forgot to correct his notes.

Still, the possibility that Fermat had a proof raises an interesting question: is there a way to simplify Wiles' result? Is it possible to derive a result from Wiles' proof that corresponds to the mathematics of Fermat?

The information in this blog was taken from the following sources:

• MacTutor
• Simon Singh's Web Site (author of Fermat's Enigma)
• Wikipedia

The Problem

For this blog, I assume that everyone is familiar with exponents. If you are not, here is an introduction.

Fermat's Last Theorem states for the equation: xⁿ + yⁿ= zⁿ, there are no whole number solutions where x * y * z ≠ 0 and n > 2.

If x * y * z = 0, then it is easy to find a solution. For example (5)ⁿ + (0)ⁿ = (5)ⁿ.

Likewise, if we consider real numbers, then the solution is straight-forward algebra:
z = (xⁿ + yⁿ)^(1/n).

Finally, if n = 2, then we have the Pythagorean Theorem a² + b² = c². This is solveable by any Pythagorean Triple such as 3,4,5 (3² + 4² = 5²) .

I think that this is the real appeal of the problem. It is easily stated and on its surface looks like it shouldn't be too difficult to resolve one way or the other.

Pierre de Fermat rarely published any of his results. He prefered to describe the problem and claim that he had found a solution. This has made the problem even more appealing: did Fermat actually have a proof?

The theorem itself became public without proof in 1670 when Fermat's son, Clement-Samuel published his father's notes. Unfortunately, Fermat was not around to explain his famous theorem because he had died in 1665. Instead, the reader was left with the famous statement of the problem:

"It is impossible for a cube to be written as a sum of two cubes or a fourth power to be written as the sum of two fourth powers or, in general, for any number which is a power greater than the second to be written as a sum oftwo like powers."

And this very mysterious statement about the proof:

"I have a truly marvelous demonstration of this proposition which this margin is too narrow to contain." (both quotes are from Fermat's Engima)

For over 350 years, this problem remained unsolved. Many of the greatest mathematicians were able to make progress on the problem including Leonhard Euler, Carl Friedrich Gauss, and Ernst Kummer but none of these great minds offered a solution.

The solution had to wait until 1995.

Popular Books on Fermat's Last Theorem

Fermat's Last Theorem is one of the most famous math problems of all time. When Andrew Wiles proposed that he had a solution, it was front page news.

There are numerous books which provide a general introduction to the problem and the history of its solution. The most popular one seems to be Simon Singh's Fermat's Enigma. Simon Singh also has a web site.

Additionally, PBS dedicated an entire episode of Nova to the investigation of the theorem and its solution by Andrew Wiles. The Nova episode is called The Proof. PBS has a web site.

The popular books and the Nova episode provide great background on the story and the significance of Wiles' proof. Unfortunately, they provide only a high level coverage of the proof's major ideas.

Many books have claimed to provide details for amateurs. I find all of these books to be very difficult. They are great for more advanced math students, but for most others, they can be quite a challenge.

• Fermat's Last Theorem for Amateurs - Explores the more elementary proofs.
  
• Notes on Fermat's Last Theorem - Great snippets. More a set of hints than actual proofs.
  
• Fermat's Last Theorem: A Genetic Introduction - Very in depth. Goes up to Kummer.
• Algebraic Number Theory and Fermat's Last Theorem - Very in depth. Broad coverage of Algebraic Number Theory.

In this blog, I will try to present details that bridge the gap between these works and Wiles' proof.

Purpose

I have long been fascinated by Fermat's Last Theorem and greatly excited by Andrew Wiles' proof. I have wanted for a long time to explore the history of the problem and also the amazing proof by Wiles.

Just this week, for example, I learned that another mathematician is about to provide a very significant result that builds on the work done by Wiles. For those interested this mathematician, Chandrashekhar Khare, has a home page here: http://www.math.utah.edu/~shekhar/. After I get to Wiles' proof, I will attempt to also analyze Khare's result.

This blog will be an effort to trace the history of Fermat's Last Theorem and work through the details of the second version of Wiles's proof. Fermat has been called the Prince of Amateurs (Bell, Men of Mathematics) so I hope it is appropriate that this blog is also run by a mathematical amateur.

My goal in this blog is to present a set of proofs in a style very much like Euclid. Each proof presented will either rely on a previous blog or rely on a set of identified postulates. I am not a professional mathematician so I may make mistakes. I am hopeful that other participants will provide useful comments to keep the quality of this blog up.

The solution to Fermat's Last Theorem embodies a large part of the history of mathematics including many of its major results and many of its most famous persons. This blog will start with Fermat. For the most part, it will follow a historical flow, going from Fermat to Euler to Gauss to Kummer, etc.

It would be a shame to trace the history of the proof without also looking at the lives of the mathematicians involved. Some blog topics will explore the life and achievements of the participants of this story.

I look forward to people adding their comments and corrections to the biographies and the proofs. I think that this is one of the most exciting stories in all of mathematics and a blog is a great way to explore it.

Cheers,

-Larry