Blaise Pascal

Blaise Pascal was born in Clement, France in 1623. From an early age, he was identified as a prodigy. When he was eleven, he wrote an essay on the sounds of vibrating bodies. His father did not take well to this and forbid him from studying mathematics. At 12, he came up with his own proof that the sum of the angles of a triangle add up to sum of two right angles. His was father was impressed with this and decided that it was ok if Pascal read Euclid's Elements. At the age of 16, Pascal wrote an essay on conic sections which included a result known today as Pascal's Theorem. At age 18, he created a mechanical calculator capable of addition and subtraction (he was the second person to do this; Wilhelm Schickard had built the first one in 1624).

Pascal's father, Etienne, was a local magistrate and was an acquaintance of many of the most famous intellectuals of his day including Rene Descartes, Pierre Gassendi, and Girard Desargues. It is said that when Descartes saw Pascal's writing on conic sections, he refused to believe that the boy had written it. Pascal's mother died when he was three years old so he was raised and educated by his father.

In 1653, Pascal wrote his famous treatise on what is today known as Pascal's Triangle. This is a method for determining the binomial coefficients for a given value of (a + b)ⁿ. Interestingly, Pascal was not the first to come up with this method. It had been discovered 400 years earlier by the Chinese mathematician Yang Hui.

In 1654, Pascal became interested in figuring out the odds associated with gambling. He wrote some letters to Pierre de Fermat seeking his opinion and their correspondences became the foundation for Probability Theory.

Pascal made numerous contributions to science. He studied the effects of air pressure on fluids and correctly proposed that a vacuum explained the the dynamics of mercury in a barometer (at the time, it was not clear how mercury-based barometers worked). His experiments with barometers attracted attention across Europe. At the time, Descartes argued that a vacuum could not exist.

In 1654, Pascal's carriage almost fell off a bridge. He was crossing the bridge when his horses lost their step and went over. The carriage would have followed but the reins broke and the carriage balanced over the edge of the bridge. Pascal and his passenger managed to get back to the bridge but fifteen days later, Pascal had a religious experience. After this, he began writing works of a philosophical and religious nature. His Provincial Letters and Pensees became very famous. The Pensees was released after his death.

In 1659, Pascal grew seriously ill; through out his life, his health had been bad but this time, he did not recover. In 1662, when he was 39, he died.

Today Pascal stands as one of the giants of his age. The pascal is the SI unit of pressure and his literary works are considered among the greatest of his time. He is seen as one of the fathers of the computer and his work with Fermat on probability theory is at the foundation of economics and the social sciences.

References

• Blaise Pascal, Wikipedia
• Blaise Pascal, MacTutor
  

Fermat's Last Theorem: n = 5 : Overview

The proof for Fermat's Last Theorem: n=5 turns out to be more complicated and more difficult than any of the previous proofs. This is perhaps not too surprising since otherwise, Euler or Gauss would have most likely found the proof.

The final proof was the work of two very talented mathematicians: Adrien-Marie Legendre and Johann Peter Gustav Lejeune Dirichlet. The solution rested on the work done on the Binomial Theorem and Continued Fractions.

Since the purpose of this blog is to present a complete set of proofs, I will need to take a step back from Fermat's Last Theorem and review the developments of these two ideas. The Binomial Theorem was the result of three major mathematicians: Pierre de Fermat, Blaise Pascal, and Sir Isaac Newton. Major work on continued fractions was done by Joseph Louis Lagrange.

I will not attempt to do a complete survey of either of these very broad fields. Instead, I will review the lives of the mathematicians mentioned above and go through the proofs of the major results which are used later in the proof for n=5.

Sophie's Proof

Sophie's Proof is the proof discovered by Sophie Germain that later led to the proof of Fermat's Last Theorem for n =5. The proof is very important historically in that it represents a new approach to the problem. Rather than demonstrating that no solution exists, Sophie shows for certain values of n, n must divide x,y, or z.

After this proof, it became traditional to divide Fermat's Last Theorem into two parts:

Case I: None of the three values x,y,z is divisible by n.
Case II: n divides 1 of the three values x,y,z.

Here is Sophie Germain's Theorem:

Theorem: If xⁿ + yⁿ = zⁿ and n is a prime ≥ 3 and 2n+1 is a prime, then n must divide xyz.

(1) We can assume that x,y,z are relatively prime to each other. [See here]

(2) So, let's assume that n doesn't divide xyz.

(3) Since n is odd, we can set z' to -z and get:
xⁿ + yⁿ + (z')ⁿ = 0.

(4) We can rewrite this to be [see here for details]
-xⁿ = (y+z)(y^(n-1) - y^(n-2)z + ... - yz^(n-2) + z^(n-1))

(5) From this, we can show that:
y+z and y^(n-1) - y^(n-2)z + ... - yz^(n-2) + z^(n-1) are relatively prime since:

(a) Assume that they are not relatively prime.
(b) Then, there exists a prime p such that p divides y+z and p divides y^(n-1) - y^(n-2)z + ... - yz^(n-2) + z^(n-1).
(c) from this, we know that z ≡ -y (mod p).
(d) Using the fact that they are congruent (see here), we get:
y^(n-1) - y^(n-2)z + ... - yz^(n-2) + z^(n-1) ≡
y^(n-1) - y^(n-2)(-y) + ... -y(-y)^(n-2) + (-y)^(n-1) ≡
y^(n-1) + y^(n-1) + ... + y^(n-1) + y^(n-1) ≡ (n)y^(n-1)
(e) So from Euclid's Lemma, p either divides n or it divides y^(n-1).
(f) But p doesn't divide n since:

(i) n is prime so p would equal n (a prime is only divisible by itself or 1)
(ii) And then n would divide (-x)ⁿ
(iii) Which means that it would divide x (from Euclid's Generalized Lemma)
(iv) But this impossible by step #2.

(g) And p doesn't divide y^(n-1) because then p would divide y (Euclid's Generalized Lemma) and p would also divide z (since it divides y and it divides y+z), but this is impossible by step #1.
(h) So we have a contradiction and we reject our assumption.

(6) From Step #5 (see here), we can conclude that there exists a such that:
y + z = aⁿ

(7) But we can make this same argument for z+x and x + y where:
-yⁿ = (x + z)(x^(n-1) - x^(n-2)z + ... -xz^(n-2) + z^(n-1))
-zⁿ = (x + y)(x^(n-1) - x^(n-2)y + ... -xy^(n-2) + y^(n-1))

(8) So we know that there exists b,c where:
z + x = bⁿ
x + y = cⁿ

(9) We also know that any value uⁿ ≡ ±1 or 0 (mod 2n + 1) since:

(a) Assume that 2n+1 doesn't divide uⁿ [ we only need to consider the case where it is not congruent to 0]
(b) Since 2n+1 is a prime, we can apply Fermat's Little Theorem to get:
u²ⁿ ≡ 1 (mod 2n+1)
(c) And this means that:
(uⁿ)² ≡ 1 (mod 2n + 1)
(d) So that uⁿ ≡ ± 1 (mod 2n+1) [See here for review of modular arithmetic]

(10) From this, we can conclude that (2n+1) divides xyz' since:

(a) Assume 2n+1 doesn't divide xyz'.
(b) Then, from step #9, we have:
xⁿ ≡ ± 1
yⁿ ≡ ± 1
zⁿ ≡ ± 1
(c) But this is impossible since:
xⁿ + yⁿ + (-z)ⁿ ≡ 0 (mod 2n+1)
but
± 1 ± 1 ± 1 cannot be congruent to 0 (mod 2n+1)
(d) So, we reject(a).

(11) Let's assume that (2n+1) divides x. (We will be able to make the same type of argument if it divides y or z')

(12) Then 2x = bⁿ + cⁿ + (-a)ⁿ since from step #6 and step #8, we know that:
2x = z + x + x + y -y - z = bⁿ + cⁿ + (-a)ⁿ

(13) And we can conclude that (2n+1) divides acb since;

(a) We assumed that 2n+ 1 divides x. This gives us:
bⁿ + cⁿ + (-a)ⁿ ≡ 0 (mod 2n+1)

(b) Now, we can apply the same argument as Step #10 since:

(i) Assume that 2n+1 doesn't divide acb

(ii) From step #9 and (i) above, we can conclude that bⁿ, cⁿ, (-a)ⁿ ≡ ± 1.

(iii) But this is impossible since:
± 1 ± 1 ± 1 cannot be congruent to 0 (mod 2n+1)

(iv) So we can reject our assumption in (i) and conclude that 2n+1 divides acb.

(14) But (2n+1) can't divide abc since:

(a) (2n+1) can't divide b since:

(i) 2n+1 divides b → 2n+1 divides bⁿ → 2n+1 divides z + x [since by step #8, z + x = bⁿ.]
(ii) We know that 2n+1 divides x, so if it divides z+x, it must also divide z.
(iii) But this is impossible since x,z have no common factors by step #1.

(b) (2n+1) can't divide c for the same reason as above. That is, x,y have no common factors by step #1.

(c) (2n+1) can't divide a since:

(i) Assume (2n+1) divides a.
(ii) Then (2n+1) divides y + z which means that z ≡ -y (mod 2n+1)
(iii) We know from Step #5, there exists a value d, e such that:
dⁿ = y^(n-1) - y^(n-2)z + ... - yz^(n-2) + z^(n-1).
eⁿ = x^(n-1) - x^(n-2)y + ... -xy^(n-2) + y^(n-1).

(iv) Applying (ii) gives us:
dⁿ ≡ (n)y^(n-1) (mod 2n+1) [See Step #5d]

(v) Since we know that x ≡ 0 (mod 2n + 1) [from Step #11], we get:
eⁿ ≡ (0)^(n-1) - (0)^(n--2)y + ... -(0)y^(n-2) + y^(n-1) ≡ y^(n-1) (mod 2n+1)

(vi) Appyling (iv), this gives us:
dⁿ ≡ (n)eⁿ (mod 2n+1)

(vii) From Step #9, we know that dⁿ,eⁿ are 0,1, or -1. This means that they must both be equal to 0 since this is the only way that 0 ≡ n*0.

(viii) But if 2n+1 divides both dⁿ,eⁿ, then it divides both d,e from Euclid's Lemma

(ix) But 2n+1 divides e implies that it also divides y (from (v).)

(x). But then it divides both x,y which goes against step #1.

(xi). So we have a contradiction and we reject (i).

(15) And this means that step #14 contradicts step #13 so we reject Step #2.

QED

Corollary: x⁵ + y⁵ = z⁵ → 5 divides xyz.

This is true by above since 5 is a prime and 2*5+1 = 11 is a prime.

QED

References

• Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.
  

Fermat's Little Theorem

Despite its name, Fermat's Little Theorem is one of Pierre de Fermat's most important theorems. Fermat first presented it without proof in one of his letters in 1640. Leonhard Euler provided the first published proof in 1736. The theorem is very useful as a way of testing very large primes.

In a nutshell, the proof states that given an integer a and a prime p that does not divide a, then we know that a^(p-1) has a remainder of 1 when divided by p.

From this theorem, then we know that:
47² ≡ 1 (mod 3).
110⁶ ≡ 1 (mod 7).
etc.

From the basic ideas of modulus (see here for review), we also know that:
(47²)ⁿ ≡ 1 (mod 3)
(110⁶)ⁿ ≡ 1 (mod 7)

This theorem is also the foundation of Sophie's Proof (see here for the details), an important result from the mathematician Sophie Germain.

Theorem: If an integer a is not divisible by a prime p, then a^p-1 ≡ 1 (mod p).

(1) From the Division Algorithm, we know that any number divided by an integer p has a remainder between 0 and p-1.

(2) This means that for any value a there is exactly p-1 possible congruences.
a ≡ 0 (mod p) or
a ≡ 1 (mod p) or
...
a ≡ p-1 (mod p)

(3) Consider the values 0a, 1a, 2a,... (p-1)a.

(4) It turns out that each of these values (0a, 2a, ...) is equal to a unique congruence when divided by p since:

(a) Assume that 2 of the values are equal to the same congruence: let's say ia, ja
(b) So ia ≡ ja (mod p)
(c) This tells us that p divides ia - ja (from the definition of ≡)
(d) Which means that p divides a(i-j).
(e) From Euclid's Lemma, we know that p must divide i-j (since a is not divisible by p)
(f) But then i = j since both values are between 0 and p.

(5) So, this tells us that:
(1a)(2a)...(p-1)a ≡ (1)(2)...(p-1) (mod p). [Applying the principles of modular arithmetic]

(6) We also know that (1a)(2a)*...*(p-1)a = a^(p-1)(p-1)! [See here for a review of factorials such as (p-1)!]

(7) Putting (5) and (6) together gives us:
a^(p-1)(p-1)! ≡ (p-1)! (mod p).

(8) Since (p-1)! is not equal to 0, we can divide this from both sides of the modulus (see Lemma 4, here) and get:
a^(p-1) ≡ 1 (mod p).

QED

Sophie Germain

Sophie Germain was born on April 1, 1776 at the dawn of the French Revolution. Her father was a very successful merchant who was able to avoid the great unrest of the revolution.

When Sophie was 13, she discovered a book on the history of mathematics in her father's library. She became fascinated with the story of Archimedes who was killed by a Roman soldier while he was working on a geometry problem in the sand. Later, she became interested in the works of Leonhard Euler and Sir Isaac Newton. Her parents thought it was inappropriate for a young girl to be so interested in mathematics and tried to discourage her. There is a story where they tried to hide her candles and her warm clothes so that she would not be able to study math at night.

In 1794, when Sophie was 18, the Ecole Polytechnique opened in Paris. This was a school founded upon the principle of training France's most talented mathematicians and scientists. Unfortunately for Sophie, no women were allowed to enroll. Sophie found a way around this restriction. She learned of a Monsieur LeBlanc who had enrolled in the Polytechnique but was now traveling out of Paris. She took on his identity, picking up his assignments, and handing in his homework. The plan went very well until some of her math work caught the attention of the mathematician Joseph Louis Lagrange.

Lagrange was one of the most famous mathematicians of his day. He had worked closely with M. LeBlanc and was now surprised to see a student which had showed little aptitude for mathematics previously was now doing outstanding work. He requested to set up a meeting. It was at this point that Sophie admitted her ruse. Lagrange was quite impressed.

It was perhaps this experience with Lagrange that led Sophie to start a correspondence with the most famous mathematician of the day, Carl Friedrich Gauss in 1804. Uncertain if Gauss would respond well to a woman, she again used the identity of M. LeBlanc. Gauss was quite impressed with the content of the letter and from this point on, Gauss and Sophie continued to exchanged letters on a regular basis.

It was because of Napoleon in 1807 that Gauss found out about Sophie's true identity. Napoleon's army was heading into the Brunswick where Gauss lived. The invading force was led by General Pernety who was a personal friend of Sophie's. She asked the General to ensure Gauss's safety. The general told Gauss that he was under the protection of one Sophie Germain who Gauss had never heard of. After this, Sophie admitted her true identity.

The response by Gauss is often quoted (the translation below is taken from Wikipedia):

But how to describe to you my admiration and astonishment at seeing my esteemed correspondent Monsieur Le Blanc metamorphose himself into this illustrious personage who gives such a brilliant example of what I would find it difficult to believe. A taste for the abstract sciences in general and above all the mysteries of numbers is excessively rare: one is not astonished at it: the enchanting charms of this sublime science reveal only to those who have the courage to go deeply into it. But when a person of the sex which, according to our customs and prejudices, must encounter infinitely more difficulties than men to familiarize herself with these thorny researches, succeeds nevertheless in surmounting these obstacles and penetrating the most obscure parts of them, then without doubt she must have the noblest courage, quite extraordinary talents and superior genius. Indeed nothing could prove to me in so flattering and less equivocal manner that the attractions of this science, which has enriched my life with so many joys, are not chimerical, the predilection with which you have honored it.

Sophie became fascinated in particular with Fermat's Last Theorem. She developed a new line of attack. Rather than proving that there were no solutions to a given value n, she showed that if there was a solution, a certain condition would have to apply. This insight would later lead to the proof for Fermat's Last Theorem where n=5.

In 1811, Sophie submitted an entry to a contest sponsored by the French Academy of Sciences. The goal was to provide a solution to a fundamental mathematical problem in the physics of elasticity. Sophie's was the only submission for that year. Unfortunately, her solution was not accepted. In 1816, on her third submission, Sophie's solution was accepted. This was a very important achievement in establishing Sophie Germain as a professional mathematician. Her paper would later become one of the foundations of the modern theory of elasticity.

Throughout her life Sophie Germain never married. She relied on the support of her father in order to continue her studies in mathematics. She became the first woman invited to attend the Academy of Sciences sessions. In 1830, she was awarded an honorary PhD.

Sophie Germain died in 1831 at the age of 55 after a two year struggle with breast cancer. Today, Sophie Germain is considered one of the most talented mathematicians of all time.

References:

• Sophie Germain, Wikipedia
• Sophie Germain, Biographies of Women's Mathematicians Web Site
• Math's Hidden Woman, PBS Web Site

Another False Proof: Russian Professor claims to have solved Fermat's Last Theorem

Today, I learned of a news story that a professor in Russia claims to have solved Fermat's Last Theorem in 3 lines. The story has also been covered in Italy here and here. (For those who don't speak Russian, here is an English translation. For those who don't speak Italian, here and here are English translations of the Italian articles).

It looks like the Russian article is the only one that gives the details of the proof. So, I am relying on this translation.

Let me start by saying that it is very important to be skeptical of anyone claiming to have a solution. In May, the Manilla Times stated that a local man found a mistake in the proof by Andrew Wiles. This of course turned out to be a hoax.

There are two tell-tales sign that the proof is most likely flawed. First, the professor who claims to have found the proof is not an expert in number theory. Second, the proof has not been reviewed by anyone who is an expert of number theory. From the mathematician Gabriel Lame to the very famous Leonhard Euler, it is very easy to make mistakes when it comes to number theory. Even Fermat made a mistake when it came to his conjecture about what are today known as Fermat primes.

Now, onto his proposed proof. After reviewing an English translation of the article, I found (no surprise) that the proof is not a valid demonstration of Fermat's Last Theorem.

First, let me summarize the proof:

(1) Assume that Fermat's Last Theorem is true.
(2) Then, there exists xⁿ + yⁿ = zⁿ with n ≥ 3.
(3) For any x,y, we can create a right triangle and so we can suppose a value r such that:
r² = x² + y².
(4) Since n ≥ 3, we know that z is less than r. [I will add the details for this later]
(5) We can construct a triangle based on z,x,y. Since z is less than r, we know that the angle opposite z (let's call it angle B) must be less than 90 degrees (and of course, greater than 0 degrees).
(6) Now, using the Law of Cosines, we know that:
z² = x² + y² - 2xycosB.
(11) Since B is greater than 0 and less than 90, we know that cosB cannot be a whole number. [See here for a Cosine look up table]

QED

OK, so if the steps are valid, is Fermat's Last Theorem proven? Not at all. Just because cos B is not a whole number doesn't prove that z is not a whole number. That is ultimately what must be proven.

The professor must either prove that cos B is irrational or he must show that based on 2xycosB, z cannot be a whole number.

Since cos B is a continuous function, it necessarily covers rational values and therefore it is quite possible that 2xycosB is a whole number.

Thanks go out to David (see the comments section) for providing his thoughts.

Proof for n=3: Eistenstein Integers: Step Two

If you are not familiar with Eisenstein Integers, please start here. The details of today's blog are based on an English translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.

Lemma: There is no Eisenstein integer whose norm is 2.

(1) The norm of a G-number aJ + bO is a² + b² - ab.

(2) We note first that both abs(a) and abs(b) need to be greater than 1.

Case I: 0,0 --> 0² + o² - (0)(0) = 0

Case II: 1,0 --> 1² + o² - (1)(0) = 1

Case III: 0,1 --> 0² + 1² - (0)(1) = 1

(3) There is no solution where the min(a,b) ≥ 2.

(a) In all cases, a² + b² - ab ≥ min(a,b)²

Case I: a=b: a² + b² - ab = a² + a² - a² = a²

Case II: b greater than a: this implies that b² is greater than ab and therefore: a² + (b² - ab) is at least equal or greater than a²

Case III: The same argument applies only swap a for b.

(b) So, in this case, if min(a,b) ≥ 2 → Norm ≥ 4.

(4) But there is also no solution for max(a,b) ≥ 2.

Assume b greater than a: this implies that 2b + 1 is greater than a which implies that 2b + 1 - a is greater than 0. Adding the equation to both sides, this gives us that a² + (b+1)² - a(b+1) is greater than a² + b² - ab.

So all we need to do is to show that the equation doesn't hold for b=2, a=0 and b=2,a=1. We know that any other value of b will be greater than our result here and a must be less than 2.

Case I: b=2,a=0 (0)² + (2)² - (0)(2) = 4.

Case II: b=2,a=1 (1)² + (2)² - (1)(2) = 3.

QED

Lemma: If J - O divides αβγ where αβγ ≠ 0, then we have reached a position of infinite descent.

(1) Let us suppose that J-O divides α. [We can apply the same argument where it divides β or γ since we are assuming a form where α³ + β³ + γ³ = 0]

(2) We can assume that J-O does not divide β nor γ since all three are relatively prime to each other.

(3) For purposes of clarity, let's let ζ = J - O.

(4) So, α³ ≡ 0 (mod ζ³). [J-O divides ζ implies (J-O)³ divides ζ³; Start here if you unfamiliar with the '≡' symbol.]

(5) We also know that β³ + γ³ ≡ 0 (mod ζ³) [See Definition 1, here if this point is not clear]

(6) Since J-O doesn't divide β³ or γ³, we can conclude that:

β³ ≡ e (mod 9), γ³ ≡ f (mod 9) where e=±1 and f= ±1. [See Lemma 8, here for proof]

(7) Now since ζ³ divides 9 (see Corollary 9.1 here), we also have:

β³ ≡ e (mod ζ³), γ³ ≡ f (mod ζ³) [From step #6 above]

(8) Now e + f ≡ 0 (mod ζ³). [From Step #5 above since f + e ≡ β³ + γ³ ≡ -α³ ≡ 0 (mod ζ³)]

(9) So, f = -e . [From step #8 above and step #6]

(10) But then e + f = 0 and e + f ≡ 0 (mod 9); and therefore β³ + γ³ ≡ 0 (mod 9). [Since from step #6, e + f ≡ β³ + γ³ (mod 9)

(11) Which gives us that α ≡ 0 (mod 9) [since -α = β³ + γ3] and since 9 = ζ⁴:
α³ ≡ 0 (mod ζ⁴). [See here for proof of 9 = (J-O)⁴]

(12) And since J-O divides α, we know that:
α ≡ 0 (mod ζ²).

(13) Let us now define 3 values: η, μ, χ where:
η = βJ + γO
μ = βO + γJ
χ = β + γ

(14) Now, it follows that: β³ + γ³ = ημχ since:

(a) (βJ + γO)(βO + γJ)(β + γ) =
(β²OJ + βγJ² + βγO² + γ²OJ)(β + γ) = [β² + γ² + βγ(-O) + βγ(-J)](β + γ) =
(β² + γ² - βγ)(β + γ)

(b) (β² + γ² - βγ)(β + γ) =
β³ + β²γ + βγ² + γ³ - β²γ - βγ² = β³ + γ³

(15) Now, it also follows that each value η, μ, χ is divisible by J-O since:

(a) (J-O)³ divides ημχ. [Since J-O divides α and α³ = ημχ]

(b) J - O divides μ - η since μ - η = ( βO + γJ) - (βJ + γO)=(J-O)(γ - β)

(c) Also note that μ + η = βO + γJ + βJ + γO = β + γ = χ

(d) From (a), we know that J-O divides μ, η, or χ. [See Euclid's Generalized Lemma for details since J-O is a prime.]

(e) This is enough to show that J-O divides all three since:

Case I: J-O divides μ

(i) Step 15b gives us that J - O divides μ - η

(ii) Since J-O divides μ, it must also divide η.

(iii) Then J-O divides χ, since χ = μ + η from step #c above.

Case II: J-O divides η,

(i) Step 15b gives us that J - O divides μ - η

(ii) Since J-O divides η, it must also divide μ.

(iii) Then J-O divides χ, since χ = μ + η from step #c above.

Case III: J-O divides χ

(i) Then J-O divides μ + η since χ = μ + η from step #c above.

(ii) But J-O divides μ - η from step 15b above.

(iii) This means that J-O divides both 2μ = (μ + η) + (μ - η) and 2η = (μ + η) - (μ - η).

(iv) Now, J-O is prime (see Lemma 5, here), we can apply Euclid's Lemma for Gaussian Integers (see Corollary 3.3, here) to establish that for each J-O divides 2 or J-O divides both μ and η.

(v) J-O does not divide 2.

In order for J-O to divide 2, then N(J-O) must divide N(2) [This is the same as the argument for norms for Gaussian Integers, see Corollary 7.1, here ]. N(J-O) = 3 [See Lemma 5, here] and N(2) = a² + b² - ab = 2² + 0 - 0 = 4 [See Lemma 3, here] And clearly, 3 does not divide 4.

(16) So, there exists: μ', η', χ' such that:
μ = μ'(J-O)
η = η'(J-O)
χ = χ'(J-O)

(17) And we can further show that μ', η', χ' are all coprime since:

(a) If any two have a common divisor, then μ', η' would have a common divisor.

Case I: μ', χ' have common divisor δ: μ' = μ''δ, χ' = χ''δ. (η')(J-O) + (μ')(J-O) =(χ')(J-O) so η' + μ' = χ' so that η' = χ' - μ' = δ(χ'' - μ'').

Case II: η', χ' have common divisor. You can apply the same argument as above.

(b) Assume η', μ' have a common divisor δ which is not a unit.

(c) Then δ divides β - γ since:
μ' - η' = (β - γ) [From 13b above]

(d) And δ divides β + γ since:
μ + η = β + γ [From 13c above and the fact that δ divides μ + η]

(e) So, δ divides both 2β and 2γ since:
2β = (β + γ) + (β - γ)
2γ = (β + γ) - (β - γ)

(f) Now, since β, γ are relatively prime and 2 is an Eisenstein prime (see below), we can conclude that δ = 2.

We know that any number that is only divisible by itself and 1 is a prime. We also know that if a number is divisible by another number, then its norm is likewise divisible by the norm of the other number.

The norm(2) = 2J+2O = 2² + 2² - (2)(2) = 4.

4 is divisible by 1,2,4. So, to prove that 2 is a prime, we only need to prove that there is no Eisenstein integer whose norm is 2.

We already proved this in the lemma above.

(g) But the we have a contradiction since η is not divisible by 2. Here is the details:

Since δ = 2, both β + γ and β - γ are divisible by 2.

This means that there are four possible states where λ, θ are half the value - a unit.:

Case I: β has a form 2λ + unit, γ has a form 2θ + unit.
Case II: β has a form 2λ + unit, γ has a form 2θ - unit.
Case III: β has a form 2λ - unit, γ has a form 2θ + unit.
Case IV: β has a form 2λ - unit, γ has a form 2θ - unit.

Let's look at η = βJ + γO. So:

CaseI: μ = (2λ + unit)J + (2θ + unit)O = 2λJ + (unit)*J + 2θO + (unit)*O = 2(λJ + θO) + unit.

Case II: gets us: 2λJ + unitJ + 2θO = unitO = 2(λJ + θO) + unit(J-O)

Case III: gets us: 2λJ - unitJ + 2θO = unitO = 2(λJ + θO) - unit(J-O).

Case IV. gets us: 2(λJ + θO) - unit.

(h) So we reject our assumption (b).

(18) Since J-O divides α, we know that there exists a value ω such that:

α = (J-O)*ω

Which also gives us that:

(α)³ = (J-O)³(ω)³ = -ημχ

And dividing both sides by (J-O)³ gives us:

(ω)³ = -η'μ'χ'

(19) Since η', μ', and χ' are coprime, we can know that they are equal to a cube multiplied to a unit. [See here for the proof on this.]

So we could suppose that there exist: ι, ν, ξ which are Eisenstein Integers and ο, ρ, τ which will represent units. With this, we can assume:

μ' = ο * ι³
η' = ρ * ν³
-χ' = τ * ξ³

(20) Putting it all together, gives:

ω³ = ορτ*ι³ν³ξ³ where ορτ = a unit.

(21) Since μ' + η' = χ', we also have:

ο * ι³ + ρ * ν³ + τ * ξ³ = 0.

(22) Since μ', η', and χ' are coprime so: ι, ν, ξ must be coprime.

(23) Now, we also know that there exists κ such that:
ω = κ * ινξ.

So that κ³ = ορτ = a G-unit. Since J-O is not a unit, we know it doesn't divide κ and we have κ ≡ E (mod 9) which means that ορτ ≡ E (mod 9).

So that ορτ = E and E² = 1.

(24) Since we know that from step (10) that α ≡ 0 (mod ζ²), we know that:

ω ≡ 0 (mod J-O).

So we know that J-O divides ι, ν, or ξ. Let us suppose that it divides ι so that we have:

ι ≡ 0 (mod J-O)
ν not ≡ 0 (mod J-O)
ξ not ≡ 0 (mod J-O)

So that we have:

ν³ ≡ e (mod 9) where e² = 1
ξ³ ≡ f (mod 9) where f² = 1.

And then since ρ * ν³ + τ * ξ³ ≡ 0 (mod [J-O]³), and since (J-O)³ equals 9, we know that:

ρ *ν³ + τ * ξ³ ≡ 0 (mod 9) and applying above, we he have: e*ρ + f*τ ≡ 0 (mod 9) and since e*ρ + f*τ is less than 9, we know that e*ρ + f*τ = 0.

If we set F = -e/f then we have:
τ = Fρ

Combining this with E = ορτ, we get:
E = ορ²F.

And we know that F² = (-e/f)² = e²/f² = 1.

Multiplying Fρ² to both sides of ο * ι³ + ρ * ν³ + τ * ξ³ = 0, we get:

E* ι³ + Fρ³* ν³ + (F²)τ³ * ξ³ = E* ι³ + Fρ³* ν³ +τ³ * ξ³ = 0

(25) Now, if we set α' = Fρν, β' = τξ, γ' = Eι, then we get:

α'³ + β'³ + γ³ = (F²)Fρ³* ν3 + τ³ * ξ³ + (E²)Eι^3.

(26) But, we know that α'*β'*γ' is divisible fewer times by J-O than α*β*γ is since:

α³ = (J-O)³(μ'η'χ') and
μ'η'χ' = ε(α'³*β'³*γ'³) where ε is a unit.

(27) We can then apply the same reasoning to these values which prove infinite descent.

QED

Proof for n=3: Eisenstein Integers: Step One

If you are not familiar with Eisenstein Integers, please start here. The details of today's blog are based on an English translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.

Lemma: α³ + β³ + γ³ = 0 → that J-O divides one and only one of α, β, or γ.

(1) Assume that J-O doesn't divide any of them.

(2) We know that there exists e,f,g such that:

α³ ≡ e (mod 9)
β³ ≡ f (mod 9)
γ³ ≡ g (mod 9)

(3) e² = f² = g² = 1. [See here for proof]

(4) But then e + f + g is not ≡ 0 (mod 9) since: [See here for a review of modular arithmetic]

Case I: 1,1,1 : e + f + g ≡ 3 (mod 9)
Case II: -1,1,1 : e + f + g ≡ 1 (mod 9)
Case III: -1,-1,1 : e + f + g ≡ -1 (mod 9)
Case IV: -1,-1,-1 : e + f + g ≡ -3 (mod 9)

(5) And yet this contradicts α³ + β³ + γ³ ≡ 0 (mod 9).

(6) So we reject our assumption at step 1.

QED

Corrolary: Only 1 of the three values is divisible by J - O

(1) We know this since α, β, γ are relatively prime to each other.

Proof for n=3: Using Eisenstein Integers

If you are not familiar with Eisenstein Integers, please start here. The details of today's blog are based on an English translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.

For those who would like to see this proof done in terms of rational integers only, check out this previous blog.

As in previous blogs, I will use Greek letters to represent quadratic integers and Latin letters to represent rational integers.

Theorem: The equation α³ + β³ = γ³ does not have any integer solutions where
α, β, γ, are Eisenstein integers and α * β * γ ≠ 0.

(1) Since Eisenstein Integers are Euclidean (proof), we know that they are characterized by a Division Algorithm (proof), Bezout's Identity (proof), and Unique Factorization (proof).

(2) We can assume that α, β and γ are coprime. [See here for proof]

(3) If we set ζ = -γ, then we have:
α³ + β³ + ζ³ = 0.

(4) Let:

J = (1 + i√3)/2
O = (1 - i√3)/2

(5) Then, J - O is an Eisenstein prime number (proof ) which divides α*β*ζ [See here for proof.]

(6) But if α * β * γ ≠ 0, then we have an infinite descent. [See here for proof.]

QED

Eisenstein Integers

Eistenstein integers are quadratic integers of the form Z[(-1 + √-3)/2]. They are named in honor of Ferdinand Eisenstein. For those, who need a review of quadratic integers, start here.

In today's blog, I will go over some basic properties of Eisenstein Integers. Later on, I will show how they can be used to prove that Fermat's Last Theorem has no solutions for n=3.

The details of today's blog are based on an English translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.

For purposes of the proof, Dorrie proposed the idea of G-numbers. There are numbers that are based on two values: J,O where

J = (1 + i√3)/2
O = (1 - i√3)/2.

NOTE: In case this is not clear to anyone, i√3 = √-3.

Lemma 1: Every integer g is also a G-number.

g = gJ + gO = (g + gi√3)/2 + (g - gi√3)/2 = 2g/2 = g.

QED

Lemma 2: The following properties are true:

(a) J + O = 1

From the previous lemma.

(b) JO = 1

JO = (1 - [(-1)(3)])/4 = 4/4=1

(c) J² + O = 0

J² = (1 + i√3)²/4 = (1 + 2i√3 + (-1)(3)) /4 = (-2 + 2i√3)/4 =
(-1 + i√3) /2

(d) O² + J = 0

O² = (1 - i√3)²/4 = (1 - 2i√3 + (-1)(3))/4 = (-2 - 2i√3)/4 =
(-1 - i√3) /2

(e) J³ = -1

J² = -O [From (c)]
So, J³ = -JO = -1 [From (b)]

(f) O³ = -1

O² = -J [From (d)]
So, O³ = -OJ = -1 [From (b)]

QED

Lemma 3: The norm of a G-number aJ + bO is a² + b² - ab

(1) aJ + bO = (a + ai√3)/2 + (b - bi√3)/2 = (a + b)/2 + [(a - b)i√3]/2

(2) Norm(aJ + bO) = [(a + b)/2 + [(a - b)i√3]/2][(a + b)/2 - [(a - b)i√3]/2] =
(a + b)²/4 - (a - b)²(-1)(3)/4 =
[a² + 2ab + b²]/4 - (-3)[a² - 2ab + b²]/4 =
(a² + 3a² + b² + 3b² + 2ab - 6ab)/4
= a² + b² - ab.

QED

Corollary 3.1: The norm is always greater or equal to 0.

In other words, we can prove that a² + b² ≥ ab.

Case I: a=b

a² + b² = 2a² which is ≥ ab = a².

Case II: a is greater than b

In this case: a² is greater than ab.

Case III: a is less than b

Same argument using b instead of a

QED

Lemma 4: The following numbers are units: J,-J, O, -O, 1,-1

(a) The norm of each of these values is 1. [See here for the explanation of why the norm of a unit is one]

J = 1(J) + 0(O) = 1 + 0 + 0 = 1
-J = (-1)J + 0(O) = 1 + 0 + 0 = 1
O = 0J + 1(O) = 0 + 1 + 0 = 1
-O = 0J + (-1)O = 0 + 1 + 0 = 1
1 = (1)J + (1)O = 1 + 1 - 1 = 1
-1 = (-1)J + (-1)O = 1 + 1 -1 = 1

(b) There are no other units


Norm(Z[(-1 + i√3)/2]) =[(a + bi√3)/2][(a - bi√3)/2] = (a² - b²(-3) )/4= ±1

This means that: (since the norm in this case must be positive)

a² + 3b² = 4.

We know that absolute(b) has to be less than 2. Otherwise, 3b² is greater than 4.

This means that we only need to consider b=-1,0,1.

If b = 0, a = ±2.
If b = 1 or b = -1 a = ±1

So, the only units are:

a=2,b=0: (2+0i√3)/2 = 1
a=-2,b=0 (-2+0i√3)/2=-1
a=1,b=1 (1 + i√3)/2 = J
a=-1,b=1 (-1 + i√3)/2 = -O
a=1, b=-1 (1 - i√3)/2 = O
a=-1, b=-1 (-1 - i√3)/2 = -J

QED

Lemma 5: J-O is a prime.

(1) Norm(J-O) = (1)² + (-1)² - (1)(-1) = 3.
(2) Only primes have norms that are rational integer primes. [The argument is the same as the one for Gaussian Integers, see here]

QED

Lemma 6: If aJ+bO is divisible by J-O, then a + b is divisible by 3.

(1) Since aJ + bO is divisible by J-O, there exists a value mJ + nO such that:

aJ + bO = (mJ + nO)(J - O) = mJ² - mJO + nJO - nO²

(2) From the properties of G-Numbers (see above), we know that:

J² = -O
O² = -J
OJ = 1
-m = -mJ -mO
n = nJ + nO

(3) Applying these properties to (1), gives us:

aJ + bO = m(-O) -mJ - mO +nJ + nO -n(-J) = (2n-m)J + (n - 2m)O

(4) So that:
a = 2n-m
b = n-2m

(5) a + b = 2n -m + n - 2m = 3(n - m)

QED

Lemma 7: If 3 divides a + b, then J - O divides aJ + bO.

(1) So there exists g such that a + b = 3g
(2) We also support that there exists m,n such that n - m = g and 2n -m = a.
(3) This gives us that b = 3g -a = 3n - 3m - 2n + m = n - 2m.
(4) aJ + bO = (2n - m)J + (n - 2m)O
(5) Applying the reverse logic from above, we get:
(2n - m)J + (n - 2m)O = m(-O) -mJ - mO +nJ + nO -n(-J) =
=mJ² - mJO + nJO - nO² = (mJ + nO)(J - O)

QED

Lemma 8: If J - O doesn't divide aJ + bO, then (aJ + bO)³ ≡ ±1 (mod 9)

(1) Since J - O doesn't divide aJ + bO, we know that 3 doesn't divide a + b since:

(a) Assume that J-O doesn't divide aJ + bO
(b) Assume that 3 divides a + b
(c) Then by the Lemma above J-O divides aJ + bO
(d) But this is not true by the assumption in (a) so we have a contradiction and we reject the assumption in (b) and conclude that 3 doesn't divide a+b.

(2) This means that we have the following possible cases:

(a) a = 3g, b = 3k ±1
(b) a = 3g ±1, b = 3k
(c) a = 3g ±1, b = 3k ±1

(3) Let λ = gJ + kO. Then in each of the three cases we have:

(a) (3g)J + (3k ±1)O = 3λ ±O
(b) (3g ±1)J + (3k)O = 3λ ±J
(c) (3g ±1)J + (3k ±1)O = 3λ ±1

(4) We can rewrite this as 3λ + ε where ε = ±O, ±J, or ±1.

(5) We can cube this result to get:
(3λ + ε)³ = (9λ² + 6λε + ε²)(3λ + ε) =
=27λ³ + 18λ²ε + 3λε² + 9λ²ε + 6λε² + ε³ =
= 9 (3λ³ + 3λ²ε + λε²) + ε³

(6) Additionally, we note that ε³ = ±1 since:

(a) O³ = -1
(b) (-O)³ = 1
(c) J³ = -1
(d) (-J)³ = 1
(e) 1³ = 1
(f) (-1)³ = -1

(7) Hence, we have shown that:
(aJ + bO)³ ≡ ±1 (mod 9).

QED

Lemma 9: J - O divides 3.

(1) 3 = 3J + 3O
(2) Since 3 + 3 is divisible by 3, then 3J + 3O is also divisible by J - O. [See above]

QED

Corollary 9.1: (J-O)³ divides 9.

(1) (J - O)³ = (J² - 2JO + O²)(J - O) =
J³ - 2J²O + JO² - J²O +2JO² - O³ =
-1 -3J²O + 3JO² + 1 =
3(JO² - J²O)

(2) Since O² = -J and J² = -O, we get:
3(O² - J²) = 3(O - J)(O + J) = 3(-1)(1)(J - O)

QED

Lemma 10: (J - O)² = -3

(1) (J - O)² = J² - 2JO + O²

(2) Since OJ = 1, -2JO = -2.

(3) Since O² = -J and J² = -O, we get J² + O² = -O + -J = -1(O + J) = -1.

QED