School of Computing, Engineering and Mathematics

AN INTRODUCTION TO FERMAT’S LAST THEOREM

Yogesh Warren Karunavannan

May 2016

DECLARATION

I declare that no part of the work in this report has been submitted in support of an application

for another degree or qualification at this or any other institute of learning.

Yogesh Warren Karunavannan

ACKNOWLEDGEMENTS

I am grateful to have had John Taylor supervise the typing of this dissertation and pushing me in

the right direction towards its completion. I would like thank Wajid Mannan for providing me with

essential resources that I needed to type this dissertation. I am also indebted to my friends for

using their spare time to read my work and provide me with valuable suggestions that aided in

the completion of this project.

ABSTRACT

Declared as the ‘most difficult mathematical problem’ in the Guinness Book of World Records,

and only being successfully proven by Andrew Wiles in 1994, Fermat’s Last Theorem was

conjected by Pierre de Fermat in 1637 and has created a great deal of advancements in number

theory. This paper has explored the major steps taken to prove the theorem and focus on the

mathematical techniques used.

Supervisor: John Taylor

CONTENTS

1 A brief introduction to Fermat’s Last Theorem ........................................................................... 1

1.1 Creating a general proof .................................................................................................... 1

1.2 The Pythagorean Theorem ................................................................................................. 2

2 The Biquadratic Equation ............................................................................................................ 3

2.1 The Method of Infinite Descent ......................................................................................... 3

2.2 The case of 𝒏 = 𝟒 ............................................................................................................. 4

2.3 Euclid’s Lemma and the proof of 𝒏 = 𝟒 ........................................................................... 5

3 The case of 𝒏 = 𝟑 ....................................................................................................................... 7

Case 1: 𝒛 is even ........................................................................................................................... 8

Case 2: 𝒙 is even .......................................................................................................................... 9

Case 1: 3 does not divide into 𝒑 ................................................................................................ 11

Case 2: 3 divides into 𝒑 .............................................................................................................. 13

4 Ernst Kummer and the proof for regular primes ....................................................................... 15

4.1 The Cyclotomic ring/field and Root of unity .................................................................... 15

4.2 Gabriel Lamé’s Proposed Proof ........................................................................................ 16

4.3 Ideal Numbers .................................................................................................................. 17

4.4 The proof for Regular Primes ........................................................................................... 19

Case 1: not divisible by 𝒑 .................................................................................................... 19

Case 2: divisible by 𝒑 ........................................................................................................... 20

5 Conclusion.................................................................................................................................. 21

6 Evaluation .................................................................................................................................. 22

7 Bibliography ............................................................................................................................... 24

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1 A brief introduction to Fermat’s Last Theorem

For almost four centuries, many mathematicians tried to prove Fermat’s famous conjecture, or

otherwise known as his Last Theorem. Pierre de Fermat studied the book Arithmetic by Diophantus in

the 1630’s. During his time, the French lawyer made several notes on the margin of the book. Of all the

notes he made, one of these stood out more than the rest, and translated to English it states, as shown

in Paulo Ribenboim’s book; Fermat's Last Theorem for Amateurs (2013):

“It is impossible to separate a cube into two cubes, or a biquadrate into two biquadrates, or in general

any power higher than the second into powers of like degree; I have discovered a truly marvellous proof,

which this margin is too small to contain”.

Fermat had a habit of claiming theorems without the proofs, and though all of his conjectures were

proven, the Last Theorem remained unsolved until 1994 by Andrew Wiles. Fermat’s statement and the

theorem, written in a much more visible and familiar way, is as such:

Theorem. If 𝑛 > 2 is any integer. The equation 𝑋𝑛 + 𝑌𝑛 = 𝑍𝑛 has no solution in positive whole

numbers.

Though Fermat claimed to have a ‘marvellous proof’, it was a theorem that many mathematicians failed

to prove for hundreds of years (Singh, 2002, Page 6). Even though the proof exists due to Andrew Wiles,

it inherited many mathematical concepts and techniques that was unavailable in Fermat’s time. We will

explore the introduction and development for the proof of the Last Theorem, such as the proof for

regular primes by Ernst Kummer and the failed proof by Gabriel Lamé, but not include Andrew Wiles’s

proof.

The proofs for this chapter will be largely based on Paulo Ribenboim’s book; Fermat’s Last Theorem for

Amateurs (2013).

1.1 Creating a general proof

To understand Fermat’s Last Theorem in detail, we must focus on the necessary components of the

theorem. In order to obtain a general proof, we must begin with the following:

Theorem 2. To prove Fermat’s Last Theorem for all integers greater than 2, proving that there cannot be

any integer solutions for 𝑛 = 4 and all odd prime numbers, p, is sufficient.

Proof. Let 𝑋𝑛 + 𝑌𝑛 = 𝑍𝑙 have solutions for positive integers 𝑋, 𝑌, 𝑍. Since 𝑛 is composite, meaning that

it can be divided by numbers other than one and itself, we can confirm that 𝑛 > 2.

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Subsequently we can assume it has a factor 𝑚 such that 𝑚 is 4 or an odd prime number, 𝑝. Hence there

can exist an exponent 𝑛 such that 𝑛 = 𝑚𝑙, whereby 𝑙 is an exponent smaller than 𝑛. The conclusive

equation is therefore:

𝑋𝑛 + 𝑌𝑛 = 𝑍𝑛,

(𝑋𝑙)𝑚 + (𝑌𝑙)𝑚= (𝑍𝑙)

𝑚.

Thus 𝑋𝑙 , 𝑌𝑙 and 𝑍𝑙 are solutions to the equation with power 𝑚, which equals 4 or an odd prime number,

𝑝 (Ribenboim, 2013, Page 1-2).

1.2 The Pythagorean Theorem

Fermat’s Last Theorem states that there are no integer solution to the equation 𝑋𝑛 + 𝑌𝑛 = 𝑍𝑛 when

𝑛 > 2. Yet we know that there are integer solutions to the equation 𝑋2 + 𝑌2 = 𝑍2 or otherwise known

as the Pythagoras equation/theorem. The three positive integers 𝑋, 𝑌, 𝑍 that satisfy the equation are

known as the Pythagorean Triples. For example, a triple (𝑥, 𝑦, 𝑧) such as (3, 4, 5) can exist since 32 +

42 = 52. In order to create a proof for 𝑛 = 4, we must factorize the equation so that the integers of the

factor are in the form of a Pythagorean Triple whereby 𝑔𝑐𝑑(𝑋, 𝑌, 𝑍) = 1; this form of factorization is

known as the unique factorization. The abbreviation 𝑔𝑐𝑑 is used to define the greatest common divisor

(Ribenboim, 2013, Page 3-4).

Theorem 3. If 𝑎, 𝑏 are integers such that 𝑎 > 𝑏 > 0, 𝑔𝑐𝑑(𝑎, 𝑏) = 1. Then the triple (𝑥, 𝑦, 𝑧), given by:

{𝑥 = 2𝑎𝑏

𝑦 = 𝑎2 − 𝑏2

𝑧 = 𝑎2 + 𝑏2,

is a primitive solution to the equation 𝑋2 + 𝑌2 = 𝑍2 and 𝑔𝑐𝑑(𝑥, 𝑦, 𝑧) = 1.

Proof. If 𝑎, 𝑏 are integers that satisfy the above equation, then:

𝑥2 + 𝑦2 = 4𝑎2𝑏2 + (𝑎2 − 𝑏2)2 = (𝑎2 + 𝑏2)2 = 𝑧2.

Since 𝑎 and 𝑏 are different in parity, meaning either 𝑎 or 𝑏 is positive or negative and 𝑎 > 𝑏 > 0, then in

relation to the above equation 𝑥 > 0, 𝑦 > 0 and 𝑧 > 0 where 𝑥 is even and both 𝑦, 𝑧 are odd. Now let

𝑑 = 𝑔𝑐𝑑(𝑥, 𝑦, 𝑧), if 𝑑 divides into 𝑥, 𝑦 and 𝑧 then 𝑑 must divide into 2𝑎2 and 2𝑏2 since 𝑦 + 𝑧 = 2𝑎2 and

𝑧 − 𝑦 = 2𝑏2. Therefore 𝑑 = 1 since 𝑔𝑐𝑑(𝑎, 𝑏) = 1 and 𝑑 divides into 𝑎 and 𝑏.

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2 The Biquadratic Equation

Fermat created his Last Theorem whilst attempting to solve the equation 𝑋4 − 𝑌4 = 𝑍2. He had the

desire to see if the area of the Pythagorean triangle, would be equal to the square of an integer.

In the margin of Arithmetic, Fermat wrote: "If the area of a right-angled triangle were a square, there

would exist two biquadrates that difference of which would be a square number” (Edwards, 1977, Page

11)

A biquadrate is a value to the fourth-power, therefore, the biquadrate of 2 is 24 which is 16. This, in

turn, leads to the equation stated above. By showing that there are no right-angle triangles that have an

area equal to a square, it will, in turn, prove Fermat's Last Theorem for 𝑛 = 4, which will be further

explained in the latter parts of this project.

2.1 The Method of Infinite Descent

Fermat’s infamous method of infinite descent, one which he was greatly proud of suggests that if there

are one or more solutions to an equation, then there can be smaller solutions to that same problem and

so forth. Therefore, if we prove that the smallest solution exists, then it implies that there must be an

even smaller solution. Yet this is impossible, as a sequence of whole positive numbers cannot descent

indefinitely (Edwards, 1977, Page 8). The proof for the method of infinite descent will be based on the

same the proof shown by Carl Eberhart in his paper; Fermat and His Method of Infinite Descent (1999).

For example, let's prove that √2 is irrational. In order to apply the method of infinite descent, we must

first assume that √2 has a rational solution. By applying this assumption, we can show that √2 as a

quotient of two positive integers:

√2 =𝑎1

𝑏1, where𝑎1 > 𝑏1.

Now let’s assume the equation:

1

√2 − 1= √2 + 1.

By replacing the √2 on the left-hand side with 𝑎1

𝑏1, we acquire:

1

√𝑎1𝑏1

− 1

= √2 + 1,

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therefore, by solving for √2, we derive the following equation:

√2 =2 −

𝑎1𝑏1

𝑎1𝑏1

− 1.

After further simplifying the above equation we obtain:

√2 =2𝑏1 − 𝑎1

𝑎1 − 𝑏1.

Since we know that 1 <𝑎1

𝑏1< 2, due to the above equation, we can denote that both the numerator and

denominator are positive integers. Consequently, we can replace the numerator with the reprehensive

positive integer, 𝑎2 and the denominator with 𝑏2.

Hence, we have the new quotient of √2:

√2 =𝑎2

𝑏2, where 𝑎2 > 𝑏2.

We can derive the equation above, since clearly, we know that 𝑎2 < 𝑎1 and 𝑏2 < 𝑏1 due to 𝑎1 > 𝑏1.

We can further repeat this process such that with an infinite amount of repeats, there will exist an

𝑎(𝑛 + 1), which is less than 𝑎𝑛, and a 𝑏(𝑛 + 1), which is less than 𝑏𝑛. Henceforth, there is an infinite

descent for all positive integers that satisfy √2 =𝑎

𝑏, but this is impossible due to the fact that a

sequence of infinitely descending positive integers cannot exist. This, by contradiction, proves that √2 is

irrational and subsequently does not have a quotient of positive integers. As Fermat stressed upon, ‘the

method of infinite descent proves that certain things are impossible’ (Edwards, 1977, Page 8), therefore,

it is also known as the proof by contradiction.

2.2 The case of 𝒏 = 𝟒

Now we will take up the case of 𝑛 = 4. In order to show a proof for 𝑛 = 4, we must apply the method of

infinite descent and the method of generating a Pythagorean triple as shown in Theorem 3. The proofs

for 𝑛 = 4 will be largely based on the proofs shown by Larry Freeman in his blog; Fermat’s Last Theorem

(2005) and Harold M. Edwards in his book; Fermat’s Last Theorem: A Genetic Introduction to Algebraic

Number Theory (1977).

Theorem 4. The equation, 𝑋4 + 𝑌4 = 𝑍4 has no positive integer solution 𝑋, 𝑌, 𝑍.

Proof. Using the method to generate a Pythagorean triple, as explained in 1.2, we can assume that

𝑋, 𝑌, 𝑍 have no common divisor greater than 1 and as such, we say that 𝑋, 𝑌, 𝑍 are co-prime. The

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equation above can be written as (𝑋2)2 + (𝑌2)2 = (𝑍2)2, therefore, 𝑋2, 𝑌2 and 𝑍2 are the primitive

Pythagorean triple. Written in Pythagorean form:

𝑋2 = 2𝑝𝑞,

𝑌2 = 𝑝2 − 𝑞2,

𝑍2 = 𝑝2 + 𝑞2,

where 𝑝, 𝑞 are relatively prime (or co-prime) of opposite parity and 𝑝 > 𝑞 > 0. Out of the three

equations above, we can rearrange 𝑌2 = 𝑝2 − 𝑞2 into 𝑌2 + 𝑞2 = 𝑝2. Since 𝑝, 𝑞 are relatively prime, we

have another set of primitive Pythagorean triple which are 𝑝, 𝑞, 𝑌. Additionally, 𝑝, 𝑞 are also opposite in

parity and 𝑝 > 𝑞 > 0, where 𝑝 is odd and 𝑞 is even and so the triple can be written out as follows:

𝑞 = 2𝑎𝑏,

𝑌 = 𝑎2 − 𝑏2,

𝑝 = 𝑎2 + 𝑏2,

whereby 𝑎 and 𝑏 are relatively prime of opposite parity and 𝑎 > 𝑏 > 0. Henceforth we acquire the

following:

𝑋2 = 2𝑝𝑞 = 2(𝑎2 + 𝑏2)(2𝑎𝑏) = 4𝑎𝑏(𝑎2 + 𝑏2),

𝑋2

22= 𝑎𝑏(𝑎2 + 𝑏2).

This shows the 𝑎𝑏(𝑎2 + 𝑏2) is the square of half of 𝑋, where 𝑋 is an even integer (Edwards, 1977, Page

9 – 10).

2.3 Euclid’s Lemma and the proof of 𝒏 = 𝟒

Euclid of Alexandria (325BC – 265BC) is a renowned mathematician known for his treatise, The Elements

(O’Connor and Robertson, 1999). Euclid’s Lemma states that if a prime number divides the product of

two numbers, it must divide at least one of these numbers but not both (Freeman, 2005).

Lemma. If any prime, 𝑃, divides 𝑎𝑏. Then 𝑃 must divide either 𝑎 or 𝑏.

Proof. Following the case of 𝑛 = 4, we know that 𝑎𝑏(𝑎2 + 𝑏2) is a square. Since any prime, 𝑃, can

divide 𝑎𝑏, we can say that 𝑎𝑏 and 𝑎2 + 𝑏2 are relatively prime since 𝑃 would have to divide 𝑎 or 𝑏 but

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not both, and therefore 𝑎2 + 𝑏2 is not divisible by 𝑃. Now that we know 𝑎𝑏 is a square, 𝑎 and 𝑏 must

both be square since they are relatively prime. In other words, we can write both 𝑎 and 𝑏 as follows:

𝑎 = 𝑋2,

𝑏 = 𝑌2.

By substituting the above equation into 𝑎2 + 𝑏2 we get 𝑋4 + 𝑌4. Since 𝑎2 + 𝑏2 is a square, 𝑋4 + 𝑌4 is

also a square, which leads to the fact that 𝑋4 + 𝑌4 = 𝑎2 + 𝑏2 = 𝑝 < 𝑝2 + 𝑞2 = 𝑍 < 𝑍2 = 𝑋4 + 𝑌4.

This begins a sequence of infinitely descending positive integers, which is implausible due to the same

argument as shown in section 2.1. Hence, the sum of two fourth power integers cannot equal to a

square (Edwards, 1977, Page 10).

Corollary. The equation 𝑋4 + 𝑌4 = 𝑍4 has no integer solution where 𝑥, 𝑦, 𝑧 ≠ 0.

Proof. The equation above can be written as 𝑋4 + 𝑌4 = (𝑍2)2. Subsequently the sum of two fourth

power integers cannot equal a square as shown above, therefore by contradiction, it cannot equal a

fourth power (Edwards, 1977, Page 10) thus proving the Last Theorem for 𝑛 = 4.

Due to this, we must now only prove that the theorem is true for 𝑛 as an odd prime value. By proving

that the theorem is true for a given prime exponent, then it will be true for any exponent that is divisible

by the given prime value and in turn we will have shown the proof for all exponents (Freeman, 2005). To

explain this idea, supposed the equation 𝑥𝑛 + 𝑦𝑛 = 𝑧𝑛 has solutions when 𝑛 ≥ 2. Since 𝑛 is greater

than or equal to 2, we can say that 𝑛 is divisible by 4, hence 𝑥𝑛

4 , 𝑦𝑛

4 and 𝑧𝑛

4 are solutions to 𝑥𝑛 + 𝑦𝑛 =

𝑧𝑛. Similarly, if 𝑛 is an odd prime (𝑝), then 𝑥𝑛

𝑝, 𝑦𝑛

𝑝 and 𝑧𝑛

𝑝 are also solutions to the equation. Therefore

by proving that no solutions exist for 𝑛 = 4 and 𝑛 = 𝑝, then we have proven that there are no solutions

for any exponent (Mack – Crane, 2015).

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3 The case of 𝒏 = 𝟑

Leonhard Euler (1707-1783) was a Swiss mathematician whom made important discoveries in

mathematics such as graph theory and number theory (Edwards, 1977, Page 39). Euler was also the first

mathematician to make a progress to Fermat’s Last Theorem with the proof for 𝑛 = 3, after Fermat,

who provided the proof for 𝑛 = 4 (Freeman, 2005).

Leonhard Euler had come up with two proofs for the case of 𝑛 = 3. His first proof had a mistake, which

he did not realise until trying to prove his lemma to the proof of 𝑛 = 3. In order to correct this, he gave

an alternative proof (Freeman, 2005). We will explore the details of this proof, which is greatly based on

the proofs shown by Harold M. Edwards in his book; Fermat’s Last Theorem: A Genetic introduction to

Algebraic Number Theory (1977) and by Larry Freeman in his blog; Fermat’s Last Theorem (2005).

Theorem 5. The equation 𝑋3 + 𝑌3 = 𝑍3 has no positive integer solution 𝑥, 𝑦, 𝑧.

Euler applied Fermat’s method of infinite descent to show the proof to Fermat’s proposition; ‘A cube

cannot be equal to the sum of two non-zero cubes’ (Ribenboim, 2013, Page 24). Euler showed that if it

was possible to find positive integer solutions 𝑥, 𝑦, 𝑧 for the equation 𝑥3 + 𝑦3 = 𝑧3, then one can find

smaller positive integer solutions to the equation. This would then make it possible to find even smaller

solutions, causing a continuous decreasing sequence of triples to the equation, which is impossible

(Edwards, 1977, Page 40).

Proof. In order to show this proof, as shown by both Larry Freeman and Harold M. Edwards, we must

first assume that 𝑥, 𝑦, 𝑧are co-prime. This means that the greatest common divisor of (𝑥, 𝑦), (𝑥, 𝑧) or

(𝑦, 𝑧) is 1. Following a simpler notation, gcd(𝑥, 𝑦, 𝑧) = 1. Since 𝑥, 𝑦, 𝑧 are co-prime, we know that at

least one of the integers is even. Therefore, if both 𝑥 and 𝑦 are odd, then 𝑧 would be even since

(Edwards, 1977, Page 40);

𝑜𝑑𝑑 + 𝑜𝑑𝑑 = 𝑒𝑣𝑒𝑛.

We also know that 𝑥 and 𝑦 are symmetrical, since the equation 𝑥3 + 𝑦3 = 𝑧3 implies that both 𝑥 and 𝑦

have similar properties, thus by showing the case of 𝑥 being even, it will cover the case of 𝑦 being even

(Freeman, 2005). Due to this, we can now split the proof into two cases, with case 1 showing if 𝑧 is even

and case 2 showing if 𝑥 is even.

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Case 1: 𝒛 is even

Lemma 2. If 𝑧 is even, then 𝑝, 𝑞 exists such that:

1. gcd(𝑝, 𝑞) = 1.

2. 𝑝, 𝑞 are positive.

3. Either 𝑝 or 𝑞 is odd, therefore, have opposite parity.

4. 𝑥3 + 𝑦3 = 𝑧3 only has a solution if 𝑝, 𝑞 exists with the above properties.

Proof. Since we have assumed that 𝑥, 𝑦, 𝑧 are co-prime and we have assumed that 𝑧 is even for this case,

then 𝑥 and 𝑦 are odd such that (Edwards, 1977, Page 40 – 41):

𝑥 + 𝑦 = 𝑒𝑣𝑒𝑛,

𝑥 − 𝑦 = 𝑒𝑣𝑒𝑛.

With the above in mind, let:

2𝑝 = 𝑥 + 𝑦,

2𝑞 = 𝑥 − 𝑦.

Since any number 𝑝, 𝑞 would be even if in the form 2𝑝, 2𝑞. We will then have 𝑥 and 𝑦 such that:

𝑥 =1

2(2𝑝 + 2𝑞) = 𝑝 + 𝑞,

𝑦 =1

2(2𝑝 − 2𝑞) = 𝑝 − 𝑞.

This can be worked out by applying the method of solving simultaneous equations to 2𝑝 = 𝑥 + 𝑦 and

2𝑞 = 𝑥 − 𝑦. We can now express 𝑥3 + 𝑦3 = 𝑧3 in terms of 𝑝, 𝑞:

𝑧3 = 𝑥3 + 𝑦3,

= (𝑥 + 𝑦)(𝑥2 − 𝑥𝑦 + 𝑦2),

= (𝑝 + 𝑞 + 𝑝 − 𝑞)[(𝑝 + 𝑞)2 − (𝑝 + 𝑞)(𝑝 − 𝑞) + (𝑝 − 𝑞)2],

= 2𝑝(𝑝2 + 3𝑞2).

Both 𝑝 and 𝑞 have opposite parity since both 𝑝 + 𝑞 = 𝑥 and 𝑝 − 𝑞 = 𝑦 are odd. Equally 𝑝 and 𝑞 also

have gcd(𝑝, 𝑞) = 1 since any factor that divides into 𝑝 and 𝑞 would divide into 𝑥 and 𝑦, but that is

9

impossible as 𝑥 and 𝑦 are co-prime, hence it would be contradictory if that was the case. With this in

mind, it can be said that there can exist a solution 𝑥, 𝑦 to 𝑥3 + 𝑦3 = 𝑧3 due to the co-prime integers,

𝑝, 𝑞, existing such that (Freeman, 2005):

𝑧3 = 2𝑝(𝑝2 + 3𝑞2) = 𝑐𝑢𝑏𝑒.

Case 2: 𝒙 is even

The same argument can be applied to case 2. In order to show the proof, we will be using Lemma 2 as

shown in case 1, but with the difference in being 𝑥 is even rather than 𝑧.

Proof. As we know, 𝑥, 𝑦, 𝑧 are co-prime, therefore, 𝑦 and 𝑧 are both odd. In turn 𝑧 + 𝑦 and 𝑧 − 𝑦 are

both even, therefore there exists 𝑝, 𝑞 such that we let (Freeman, 2005):

2𝑝 = 𝑧 − 𝑦,

2𝑞 = 𝑧 + 𝑦,

due to the fact that the equation will be rearranged to 𝑥3 = 𝑧3 − 𝑦3 since 𝑥 is now even. Using

simultaneous equations, we have 𝑥, 𝑦 such that:

𝑧 = 𝑞 + 𝑝,

𝑦 = 𝑞 − 𝑝.

Since 𝑧, 𝑦 are both odd, 𝑝 and 𝑞 have opposite parity, and using the same argument as case 1,

gcd(𝑝, 𝑞) = 1. Therefore, there exists 𝑝, 𝑞 such that:

𝑥3 = [𝑞 + 𝑝 − (𝑞 − 𝑝)][(𝑞 + 𝑝)2 + (𝑞 + 𝑝)(𝑞 − 𝑝) + (𝑞 − 𝑝)2]

= 2𝑝(𝑝2 + 3𝑞2) = 𝑐𝑢𝑏𝑒.

This results in the same conclusion as case 1 (Freeman, 2005).

Now that we have shown that regardless of either 𝑥, 𝑦, 𝑧 being even, it will conclude with coprime

positive integers 𝑝, 𝑞 such that 2𝑝(𝑝2 + 3𝑞2) is a cube. The next step is to show that 2𝑝 and 𝑝2 + 3𝑞2

are co-prime and ‘that the only way that their product can be a cube is for each of them separately to be

a cube’ (Edwards, 1977, Page 41).

We know that 𝑝 and 𝑞 have opposite parity due to Lemma 2, therefore 𝑝2 + 3𝑞2 is odd, and we know

that 𝑝 and 𝑞 are co-prime, therefore we end up with the argument that if the above statement is true,

then gcd(2𝑝, 𝑝2 + 3𝑞2) = 1 or 3.

10

Lemma 3. If 𝑝, 𝑞 are co-prime and have opposite parity, then gcd(2𝑝, 𝑝2 + 3𝑞2) = 1𝑜𝑟3.

Proof. We begin this proof by assuming that we have any positive integer 𝑠, such that 𝑠 can divide into

2𝑝 and 𝑝2 + 3𝑞2, therefore, 𝑠 is a common factor of 2𝑝 and 𝑝2 + 3𝑞2. We also know that 𝑠 cannot be 2

since we have shown 𝑝2 + 3𝑞2 is odd, therefore we can assume that 𝑠 is greater than 3. Consequently,

𝐴, 𝐵 exists such that (Freeman, 2005):

2𝑝 = 𝑠𝐴,

𝑝2 + 3𝑞2 = 𝑠𝐵.

Even though we know 𝑠 is not 2, we know that 2 can divide into 𝐴 since 2𝑝 = 𝑠𝐴, and so there exists 𝐶

such that 𝐶 is half of 𝐴, resulting in:

𝑝 = 𝑠𝐶.

We can now substitute the value of 𝑝 into 𝑠𝐵, and acquire the following:

𝑝2 + 3𝑞2 = 𝑠𝐵,

3𝑞2 = 𝑠𝐵 − 𝑝2,

3𝑞2 = 𝑠𝐵 − (𝑠𝐶)2,

= 𝑠(𝐵 − 𝑠𝐶2).

Due to the fact that 𝑠 cannot divide into the 3 of 3𝑞2, as it is greater than 3, it must divide into 𝑞 since

we can apply Euclid’s Lemma as explained in section 2.3. However, this is contradictory since 𝑠 is a

common factor of 2𝑝 and 𝑝2 + 3𝑞2 which, in turn, is a common factor of 𝑝, 𝑝2 + 3𝑞2 and 𝑝, 3𝑞2

(Freeman, 2005).

Therefore, if 𝑠 can divide into 𝑞, then it must divide into 𝑝, but this contradicts 𝑝, 𝑞 being co-prime since

𝑠 divides into 𝑝. Therefore, we have two cases whereby we must argue that 3 does not divide into 𝑝,

gcd(2𝑝, 𝑝2 + 3𝑞2) = 1, and that 3 does divide into 𝑝, gcd(2𝑝, 𝑝2 + 3𝑞2) = 3 (Freeman, 2005).

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Case 1: 3 does not divide into 𝒑

Since we are taking up that case that gcd(2p, p2 + 3q2) = 1, we can assume that 2p and p2 + 3q2 are

cubes due to the Relatively Prime Divisors Lemma, the proof to this can be found in Larry Freeman’s

Blog; Fermat’s Last Theorem (2005). This Lemma states that if gcd(𝑎, 𝑏) = 1 and 𝑎𝑏 = 𝑧𝑛, then there

exists 𝑥, 𝑦 such that:

𝑎 = 𝑥𝑛

𝑏 = 𝑦𝑛.

Applying this to our case, we know that 𝑧3 = 2𝑝(𝑝2 + 3𝑞2) and gcd(2𝑝, 𝑝2 + 3𝑞2) = 1, then there

exists 𝑥, 𝑦 such that:

𝑥3 = 2𝑝

𝑦3 = 𝑝2 + 3𝑞2.

There forth an assumption that 2𝑝, 𝑝2 + 3𝑞2 are both cubes can be made.

Lemma 4. If gcd(2𝑝, 𝑝2 + 3𝑞2) = 1, then there must be smaller solutions to Fermat’s Last Theorem: 𝑛 =

3.

Proof. Since we know that 2𝑝, 𝑝2 + 3𝑝2 are cube, then we know that there exists 𝑎, 𝑏 such that

gcd(𝑎, 𝑏) = 1 and both 𝑎, 𝑏 have opposite parities since 𝑝, 𝑞 also have opposite parities. Let 𝑢3 = 𝑝2 +

3𝑞2, since we are applying Relatively Prime Divisors Lemma (Freeman, 2005). We know that 𝑝, 𝑞 have

opposite parities, in regards to which 𝑢 can be stated as odd and so 𝑢 will be in the form of 𝑎2 + 3𝑏2

because we know gcd(𝑎, 𝑏) = 1, which means that every odd factor in the form 𝑎2 + 3𝑏2 will have the

same form.

As a result of the above:

𝑢3 = (𝑎2 + 3𝑏2)3,

= 𝑎2 + 3𝑏2(𝑎2 + 3𝑏2)2,

= (𝑎2 + 3𝑏2)(𝑎4 + 6𝑎2𝑏2 + 9𝑏4),

= (𝑎2 + 3𝑏2)(𝑎4 − 6𝑎2𝑏2 + 12𝑎2𝑏2 + 9𝑏4),

= (𝑎2 + 3𝑏2)[(𝑎2 − 3𝑏2)2 + 3(2𝑏)2],

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= [𝑎(𝑎2 − 3𝑏2) − 3𝑏(2𝑎𝑏)]2 + 3[𝑎(2𝑎𝑏) + 𝑏(𝑎2 − 3𝑏2)]2,

= (𝑎3 − 3𝑎𝑏2 − 6𝑎𝑏2)2 + 3(2𝑎2𝑏 + 𝑎2𝑏 − 3𝑏3)2,

= (𝑎3 − 9𝑎𝑏2)2 + 3(3𝑎2𝑏 − 3𝑏3)2.

As 𝑢3 = 𝑝2 + 3𝑞2, then 𝑝2 + 3𝑞2 = (𝑎3 − 9𝑎𝑏2)2 + 3(3𝑎2𝑏 − 3𝑏3)2. This allows us to define 𝑎, 𝑏

such that :

𝑝 = 𝑎3 − 9𝑎𝑏2 = 𝑎(𝑎 − 3𝑏)(𝑎 + 3𝑏),

𝑞 = 3𝑎2𝑏 − 3𝑏3 = 3𝑏(𝑎 − 𝑏)(𝑎 + 𝑏),

where gcd(𝑎, 𝑏) = 1.

We can now substitute the value for 𝑝 into 2𝑝, therefore (Freeman, 2005):

2𝑝 = 2(𝑎3 − 9𝑎𝑏2),

= 2𝑎3 − 18𝑎𝑏2,

= 2𝑎(𝑎 − 3𝑏)(𝑎 + 3𝑏).

Since 𝑎, 𝑏 are opposite parities, we know that both 𝑎 − 3𝑏 and 𝑎 + 3𝑏 are odd. Due to this, any

common factor that divides into 2𝑎, 𝑎 ± 3𝑏 would also be a common factor of 𝑎, 𝑎 ± 3𝑏 and

subsequently, be a common factor to 𝑎,±3𝑏 (Edwards, 1977, Page 42). We know that this common

factor cannot be 2 since 2 cannot divide into 𝑎 ± 3𝑏, therefore the only common factor would be 3. But

if 3 divides into 𝑎, it would then be able to divide into 𝑝 which is impossible since gcd(2𝑝, 𝑝2 + 3𝑞2) =

1.

As a result of this, 2𝑎, 𝑎 ± 3𝑏 are co-prime and all three are cube due to the Relative Prime Divisors

Lemma. We now have 𝐴, 𝐵, 𝐶 such that (Edwards, 1977, Page 42):

𝐴3 = 2𝑎

𝐵3 = 𝑎 − 3𝑏

𝐶3 = 𝑎 + 3𝑏.

From the above equations, we can see that 𝐵3 + 𝐶3 = 2𝑎 = 𝐴3, this gives a solution smaller than 𝑥, 𝑦, 𝑧

where 𝑥3 + 𝑦3 = 𝑧3. We know this is the case since (𝐴3)(𝐵3)(𝐶3) = 2𝑎(𝑎 − 3𝑏)(𝑎 + 3𝑏) = 2𝑝,

where 𝑥3or 𝑧3 is 2𝑝(𝑝2 + 3𝑞2) as shown in case 1: 𝒛 is even and case 2: 𝒙 is even.

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Case 2: 3 divides into 𝒑

Now, our final case to show that 𝑥3 + 𝑦3 = 𝑧3 has no integer solution 𝑥, 𝑦, 𝑧 only differs slightly from

the previous case. In both cases, we have the same assumptions, therefore, 𝑝, 𝑞 are co-prime with

opposite parity and 2𝑝, 𝑝2 + 3𝑞2 are a cube. The difference, in this case, would be the fact that

gcd(2𝑝, 𝑝2 + 3𝑞2) = 3 (Freeman, 2005).

Lemma 5. If gcd(2𝑝, 𝑝2 + 3𝑞2) = 3, then there must be smaller solutions to Fermat’s Last Theorem:

𝑛 = 3.

Proof. To begin, we note that 3 divides into 𝑝 but does not divide into 𝑞, since 3 can divide into 2𝑝 as

the gcd(2𝑝, 𝑝2 + 3𝑞2) = 3 (Edwards, 1977, Page 42). But 𝑝 and 𝑞 are co-prime, therefore due to

Euclid’s Lemma, if 3 divides into 𝑝, it cannot divide into 𝑞. Since 3 divides into 𝑝, there exists 𝑓 such that

(Freeman, 2005):

𝑝 = 3𝑓.

By substituting the value of 𝑝 into 2𝑝(𝑝2 + 3𝑞2) we acquire:

2𝑝(𝑝2 + 3𝑞2) = [2(3𝑓)((3𝑓)2 + 3𝑞2)],

= 32(2𝑓)(3𝑓2 + 𝑞2).

Since 3 cannot divide 𝑞, it cannot divide 3𝑓2 + 𝑞2. As a result of this, 32(2𝑓) and 3𝑓2 + 𝑞2 are co-

prime. Consequently, 𝑓 and 𝑞 are also of opposite parity, and therefore gcd(𝑓, 𝑞) = 1 since gcd(𝑝, 𝑞) =

1. We know that 2𝑝(𝑝2 + 3𝑞2) is cube, thus 32(2𝑓), 3𝑓2 + 𝑞2 are also cubes, since 2𝑝(𝑝2 + 3𝑞2) =

32(2𝑓)(3𝑓2 + 𝑞2).

From the previous case, we calculated 𝑝2 + 3𝑞2 = (𝑎3 − 9𝑎𝑏2)2 + 3(3𝑎2𝑏 − 3𝑏3)2, therefore, as

before, there exists 𝑎, 𝑏 such that (Freeman, 2005):

𝑞 = 𝑎3 − 9𝑎𝑏2 = 𝑎(𝑎 − 3𝑏)(𝑎 + 3𝑏),

𝑓 = 3𝑎2𝑏 − 3𝑏3 = 3𝑏(𝑎 − 𝑏)(𝑎 + 𝑏),

where gcd(𝑎, 𝑏) = 1.

Due to the fact that we know 32(2𝑓) is a cube, we can define 𝑎, 𝑏 further such that:

32(2𝑓) = 32[2(3𝑎2𝑏 − 3𝑏3)],

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= 33[2(𝑎2𝑏 − 𝑏3)],

= 33[2𝑏(𝑎2 − 𝑏2)],

= 33[2𝑏(𝑎 − 𝑏)(𝑎 + 𝑏)].

Following from the fact that 32(2𝑓) is a cube, then 33[2𝑏(𝑎 − 𝑏)(𝑎 + 𝑏)] is a cube, and subsequently

2𝑏(𝑎 − 𝑏)(𝑎 + 𝑏) is also a cube. For that reason, there exists 𝐴, 𝐵, 𝐶 such that:

𝐴3 = 2𝑏

𝐵3 = 𝑎 − 𝑏

𝐶3 = 𝑎 + 𝑏.

This, in turn, gives another solution to Fermat’s Last Theorem; 𝑛 = 3 as 𝐵3 + 𝐶3 = 2𝑏 = 𝐴3, whereby

𝐴3𝐵3𝐶3 is less than 𝑓 = 3𝑎2𝑏 − 3𝑏3 = 3𝑏(𝑎 − 𝑏)(𝑎 + 𝑏) which is also less than 𝑝 = 3𝑓, which is less

than 𝑥3 or 𝑧3 = 2𝑝(𝑝2 + 3𝑞2) as well (Edwards, 1977, Page 42).

Corollary 2. 𝑋𝟑 + 𝑌3 = 𝑍3 has no integer solution 𝑥, 𝑦, 𝑧.

Proof. In both cases where gcd(2𝑝, 𝑝2 + 3𝑞2) = 1 or 3, we acquire smaller and smaller solutions to the

already existing small solutions, which is impossible. Therefore, this proves that there are no integer

solutions to the equation by the method of infinite descent.

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4 Ernst Kummer and the proof for regular primes

Ernst Kummer’s (January 29th 1810 – May 14th 1893) contribution to the proof of the Last Theorem was

of great extent, aided by Gabriel Lamé (July 22nd 1795 – May 1st 1870) failed proof, Kummer was able to

finalise his proof for regular primes (Freeman, 2006). Kummer was a 19th century German

mathematician whom made substantial progress to FLT after Leonhard Euler (Varma, 2008). Many

mathematicians have tried to prove the theorem with many special cases such as the proof of 𝑛 = 5 by

Adrien-Marie Legendre (September 18th 1752 – January 10th 1833) and Peter Gustav Lejeune Dirichlet

(February 13th 1805 – May 5th 1859), by using Sophie Germain's (April 1st 1776 – June 27th 1831)

Theorem, and the proof of 𝑛 = 7 by Lamé that was later replaced by Kummer's proof for regular primes

(Edwards, 1977, Page 59 – 60).

Kummer came across Fermat’s Last Theorem whilst generalising the law of quadratic reciprocity (Varma,

2008). He himself stated that he had no interest in the Last Theorem (Freeman, 2006), but rather

considered it to be a ‘curiosity of number theory since it was a problem that could be understood by a

12 year old’ (Varma, 2008). The proofs that will be discussed in the latter paragraphs will be largely

based on the proofs shown by Harold M. Edwards in his book; Fermat’s Last Theorem: A Genetic

Introduction to Algebraic Number Theory (1977), Larry Freeman in his blog; Fermat’s Last Theorem

(2005), Ila Varma in her article; Kummer, Regular Primes, and Fermat’s Last Theorem (2008) and by

Sander Mack – Crane in his report; Fermat’s Last Theorem for Regular Primes (2015).

4.1 The Cyclotomic ring/field and Root of unity

As explained in section 2.3, in order to acquire a proof for the general case, we will need to prove that

no solution exists when 𝑛 is any odd prime (Freeman, 2005). Before we begin with the proof for regular

primes, we must first define some facts about cyclotomic rings and algebraic number fields. After this,

we will be able to understand the core idea of the proof (Varma, 2008).

In order for these technical terms to be understood by everyone, the definition of rings and fields will be

explained as such; a ring is a set of numbers that can be added, subtracted and multiplied whilst within

the set. Subsequently, a field is a set of numbers that can be added, subtracted, multiplied and divided

whilst within the set. Rings will be defined by notation ℤ, whereas fields will use the notation ℚ. We can

see that in the definition for rings and fields, both differ by the fact that one is able to divide within the

set whilst the other does not, however this is not the case. We can say that 𝑏 divides 𝑎, or write 𝑏|𝑎 if

there exists an integer, 𝑐, in the ring, such that 𝑎 = 𝑏𝑐. Therefore in terms of the integer set ℤ, 3 divides

6 since there is an integer 𝑐 = 2 which results these numbers being in the ring and the set. Conversely,

5|6 cannot exist in the set as there is no value for the integer 𝑐; 6 = 5 × 𝑐 (Mack – Crane, 2015).

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This is not the case with fields, since all divisions can occur in the set ℚ. An example of this is as follows;

5|6 can exist in the set due to the fact that the rational umber 5 can divide into 6 such that there exists

integer 𝑐 =6

5; 6 = 5 ×

6

5. From this it can be shown that every number in the field, ℚ, can become

irreducible, given that we are able to factorize a number into smaller numbers until it cannot factor

anymore. Henceforth a number can become a product of a prime (Mack – Crane, 2015).

This can now be followed up with the explanation of algebraic number fields (we will refer to this as

number fields), which is acquired from the addition of rational numbers to a root of a polynomial.

Consider the polynomial 𝑥𝑛 − 1; this polynomial has a complex number ζ such that 𝜁𝑝 = 1, this is

known as the primitive 𝑝𝑡ℎ root of unity (Varma, 2008) as we allow 𝜁𝑝 as the primitive root to the above

polynomial. Since we are proving for regular primes, let the primitive root be 𝜁𝑝𝑝

. By the addition of this

complex number to the field ℚ, we acquire the number field ℚ(𝜁𝑝) that is written in form (Edwards,

1977, Page 80-81):

𝑎0 + 𝑎1𝜁𝑝 + 𝑎2𝜁𝑝2 +⋯+ 𝑎𝑝−1𝜁𝑝

𝑝−1.

We are able to write the above number field in the form shown, due to the fact that each element of

the field is a complex number that has the form 𝑎 + 𝑏𝑖 (all complex numbers have this form), where 𝑎, 𝑏

are the rational numbers and 𝑖 being the imaginary unit that represents 𝜁𝑝 in our case.

Kummer had used λ to denote prime numbers and α to represent the imaginary unit such that 𝛼𝜆 = 1 is

the primitive 𝜆𝑡ℎ root of unity, as opposed to the notations we have used (Freeman, 2006). Regardless

of which notation is used, Kummer regarded these numbers as a ‘special type’ of complex numbers.

These are now known as the cyclotomic integers, due to way ζ is defined as a point on “the circle, |𝑧| =

1 of a complex 𝑧 – plane” that effects the way the circle is divided into 𝑝 equal parts. The ‘cycl-’ part of

cyclotomic means circle and ‘tom-’ means divide (Edwards, 1977, Page 81).

4.2 Gabriel Lamé’s Proposed Proof

Although Kummer did not take interest into Fermat’s Last Theorem, he did make substantial steps into

proving the theorem (Freeman, 2006). This came merely a few weeks after Lamé claimed to have found

the complete proof to the theorem, which was announced in 1847 at the Paris Academy (Edwards,

1977, Page 76). Though his evidence was flawed, the concept behind it was simple and helped to

develop the general proof to the theorem. Having worked on other cases, such as the case for 𝑛 = 7;

Lamé noticed a trend whereby the difficulty to find the solutions increased as the value for 𝑛 increased

(Edwards, 1977, Page 76). As seen from previous chapters, in order to find the proof for each value of 𝑛,

we must factorise the equation. In the search to simplify these factorizations of high 𝑛 values, he

17

realised ‘that this can be overcome by decomposing 𝑥𝑛 + 𝑦𝑛 into 𝑛 linear factors’ (Edwards, 1977, Page

76). This was made possible by Lamé by introducing the complex number, 𝜁𝑝, such that 𝜁𝑝𝑝= 1 as shown

in section 4.1 (Varma, 2008). By using the algebraic identity with the root of unity gives way to the

following:

𝑥𝑝 + 𝑦𝑝 = (𝑥 + 𝑦)(𝑥 + 𝜁𝑝𝑦)(𝑥 + 𝜁𝑝2𝑦)…(𝑥 + 𝜁𝑝

𝑝−1𝑦).

Lamé had applied techniques of factorizing with complex numbers, which have been used before in

Fermat’s Last Theorem, but no one had claimed to have used it until Lamé (Freeman, 2006). However,

though he claimed to have found the proof, he did not claim the entire credit for himself since the idea

was suggested to him by his colleague Joseph Liouville (March 24th 1809 – September 8th 1882)

(Edwards, 1977, Page 77).

Explained in a more understanding way, Lamé tried to show that if each factor 𝑥 + 𝑦,𝑥 + 𝜁𝑝𝑦, … , 𝑥 +

𝜁𝑝𝑝−1

𝑦 are relatively prime then the equation 𝑥𝑝 + 𝑦𝑝 = 𝑧𝑝 tells us that each said factor must be the

𝑝𝑡ℎ power since each factor is equal to 𝑧𝑝, following this an infinite descent must be derived in order to

finalise the proof which is an impossible task. However, if all factors mentioned are not relatively prime,

then Lamé tried to show that all factors have a common factor 𝑚, such that (𝑥+𝑦)

𝑚, (𝑥+𝜁𝑝)

𝑚, … ,

(𝑥+𝜁𝑝𝑝−1

𝑦)

𝑚

are relatively prime, which again has the impossible task to derive an infinite descent (Edwards, 1977,

page 77).

Lamé had thought that this was the revolutionary step needed to generate a complete solution, but this

was far from the case since many mathematicians, including Liouville, criticised Lamé’s announcement.

The main argument begin that the unique factorisation did not hold in the cyclotomic field ℚ(𝜁𝑝)

(Varma, 2008), even though each element of the cyclotomic ring ℤ(𝜁𝑝) could be factorized into a

product of a prime (Mack - Crane, 2015). More specifically, the unique factorization did not hold for all

primes 𝑝 ≥ 23. This was proven by Kummer three years before Lamé made his claim (Freeman, 2006),

however he had published it in a ‘very obscure’ place, which unknown to many mathematicians at the

time (Edwards, 1977, Page 78).

4.3 Ideal Numbers

From Lamé’s failed proof, a new problem arised; if these cyclotomic integers, that Lamé formed for his

‘proof’, did not have the unique factorization, then we must find ‘ideal numbers’ that fitted the

properties needed to restore the unique factorization. To begin with, the basic property that these ‘ideal

numbers’ need is its divisibility. Expanding on this point, if an ideal number is able to divide an element

in the cyclotomic ring, then it should be able to divide the product of two elements in the ring. In a more

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visual form, if an ideal number divides 𝑎 in a cyclotomic ring, then it can divide 𝑎𝑏. Subsequently, if it

can divide 𝑎 and 𝑏, then it can divide 𝑎 + 𝑏 (Mack – Crane, 2015).

With the above property in mind, an ‘ideal number’ can be described by showing the many collections

of numbers it can divide. Henceforth, we can use this description to explain an ideal prime number; we

let ideal number 𝑝 to be prime, if there exists 𝑎𝑏 in the cyclotomic ring, such that if 𝑎𝑏 is in 𝑝 (therefore

allowing ideal 𝑝 in its own ring to be a subset of the cyclotomic ring) then 𝑎 or 𝑏 is in 𝑝. This allows 𝑝 to

be a cyclotomic integers, otherwise the ideal number will not exist as a cyclotomic integer without the

property shown above. By multiplying ideal numbers, we can let the product of these numbers be the

smallest kind that contains the cyclotomic integers, thus we can let 𝑝𝑏 to be the smallest ideals

containing the product of cyclotomic integers, such as 𝑎𝑏 (Mack – Crane, 2015).

Let us consider a set of ideal numbers in a cyclotomic ring (therefore these numbers are a subset of the

ring), we can say that each element (α) of the ring can produce an ideal of the element, which can be

divided by the said element (α). This element/ divisor is known as a principle (Edwards, 1977, Page 162).

By introducing these principle ideal numbers, the unique factorization is restored, since ‘every ideal in

the ring of integers can be written uniquely as a product of prime ideals’ (Mack – Crane, 2015).

To help aid us with the above idea, consider the example used in the definitions of rings in section 4.1.

We have the equation 6 = 2 × 3, now if we consider the ring ℤ(√−5) with the field ℚ(√−5) then we

can write the equation as (Varma, 2008):

6 = 2 ∗ 3 = (1 + √−5) ∗ (1 − √−5),

whereby 2, 3, (1 + √−5), (1 − √−5) are irreducible elements to ℤ(√−5). However, with the

introduction of ideal numbers, the factors shown above can now be further factorized such that (Mack –

Crane, 2015):

2 = (2, 1 + √−5)(2,1 − √−5) = 𝑝1, 𝑝2,

3 = (3, 1 + √−5)(3,1 − √−5) = 𝑝3, 𝑝4,

1 + √−5 = (2, 1 + √−5)(3,1 + √−5) = 𝑝1, 𝑝3,

1 − √−5 = (2, 1 − √−5)(3,1 − √−5) = 𝑝2, 𝑝4.

We have now further factorized the equation into ideal prime factors 𝑝1, 𝑝2, 𝑝3, 𝑝4 that are unique and

hold in the cyclotomic field ℚ(√−5). As shown above, the unique factorization does not always hold in

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the original ring, however after the addition of principle ideal numbers the unique factorization holds,

therefore it will hold for the cyclotomic integers that created the ideal numbers. From this, we can

define class numbers. Class numbers are used to measure how far a ring is from unique factorization,

since elements from cyclotomic rings can produce principle ideal numbers, class numbers tells us the

number of ideals added into order to restore the unique factorization (Edwards, 1977, Page 163).

4.4 The proof for Regular Primes

In this final chapter that concludes the proof for regular primes, we will explore the mathematics

involved in solving this case, but we will not go into the full details of the proof. This chapter will give an

idea on how to solve Fermat’s Last Theorem for𝑛 = 𝑟𝑒𝑔𝑢𝑙𝑎𝑟𝑝𝑟𝑖𝑚𝑒(𝑝).

Where Lamé failed to create a general proof, he did however succeed in establishing Fermat’s Last

Theorem. Following from the concept of class numbers, there are still a few odd primes that have class

numbers. These numbers, in the field ℚ(𝜁𝑝) are 3, 5, 7, 11, 13, 17, 19. To salvage this and solve the final

part of the proof we turn to regular primes. We call any prime, 𝑝 a regular prime if and only if it does not

divide the class number in the field ℚ(𝜁𝑝). By solving for regular primes, we are able to complete the

proof for the theorem (Mack – Crane, 2015).

Theorem 6. The equation 𝑥𝑝 + 𝑦𝑝 = 𝑧𝑝 has no integer solution 𝑥, 𝑦, 𝑧 if 𝑝 is any regular prime.

In order to show the overview of the proof, we will split the theorem into two cases, as shown for the

case of 𝑛 = 3 in chapter three. Whereby both cases show if any of the integers 𝑥, 𝑦, 𝑧 are divisible by 𝑝.

Case 1: not divisible by 𝒑

Lemma 6. The integers 𝑥, 𝑦, 𝑧 are co-prime and not divisible by 𝑝

Proof overview. We can write the equation in Theorem 6. as (Edwards, 1977, Page 168 – 173):

𝑧𝑝 = 𝑥𝑝 + 𝑦𝑝 = (𝑥 + 𝑦)(𝑥 + 𝜁𝑝𝑦)(𝑥 + 𝜁𝑝2𝑦)… (𝑥 + 𝜁𝑝

𝑝−1𝑦).

The first part of this proof is to show that all factors from the above equation are all co-prime. Due to

the unique factorization of ideals, as explained in section 4.3, we can then conclude that the factors

have the 𝑝𝑡ℎ power of a principle ideal number since we have 𝑧𝑝 which is a principle to the 𝑝𝑡ℎ power.

To explain this using mathematical notations, let (𝑥 + 𝜁𝑝𝑦) = 𝛼𝑝 be the principle ideal number for ideal

α in the ring ℤ(𝜁𝑝). Therefore, 𝛼𝑝 lies in the ideal class group in the field ℚ(𝜁𝑝) (Mack – Crane, 2015).

However, if the class number of ℚ(𝜁𝑝) is not divisible by 𝑝 then we can use knowledge from group

theory to argue 𝛼 itself is the principle element to the ring ℤ(𝜁𝑝). Following this, we must continue using

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this fact to see if the factorization of the ring causes an infinite descent, thereby concluding that there

are no integer solutions to Fermat’s Last Theorem, where 𝑛a regular prime; if the integers 𝑥, 𝑦, 𝑧 is not

divisible by 𝑝 (Edwards, 1977, Page 170).

Case 2: divisible by 𝒑

Lemma 7. The integers 𝑥, 𝑦, 𝑧 are co-prime and divisible by 𝑝

Proof overview. Referring to the same equation as in Case 1:

𝑧𝑝 = 𝑥𝑝 + 𝑦𝑝 = (𝑥 + 𝑦)(𝑥 + 𝜁𝑝𝑦)(𝑥 + 𝜁𝑝2𝑦)… (𝑥 + 𝜁𝑝

𝑝−1𝑦).

If 𝑝 does divide either 𝑥, 𝑦, 𝑧, then this shows that all factors shown are not relatively prime meaning

that all factors are divisible by 𝑝. Despite this, the same case follows since the 𝑝𝑡ℎ power is principle (𝑧𝑝)

then the factors itself are principle 𝑝𝑡ℎ powers. We then factorize elements of ℤ(𝜁𝑝) and form an

infinite descent. Therefore we prove Fermat’s Last Theorem for regular primes with the conclusion that

in both cases, there are no integer solution 𝑥, 𝑦, 𝑧 (Mack – Crane, 2015).

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5 Conclusion

The general concept behind Fermat’s Last Theorem is uncomplicated, since it has been regarded as a

topic that a mere 12 year old child can understand (Varma, 2008). Andrew Wiles stated, “Here was a

problem, that I, a ten – year old, could understand and I knew from that moment that I would never let

it go. I had to solve it” (Neil Pieprzak, 2008). However, the interesting and more useful part about the

theorem is not the actual proof, but rather the mathematics used to create it. The theorem caused

substantial advancements in number theory, due to the sheer fact that the ‘simple’ equation introduced

many new mathematical ideas as well as a new branch of mathematics in order to create the proof.

Many of the concepts used, such as ideal principal numbers, showed a creative level of knowledge

needed to solve the theorem. Expanding on this point, we are led to assume many factors in each proof,

such as assuming that the integers are co-prime which shows that a lot of assumptions are created

before tackling the problem. The proofs demonstrated are by far not the most perfect. However, by

comparing all three proofs shown, one can understand the complexity to generalize Fermat’s Last

Theorem. Although all proofs follow a similar trend in that an infinite descent must be acquired, the

calculations of each are different and require some sense of imagination. What we have seen in this

project are the proofs for individual exponents to the theorem. However, Andrew Wiles’s proof used

new mathematical techniques. He used elliptical curves and modular forms rather than opting for the

more traditional and well known use of the infinite descent. The world now knows the general proof to

Fermat’s Last Theorem, but one day there may be a chance that Fermat’s general proof will be

discovered and for that I encourage the reader to pursue this knowledge.

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6 Evaluation

This dissertation has tested my mathematical knowledge and skills as well as my sanity, in the sense that

all effort was put into this project. However, as much as this topic has been very challenging for me, it

was one of the main reasons why I decided to study this course, especially after reading Simon Singh’s

book ‘Fermat’s Last Theorem (2002)’ whilst studying for my GCSEs. Having said that this project tested

my sanity, I did enjoy the research and understanding how the mind of all the mathematicians

essentially worked together to find the proof to the theorem, whilst learning the actual mathematics

involved to the best of my ability.

The research aspects for my project was challenging due to the difficult mathematics involved with the

‘simple’ looking equation. To put it into perspective, it did take almost 4 centuries to solve, whereby

Andrew Wiles only managed to solve it 31 years after learning about the conjecture. Many of the books

and online resources, such as blogs and PDFs that I found provided the proofs I needed, but with the

drawback that each proof for each exponent of 𝑛 was slightly different when comparing resources; such

as the proof for 𝑛 = 4, when comparing the books ‘Fermat’s Last Theorem for Amateurs’ by Paulo

Ribenboim to ‘Fermat’s Last Theorem: A Genetic Introduction to Algebraic Number Theory’ by Harold M.

Edwards, had some minor differences. Therefore in order to understand each proof I had to constantly

read the same line or paragraph over and over.

Expanding on this issue, many of the proofs from the resources I used had different notations to the

same proof. An example of this is the use of 𝑟𝑛 = 1 for roots of unity in Harold M. Edwards’s book,

rather than 𝜆𝑛 = 1 which is used by Ila Varma in her paper and Ernst Kummer himself, whom

introduced the complex number. Even though this example is not as extreme, I decided to stick to the

resources which gave me more understanding to the problem at hand.

After understanding each theorem and proof, the next difficulty was making sure that each chapter I

wrote flowed rather than having a completely different topic. To achieve this I had to write many

subchapters, with each explaining the method I would utilize before explaining the proof. An example of

this was when I explained the biquadratic equation and method of infinite descent before giving the

proof of 𝑛 = 4, whilst introducing the Euclid’s Lemma in-between the proof.

My goal for this project was to ultimately create the proof that Fermat had claimed to have had. From

my research I learnt that the mathematics used by Andrew Wiles, in his general proof, was not available

in Fermat’s time and that it was only achievable due to the ever-growing knowledge and input by

numerous mathematicians. I wanted to see if I could create this lost proof, but as my supervisor pointed

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out we are uncertain if Fermat ever created this proof. Due to my lack of knowledge in number theory

as well as my mathematical capability, this was much further from achievable than I first anticipated.

Therefore I decided to leave out my creation of the general proof and focus on introducing the theorem

and exploring the mathematics involved in the finding the proof for important special cases such as 𝑛 =

4, 3 and regular primes. This again, was proven to be difficult, especially with regular primes since I had

to understand other mathematical theorems and proofs, such as cyclotomic integers and class numbers.

In terms of writing the content of this dissertation, my biggest problem, was writing it in order for level 6

students to understand. Initially I had written the proofs based on what I learnt and understood from

the resources I had used, but I had only realised that these resources assume that the reader would

have background knowledge after my supervisor told me to explain what terms such as co-prime meant.

This in turn would mean that I had to expand on the proofs and fill in the gaps in knowledge that was

left out by the resources I used, meaning that the proofs I have shown in this dissertation are longer

than those in online resources and books.

My overall aim of the project was to introduce, explore and explain the mathematics involved from the

creation of Fermat’s Last Theorem through to, but not including, Andrew Wiles’s general proof and

explain the theorem, with the added aim to make it more accessible to students that do not have prior

knowledge of number theory. However, this aim changed due to my initial lack of research. I did not

realise that the theorem was more in-depth and much harder until it was almost too late. Having wasted

a lot of time in reading books that provided more information on the historical events rather than the

mathematical events, I had to quickly change my goals. For example, I have been pushing myself to

write the full detailed proof for regular primes but this became an unrealistic goal since I was not able to

understand the theorem enough to write a lot about it. The detailed proof was arduous for me to

understand, since I do not have any knowledge on number theory, therefore in writing this project, I

have taught myself the basics of number theory in terms of Fermat’s Last Theorem.

I believe that I have achieved my aims and provided that correct information to allow level 6 students or

any other reader to understand the theorem, and how it aided the advancements of number theory.

Knowing that I was going to write my dissertation on Fermat’s Last Theorem ever since starting

university, I wish I had prepared myself by doing the necessary research that would have allowed me to

complete this project with ease, and given me more chances to generate my own proof. Nevertheless, I

have enjoyed reading and learning about Fermat’s Last Theorem and the evolution of number theory.

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7 Bibliography

Eberhart, C. (1999) Fermat and His Method of Infinite Descent [Online], Available:

<http://www.ms.uky.edu/~carl/ma330/html/bailey21.html> [Accessed 23rd November 2015]

Edwards, H. M. (1977) Fermat’s last theorem: A genetic introduction to algebraic number theory,

Germany: Springer-Verlag Berlin and Heidelberg GmbH & Co. K

Freeman, L. (2005) Fermat’s last theorem [Online], Available:

<http://fermatslasttheorem.blogspot.co.uk/> [Accessed 10th November 2015]

Mack-Crane, S. (2015) Fermat’s last theorem for regular primes, [Online] . Available:

<https://math.berkeley.edu/~sander/speaking/22September2015%20WIM%20Talk.pdf> [Accessed 4th

May 2016]

O’Connor, J. J. and Robertson, E. F. (1999) Euclid biography [Online], Available: <http://www-

groups.dcs.st-and.ac.uk/~history/Biographies/Euclid.html> [Accessed 3rd May 2016]

Pieprzak, N. (2008) Fermat’s last theorem and Andrew wiles [Online], Available:

<https://plus.maths.org/content/fermats-last-theorem-and-andrew-wiles> [Accessed 18th April 2016]

Ribenboim, P. (2013) Fermat’s last theorem for amateurs, United States: Springer-Verlag New York

Singh, S. (2002) Fermat’s last theorem: The story of a riddle that confounded the world's greatest minds

for 358 years, London: Fourth Estate

Singh, S. and Ribet, K. A. (1997) Fermat’s Last Stand, [Online] . Available:

<http://www.fis.cinvestav.mx/~lmontano/sciam/Fermat-SC1197-68.pdf> [Accessed 25th April 2016]

Varma, I. (2008) KUMMER, REGULAR PRIMES, AND FERMAT’S LAST THEOREM, [Online] . Available:

<http://www.math.harvard.edu/~ila/Kummer.pdf> [Accessed 4th April 2016]