This document provides an introduction to Fermat's Last Theorem. It discusses how proving the theorem for specific cases of n=4 and prime numbers is sufficient to prove it generally. It also covers some of the early work done to attempt to prove the theorem, including using Pythagorean triples to represent solutions and the method of infinite descent. The document then gives proofs for n=4 using these techniques, showing there are no integer solutions to the equation.
FERMATβS LAST THEOREM PROVED BY INDUCTION (accompanied by a philosophical com...
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Fermat's Last Theorem Explained
1. School of Computing, Engineering and Mathematics
AN INTRODUCTION TO FERMATβS LAST THEOREM
Yogesh Warren Karunavannan
May 2016
2. 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
3. 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.
4. ABSTRACT
Declared as the βmost difficult mathematical problemβ in the Guinness Book of World Records,
and only being successfully proven by Andrew Wiles in 1994, Fermatβs 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
7. 2
Subsequently we can assume it has a factor π such that π is 4 or an odd prime number, π. Hence there
can exist an exponent π such that π = ππ, whereby π is an exponent smaller than π. The conclusive
equation is therefore:
π π
+ π π
= π π
,
(π π
) π
+ (π π
)
π
= (π π
)
π
.
Thus π π
, π π
and π π
are solutions to the equation with power π, which equals 4 or an odd prime number,
π (Ribenboim, 2013, Page 1-2).
1.2 The Pythagorean Theorem
Fermatβs Last Theorem states that there are no integer solution to the equation π π
+ π π
= π π
when
π > 2. Yet we know that there are integer solutions to the equation π2
+ π2
= π2
or otherwise known
as the Pythagoras equation/theorem. The three positive integers π, π, π that satisfy the equation are
known as the Pythagorean Triples. For example, a triple (π₯, π¦, π§) such as (3, 4, 5) can exist since 32
+
42
= 52
. In order to create a proof for π = 4, we must factorize the equation so that the integers of the
factor are in the form of a Pythagorean Triple whereby πππ(π, π, π) = 1; this form of factorization is
known as the unique factorization. The abbreviation πππ is used to define the greatest common divisor
(Ribenboim, 2013, Page 3-4).
Theorem 3. If π, π are integers such that π > π > 0, πππ(π, π) = 1. Then the triple (π₯, π¦, π§), given by:
{
π₯ = 2ππ
π¦ = π2
β π2
π§ = π2
+ π2
,
is a primitive solution to the equation π2
+ π2
= π2
and πππ(π₯, π¦, π§) = 1.
Proof. If π, π are integers that satisfy the above equation, then:
π₯2
+ π¦2
= 4π2
π2
+ (π2
β π2
)2
= (π2
+ π2
)2
= π§2
.
Since π and π are different in parity, meaning either π or π is positive or negative and π > π > 0, then in
relation to the above equation π₯ > 0, π¦ > 0 and π§ > 0 where π₯ is even and both π¦, π§ are odd. Now let
π = πππβ‘( π₯, π¦, π§), if π divides into π₯, π¦ and π§ then π must divide into 2π2
and 2π2
since π¦ + π§ = 2π2
and
π§ β π¦ = 2π2
. Therefore π = 1 since πππ(π, π) = 1 and π divides into π and π.
8. 3
2 The Biquadratic Equation
Fermat created his Last Theorem whilst attempting to solve the equation π4
β π4
= π2
. 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 numberβ (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 π = 4, which will be further
explained in the latter parts of this project.
2.1 The Method of Infinite Descent
Fermatβs 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 β2 is irrational. In order to apply the method of infinite descent, we must
first assume that β2 has a rational solution. By applying this assumption, we can show that β2 as a
quotient of two positive integers:
β2 =
π1
π1
, whereβ‘π1 > π1.
Now letβs assume the equation:
1
β2 β 1
= β2 + 1.
By replacing the β2 on the left-hand side with
π1
π1
, we acquire:
1
β π1
π1
β 1
= β2 + 1,
9. 4
therefore, by solving for β2, we derive the following equation:
β2 =
2 β
π1
π1
π1
π1
β 1
.
After further simplifying the above equation we obtain:
β2 =
2π1 β π1
π1 β π1
.
Since we know that 1 <
π1
π1
< 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, π2 and the denominator with π2.
Hence, we have the new quotient of β2:
β2 =
π2
π2
, where π2 > π2.
We can derive the equation above, since clearly, we know that π2 < π1 and π2 < π1 due to π1 > π1.
We can further repeat this process such that with an infinite amount of repeats, there will exist an
π(π + 1), which is less than ππ, and a π(π + 1), which is less than ππ. Henceforth, there is an infinite
descent for all positive integers that satisfy β2 =
π
π
, but this is impossible due to the fact that a
sequence of infinitely descending positive integers cannot exist. This, by contradiction, proves that β2 is
irrational and subsequently does not have a quotient of positive integers. As Fermat stressed upon, βthe
method of infinite descent proves that certain things are impossibleβ (Edwards, 1977, Page 8), therefore,
it is also known as the proof by contradiction.
2.2 The case of π = π
Now we will take up the case of π = 4. In order to show a proof for π = 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 π = 4 will be largely based on the proofs shown by Larry Freeman in his blog; Fermatβs Last Theorem
(2005) and Harold M. Edwards in his book; Fermatβs Last Theorem: A Genetic Introduction to Algebraic
Number Theory (1977).
Theorem 4. The equation, π4
+ π4
= π4
has no positive integer solution π, π, π.
Proof. Using the method to generate a Pythagorean triple, as explained in 1.2, we can assume that
π, π, π have no common divisor greater than 1 and as such, we say that π, π, π are co-prime. The
10. 5
equation above can be written as (π2
)2
+ (π2
)2
= (π2
)2
, therefore, π2
, π2
and π2
are the primitive
Pythagorean triple. Written in Pythagorean form:
π2
= 2ππ,
π2
= π2
β π2
,
π2
= π2
+ π2
,
where π, π are relatively prime (or co-prime) of opposite parity and π > π > 0. Out of the three
equations above, we can rearrange π2
= π2
β π2
into π2
+ π2
= π2
. Since π, π are relatively prime, we
have another set of primitive Pythagorean triple which are π, π, π. Additionally, π, π are also opposite in
parity and π > π > 0, where π is odd and π is even and so the triple can be written out as follows:
π = 2ππ,
π = π2
β π2
,
π = π2
+ π2
,
whereby π and π are relatively prime of opposite parity and π > π > 0. Henceforth we acquire the
following:
π2
= 2ππ = 2(π2
+ π2)(2ππ) = 4ππ(π2
+ π2),
π2
22
= ππ(π2
+ π2).
This shows the ππ(π2
+ π2
) is the square of half of π, where π is an even integer (Edwards, 1977, Page
9 β 10).
2.3 Euclidβs Lemma and the proof of π = π
Euclid of Alexandria (325BC β 265BC) is a renowned mathematician known for his treatise, The Elements
(OβConnor and Robertson, 1999). Euclidβs 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, π, divides ππ. Then π must divide either π or π.
Proof. Following the case of π = 4, we know that ππ(π2
+ π2
) is a square. Since any prime, π, can
divide ππ, we can say that ππ and π2
+ π2
are relatively prime since π would have to divide π or π but
11. 6
not both, and therefore π2
+ π2
is not divisible by π. Now that we know ππ is a square, π and π must
both be square since they are relatively prime. In other words, we can write both π and π as follows:
π = π2
,
π = π2
.
By substituting the above equation into π2
+ π2
we get π4
+ π4
. Since π2
+ π2
is a square, π4
+ π4
is
also a square, which leads to the fact that π4
+ π4
= π2
+ π2
= π < π2
+ π2
= π < π2
= π4
+ π4
.
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 π4
+ π4
= π4
has no integer solution where π₯, π¦, π§ β 0.
Proof. The equation above can be written as π4
+ π4
= (π2
)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 π = 4.
Due to this, we must now only prove that the theorem is true for π 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 π₯ π
+ π¦ π
= π§ π
has solutions when π β₯ 2. Since π is greater
than or equal to 2, we can say that π is divisible by 4, hence π₯
π
4, π¦
π
4 and π§
π
4 are solutions to π₯ π
+ π¦ π
=
π§ π
. Similarly, if π is an odd prime (π), then π₯
π
π, π¦
π
π and π§
π
π are also solutions to the equation. Therefore
by proving that no solutions exist for π = 4 and π = π, then we have proven that there are no solutions
for any exponent (Mack β Crane, 2015).
12. 7
3 The case of π = π
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 Fermatβs Last Theorem with the proof for π = 3, after Fermat,
who provided the proof for π = 4 (Freeman, 2005).
Leonhard Euler had come up with two proofs for the case of π = 3. His first proof had a mistake, which
he did not realise until trying to prove his lemma to the proof of π = 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; Fermatβs Last Theorem: A Genetic introduction to
Algebraic Number Theory (1977) and by Larry Freeman in his blog; Fermatβs Last Theorem (2005).
Theorem 5. The equation π3
+ π3
= π3
has no positive integer solution π₯, π¦, π§.
Euler applied Fermatβs method of infinite descent to show the proof to Fermatβs proposition; βA cube
cannot be equal to the sum of two non-zero cubesβ (Ribenboim, 2013, Page 24). Euler showed that if it
was possible to find positive integer solutions π₯, π¦, π§ for the equation π₯3
+ π¦3
= π§3
, 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 π₯, π¦, π§β‘are co-prime. This means that the greatest common divisor of (π₯, π¦), (π₯, π§) or
(π¦, π§) is 1. Following a simpler notation, gcd(π₯, π¦, π§) = 1. Since π₯, π¦, π§ are co-prime, we know that at
least one of the integers is even. Therefore, if both π₯ and π¦ are odd, then π§ would be even since
(Edwards, 1977, Page 40);
πππ + πππ = ππ£ππ.
We also know that π₯ and π¦ are symmetrical, since the equation π₯3
+ π¦3
= π§3
implies that both π₯ and π¦
have similar properties, thus by showing the case of π₯ being even, it will cover the case of π¦ being even
(Freeman, 2005). Due to this, we can now split the proof into two cases, with case 1 showing if π§ is even
and case 2 showing if π₯ is even.
13. 8
Case 1: π is even
Lemma 2. If π§ is even, then π, π exists such that:
1. gcd(π, π) = 1.
2. π, π are positive.
3. Either π or π is odd, therefore, have opposite parity.
4. π₯3
+ π¦3
= π§3
only has a solution if π, π exists with the above properties.
Proof. Since we have assumed that π₯, π¦, π§ are co-prime and we have assumed that π§ is even for this case,
then π₯ and π¦ are odd such that (Edwards, 1977, Page 40 β 41):
π₯ + π¦ = ππ£ππ,
π₯ β π¦ = ππ£ππ.
With the above in mind, let:
2π = π₯ + π¦,
2π = π₯ β π¦.
Since any number π, π would be even if in the form 2π, 2π. We will then have π₯ and π¦ such that:
π₯ =
1
2
(2π + 2π) = π + π,
π¦ =
1
2
(2π β 2π) = π β π.
This can be worked out by applying the method of solving simultaneous equations to 2π = π₯ + π¦ and
2π = π₯ β π¦. We can now express π₯3
+ π¦3
= π§3
in terms of π, π:
π§3
= π₯3
+ π¦3
,
= (π₯ + π¦)(π₯2
β π₯π¦ + π¦2),
= (π + π + π β π)[(π + π)2
β (π + π)(π β π) + (π β π)2],
= 2π(π2
+ 3π2
).
Both π and π have opposite parity since both π + π = π₯ and π β π = π¦ are odd. Equally π and π also
have gcd(π, π) = 1 since any factor that divides into π and π would divide into π₯ and π¦, but that is
14. 9
impossible as π₯ and π¦ 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 π₯, π¦ to π₯3
+ π¦3
= π§3
due to the co-prime integers,
π, π, existing such that (Freeman, 2005):
π§3
= 2π(π2
+ 3π2) = ππ’ππ.
Case 2: π 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 π₯ is even rather than π§.
Proof. As we know, π₯, π¦, π§ are co-prime, therefore, π¦ and π§ are both odd. In turn π§ + π¦ and π§ β π¦ are
both even, therefore there exists π, π such that we let (Freeman, 2005):
2π = π§ β π¦,
2π = π§ + π¦,
due to the fact that the equation will be rearranged to π₯3
= π§3
β π¦3
since π₯ is now even. Using
simultaneous equations, we have π₯, π¦ such that:
π§ = π + π,
π¦ = π β π.
Since π§, π¦ are both odd, π and π have opposite parity, and using the same argument as case 1,
gcd(π, π) = 1. Therefore, there exists π, π such that:
π₯3
= [π + π β (π β π)][(π + π)2
+ (π + π)(π β π) + (π β π)2]
= 2π(π2
+ 3π2) = ππ’ππ.
This results in the same conclusion as case 1 (Freeman, 2005).
Now that we have shown that regardless of either π₯, π¦, π§ being even, it will conclude with coprime
positive integers π, π such that 2π(π2
+ 3π2
) is a cube. The next step is to show that 2π and π2
+ 3π2
are co-prime and βthat the only way that their product can be a cube is for each of them separately to be
a cubeβ (Edwards, 1977, Page 41).
We know that π and π have opposite parity due to Lemma 2, therefore π2
+ 3π2
is odd, and we know
that π and π are co-prime, therefore we end up with the argument that if the above statement is true,
then gcd(2π, π2
+ 3π2) = 1 or 3.
15. 10
Lemma 3. If π, π are co-prime and have opposite parity, then gcd(2π, π2
+ 3π2) = 1β‘ππβ‘3.
Proof. We begin this proof by assuming that we have any positive integer π , such that π can divide into
2π and π2
+ 3π2
, therefore, π is a common factor of 2π and π2
+ 3π2
. We also know that π cannot be 2
since we have shown π2
+ 3π2
is odd, therefore we can assume that π is greater than 3. Consequently,
π΄, π΅ exists such that (Freeman, 2005):
2π = π π΄,
π2
+ 3π2
= π π΅.
Even though we know π is not 2, we know that 2 can divide into π΄ since 2π = π π΄, and so there exists πΆ
such that πΆ is half of π΄, resulting in:
π = π πΆ.
We can now substitute the value of π into π π΅, and acquire the following:
π2
+ 3π2
= π π΅,
3π2
= π π΅ β π2
,
3π2
= π π΅ β (π πΆ)2
,
= π (π΅ β π πΆ2).
Due to the fact that π cannot divide into the 3 of 3π2
, as it is greater than 3, it must divide into π since
we can apply Euclidβs Lemma as explained in section 2.3. However, this is contradictory since π is a
common factor of 2π and π2
+ 3π2
which, in turn, is a common factor of π, π2
+ 3π2
and π, 3π2
(Freeman, 2005).
Therefore, if π can divide into π, then it must divide into π, but this contradicts π, π being co-prime since
π divides into π. Therefore, we have two cases whereby we must argue that 3 does not divide into π,
gcd(2π, π2
+ 3π2) = 1, and that 3 does divide into π, gcd(2π, π2
+ 3π2) = 3 (Freeman, 2005).
16. 11
Case 1: 3 does not divide into π
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 Freemanβs
Blog; Fermatβs Last Theorem (2005). This Lemma states that if gcd(π, π) = 1 and ππ = π§ π
, then there
exists π₯, π¦ such that:
π = π₯ π
π = π¦ π
.
Applying this to our case, we know that π§3
= 2π(π2
+ 3π2
) and gcd(2π, π2
+ 3π2) = 1, then there
exists π₯, π¦ such that:
π₯3
= 2π
π¦3
= π2
+ 3π2
.
There forth an assumption that 2π, π2
+ 3π2
are both cubes can be made.
Lemma 4. If gcd(2π, π2
+ 3π2) = 1, then there must be smaller solutions to Fermatβs Last Theorem: π =
3.
Proof. Since we know that 2π, π2
+ 3π2
are cube, then we know that there exists π, π such that
gcd(π, π) = 1 and both π, π have opposite parities since π, π also have opposite parities. Let π’3
= π2
+
3π2
, since we are applying Relatively Prime Divisors Lemma (Freeman, 2005). We know that π, π have
opposite parities, in regards to which π’ can be stated as odd and so π’ will be in the form of π2
+ 3π2
because we know gcd(π, π) = 1, which means that every odd factor in the form π2
+ 3π2
will have the
same form.
As a result of the above:
π’3
= (π2
+ 3π2
)3
,
= π2
+ 3π2
(π2
+ 3π2
)2
,
= (π2
+ 3π2)(π4
+ 6π2
π2
+ 9π4),
= (π2
+ 3π2)(π4
β 6π2
π2
+ 12π2
π2
+ 9π4),
= (π2
+ 3π2)[(π2
β 3π2
)2
+ 3(2π)2],
17. 12
= [π(π2
β 3π2) β 3π(2ππ)]2
+ 3[π(2ππ) + π(π2
β 3π2)]2
,
= (π3
β 3ππ2
β 6ππ2
)2
+ 3(2π2
π + π2
π β 3π3
)2
,
= (π3
β 9ππ2
)2
+ 3(3π2
π β 3π3
)2
.
As π’3
= π2
+ 3π2
, then π2
+ 3π2
= (π3
β 9ππ2
)2
+ 3(3π2
π β 3π3
)2
. This allows us to define π, π
such that :
π = π3
β 9ππ2
= π(π β 3π)(π + 3π),
π = 3π2
π β 3π3
= 3π(π β π)(π + π),
where gcd(π, π) = 1.
We can now substitute the value for π into 2π, therefore (Freeman, 2005):
2π = 2(π3
β 9ππ2),
= 2π3
β 18ππ2
,
= 2π(π β 3π)(π + 3π).
Since π, π are opposite parities, we know that both π β 3π and π + 3π are odd. Due to this, any
common factor that divides into 2π, π Β± 3π would also be a common factor of π, π Β± 3π and
subsequently, be a common factor to π, Β±3π (Edwards, 1977, Page 42). We know that this common
factor cannot be 2 since 2 cannot divide into π Β± 3π, therefore the only common factor would be 3. But
if 3 divides into π, it would then be able to divide into π which is impossible since gcd(2π, π2
+ 3π2) =
1.
As a result of this, 2π, π Β± 3π are co-prime and all three are cube due to the Relative Prime Divisors
Lemma. We now have π΄, π΅, πΆ such that (Edwards, 1977, Page 42):
π΄3
= 2π
π΅3
= π β 3π
πΆ3
= π + 3π.
From the above equations, we can see that π΅3
+ πΆ3
= 2π = π΄3
, this gives a solution smaller than π₯, π¦, π§
where π₯3
+ π¦3
= π§3
. We know this is the case since (π΄3)(π΅3)(πΆ3) = 2π(π β 3π)(π + 3π) = 2π,
where π₯3
β‘or π§3
is 2π(π2
+ 3π2
) as shown in case 1: π is even and case 2: π is even.
18. 13
Case 2: 3 divides into π
Now, our final case to show that π₯3
+ π¦3
= π§3
has no integer solution π₯, π¦, π§ only differs slightly from
the previous case. In both cases, we have the same assumptions, therefore, π, π are co-prime with
opposite parity and 2π, π2
+ 3π2
are a cube. The difference, in this case, would be the fact that
gcd(2π, π2
+ 3π2) = 3 (Freeman, 2005).
Lemma 5. If gcd(2π, π2
+ 3π2) = 3, then there must be smaller solutions to Fermatβs Last Theorem:
π = 3.
Proof. To begin, we note that 3 divides into π but does not divide into π, since 3 can divide into 2π as
the gcd(2π, π2
+ 3π2) = 3 (Edwards, 1977, Page 42). But π and π are co-prime, therefore due to
Euclidβs Lemma, if 3 divides into π, it cannot divide into π. Since 3 divides into π, there exists π such that
(Freeman, 2005):
π = 3π.
By substituting the value of π into 2π(π2
+ 3π2) we acquire:
2π(π2
+ 3π2) = [2(3π)((3π)2
+ 3π2)],
= 32(2π)(3π2
+ π2).
Since 3 cannot divide π, it cannot divide 3π2
+ π2
. As a result of this, 32
(2π) and 3π2
+ π2
are co-
prime. Consequently, π and π are also of opposite parity, and therefore gcd(π, π) = 1 since gcd(π, π) =
1. We know that 2π(π2
+ 3π2
) is cube, thus 32(2π), 3π2
+ π2
are also cubes, since 2π(π2
+ 3π2) =
32(2π)(3π2
+ π2
).
From the previous case, we calculated π2
+ 3π2
= (π3
β 9ππ2
)2
+ 3(3π2
π β 3π3
)2
, therefore, as
before, there exists π, π such that (Freeman, 2005):
π = π3
β 9ππ2
= π(π β 3π)(π + 3π),
π = 3π2
π β 3π3
= 3π(π β π)(π + π),
where gcd(π, π) = 1.
Due to the fact that we know 32
(2π) is a cube, we can define π, π further such that:
32(2π) = 32[2(3π2
π β 3π3)],
19. 14
= 33[2(π2
π β π3)],
= 33[2π(π2
β π2)],
= 33[2π(π β π)(π + π)].
Following from the fact that 32
(2π) is a cube, then 33[2π(π β π)(π + π)] is a cube, and subsequently
2π(π β π)(π + π) is also a cube. For that reason, there exists π΄, π΅, πΆ such that:
π΄3
= 2π
π΅3
= π β π
πΆ3
= π + π.
This, in turn, gives another solution to Fermatβs Last Theorem; π = 3 as π΅3
+ πΆ3
= 2π = π΄3
, whereby
π΄3
π΅3
πΆ3
is less than π = 3π2
π β 3π3
= 3π(π β π)(π + π) which is also less than π = 3π, which is less
than π₯3
or π§3
= 2π(π2
+ 3π2
) as well (Edwards, 1977, Page 42).
Corollary 2. π π
+ π3
= π3
has no integer solution π₯, π¦, π§.
Proof. In both cases where gcd(2π, π2
+ 3π2) = 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.
23. 18
visual form, if an ideal number divides π in a cyclotomic ring, then it can divide ππ. Subsequently, if it
can divide π and π, then it can divide π + π (Mack β Crane, 2015).
With the above property in mind, an βideal numberβ 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 π to be prime, if there exists ππ in the cyclotomic ring, such that if ππ is in π (therefore
allowing ideal π in its own ring to be a subset of the cyclotomic ring) then π or π is in π. This allows π 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 ππ to be the smallest ideals
containing the product of cyclotomic integers, such as ππ (Mack β 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 (Ξ±) of the ring can produce an ideal of the element, which can be
divided by the said element (Ξ±). This element/ divisor is known as a principle (Edwards, 1977, Page 162).
By introducing these principle ideal numbers, the unique factorization is restored, since βevery ideal in
the ring of integers can be written uniquely as a product of prime idealsβ (Mack β 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 Γ 3, now if we consider the ring β€(ββ5) with the field β(ββ5) then we
can write the equation as (Varma, 2008):
6 = 2 β 3 = (1 + ββ5) β (1 β ββ5),
whereby 2, 3, (1 + ββ5), (1 β ββ5) are irreducible elements to β€(ββ5). However, with the
introduction of ideal numbers, the factors shown above can now be further factorized such that (Mack β
Crane, 2015):
2 = (2, 1 + ββ5)(2,1 β ββ5) = π1, π2,
3 = (3, 1 + ββ5)(3,1 β ββ5) = π3, π4,
1 + ββ5 = (2, 1 + ββ5)(3,1 + ββ5) = π1, π3,
1 β ββ5 = (2, 1 β ββ5)(3,1 β ββ5) = π2, π4.
We have now further factorized the equation into ideal prime factors π1, π2, π3, π4 that are unique and
hold in the cyclotomic field β(ββ5). As shown above, the unique factorization does not always hold in
25. 20
this fact to see if the factorization of the ring causes an infinite descent, thereby concluding that there
are no integer solutions to Fermatβs Last Theorem, where πβ‘a regular prime; if the integers π₯, π¦, π§ is not
divisible by π (Edwards, 1977, Page 170).
Case 2: divisible by π
Lemma 7. The integers π₯, π¦, π§ are co-prime and divisible by π
Proof overview. Referring to the same equation as in Case 1:
π§ π
= π₯ π
+ π¦ π
= (π₯ + π¦)(π₯ + π π π¦)(π₯ + π π
2
π¦)β¦ (π₯ + π π
πβ1
π¦).
If π does divide either π₯, π¦, π§, then this shows that all factors shown are not relatively prime meaning
that all factors are divisible by π. Despite this, the same case follows since the π π‘β
power is principle (π§ π
)
then the factors itself are principle π π‘β
powers. We then factorize elements of β€(π π) and form an
infinite descent. Therefore we prove Fermatβs Last Theorem for regular primes with the conclusion that
in both cases, there are no integer solution π₯, π¦, π§ (Mack β Crane, 2015).
26. 21
5 Conclusion
The general concept behind Fermatβs 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, βHere was a
problem, that I, a ten β year old, could understand and I knew from that moment that I would never let
it go. I had to solve itβ (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 βsimpleβ 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 Fermatβs 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 Wilesβs 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
Fermatβs Last Theorem, but one day there may be a chance that Fermatβs general proof will be
discovered and for that I encourage the reader to pursue this knowledge.
27. 22
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 Singhβs
book βFermatβs Last Theorem (2002)β 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
βsimpleβ 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 π was slightly different when comparing resources; such
as the proof for π = 4, when comparing the books βFermatβs Last Theorem for Amateursβ by Paulo
Ribenboim to βFermatβs Last Theorem: A Genetic Introduction to Algebraic Number Theoryβ 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 π π
= 1 for roots of unity in Harold M. Edwardsβs book,
rather than π π
= 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 π = 4, whilst introducing the Euclidβs 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 Fermatβs 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
28. 23
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 π =
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 Fermatβs Last Theorem through to, but not including, Andrew Wilesβs 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 Fermatβs 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 Fermatβs 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 Fermatβs Last Theorem and the evolution of number theory.
29. 24
7 Bibliography
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<http://www.ms.uky.edu/~carl/ma330/html/bailey21.html> [Accessed 23rd November 2015]
Edwards, H. M. (1977) Fermatβs last theorem: A genetic introduction to algebraic number theory,
Germany: Springer-Verlag Berlin and Heidelberg GmbH & Co. K
Freeman, L. (2005) Fermatβs last theorem [Online], Available:
<http://fermatslasttheorem.blogspot.co.uk/> [Accessed 10th November 2015]
Mack-Crane, S. (2015) Fermatβs last theorem for regular primes, [Online] . Available:
<https://math.berkeley.edu/~sander/speaking/22September2015%20WIM%20Talk.pdf> [Accessed 4th
May 2016]
OβConnor, 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) Fermatβs 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) Fermatβs last theorem for amateurs, United States: Springer-Verlag New York
Singh, S. (2002) Fermatβs 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) Fermatβs 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 FERMATβS LAST THEOREM, [Online] . Available:
<http://www.math.harvard.edu/~ila/Kummer.pdf> [Accessed 4th April 2016]