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Fermat’s Last Theorem and Quadratic Fields
Meijke Balay-Mickelson

Contents
1 Abstract

2

2 History and Background
2.1 Fermat Himself . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Hunt for the Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2
2
3

3 Pythagorean Triples
Introduction to Z[i] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Form of Primitive Pythagorean Triples . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Infinitude of PPTs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4
4
5
6

4 FLT for n = 4
Infinite Descent and Fermat’s Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Generalization to Multiples of 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7
7
8

5 Eisenstein Integers
Introduction to Z[ω] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Prime λ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9
9
10

6 FLT for n = 3
11
Proof by Infinite Ascent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
7 Conclusion

13

8 Appendix
14
8.1 Modular Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
8.2 Bezout’s Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1

Meijke Balay-Mickelson

1

Fermat’s Last Theorem

Abstract
This paper aims to give an accessible survey of the monumental mathematical
problem, Fermat’s Last Theorem, primarily through the technique of working
in quadratic fields. I provide an overview of the history of Fermat and his legacy
that propagated to mathematicians throughout the ages and subsequently introduce the methods of analyzing the solutions to Fermat’s equation by considering
primitive pythagorean triples in the Gaussian integers. This is used to prove
the case of fourth powers by way of Fermat’s infinite descent. The techniques
and principles gained from those two cases is extended to the slightly more
complicated case of Fermat’s Last Theorem for cubes, in which I work with the
Eisenstein integers.

2

History and Background

Fermat’s last theorem is one of the most famous problem in mathematics. Its deceptively and
enticingly simple statement is
Theorem 2.1 (Fermat’s Last Theorem) There are no integer solutions to
xn + y n = z n

(?)

with n > 2 and none of x, y, z equal to zero.

2.1

Fermat Himself

The original formulation of FLT goes back to the 17th century French mathematician Pierre de
Fermat. As a lawyer by vocation, Fermat was a mathematician only by avocation. However he was
anything but amateur! He collaborated with the likes of Blaise Pascal and Ren´e Descartes and is
known for his work in number theory, analytic geometry, and refraction [2]. Of course he is now
remembered most for his so-called “last theorem,” whose conjecture is an amusing piece of math
lore. Sometime in the 1630’s Fermat was reading the ancient greek text Arithmetica by Diophantus,
a treatise on integer and rational solutions to certain equations. Fermat, who liked to annotate his
copy, wrote in the margin
“It is impossible for a cube to be written as a sum of two cubes or a fourth power to be written
as the sum of two fourth powers or, in general, for any number which is a power greater than the
second to be written as a sum of two like powers. I have a truly marvelous demonstration of this
proposition which this margin is too narrow to contain.” [11]
Arithmetica would later be published with Fermat’s annotations included for other mathematicians to find, thus setting in motion the hunt for the elusive proof which would continue for over
350 years.

2

Meijke Balay-Mickelson

2.2

Fermat’s Last Theorem

Hunt for the Proof

Work on FLT extends across history, and naturally begins with proofs of particular exponents.
First Fermat proved the case of n = 4 utilizing infinite descent, which we will examine. Then the
famous mathematician Leonhard Euler gave proof of the case for n = 3 in 1770, Dirichlet and
Legendre for n = 5 in 1823, and finally Lam´e for n = 7 in 1839. The increasing complexity of these
single case proofs made generalization seem impossible until Sophie Germaine, a correspondent
of the famous Gauss proved that FLT holds for all odd primes p, where 2p + 1 is also a prime.
Unfortunately such primes are very rare [7] [12]. Soon after, the German mathematician Ernst
Kummer was able to generalize to exponents of “regular” primes, an infinite subset of the primes
which occur much more frequently.
The best mathematicians of every age from the time of Fermat to the 20th century tried their
hand at the proof, many with intense devotion and involvement. The Japanese genius Yutaka
Taniyama even killed himself after failing in his attempts, while Paul Wolfskehl claimed it had
saved him from suicide. From these efforts came substantial progress, but although Kummer came
tantalizingly close to a complete proof in the 19th century, a definitive conclusion to the problem
seemed ungraspable – so ungraspable in fact that Eric Temple Bell in his 1960 book The Last Problem predicted that it would still be unresolved when civilization eventually collapsed in a nuclear
apocalypse. [10]
Little did Bell know ten year old Andrew Wiles
would read his book and become obsessed with proving the theorem, and ultimately succeed. Thirty
years later Wiles became a prominent mathematician at Princeton University, and after working secretly for several years presented a partial proof
of the Shimura-Taniyama-Wiles Conjecture in 1995,
which was sufficient to imply Fermat’s Last Theorem [6] [9]. The complete proof was given with the
help of several other mathematicians in 2001 and is
now referred to as the modularity theorem, however
Wile’s 1995 proof alone spans about 100 pages and
Figure 2: Sir Andrew Wiles in 1998 (AP is inscrutable to all but the most specialized mathematicians. The conjecture is a profound result rePhoto/Charles Rex Arbogast)
garding elliptic curves and modular forms – needless
to say too advanced to detail here. Nevertheless, it is the consummation of centuries of beautiful
work in the wake of Fermat’s legacy and a monumental triumph of mathematics, of which we will
get a taste of by considering two particular cases.

Figure 1:
Fermat’s marginal
note in original Latin
3

Meijke Balay-Mickelson

3

Fermat’s Last Theorem

Pythagorean Triples

One reason why FLT appears innocuous on the surface is that if we take n = 2 in (?) we obtain
x2 + y 2 = z 2 , the Pythagorean Theorem with which every high schooler is familiar. Though FLT
is not explicitly concerned with n = 2, as an introduction we’ll analyze the solutions as if it were.
In doing so we’ll get a feel for FLT and working with complex numbers, as well as furnish a lemma
which will lead to a proof for the case n = 4.
Now recall that a pythagorean triple is a set of three positive integers (x, y, z) that satisfies the
pythagorean theorem. We’ll examine a subset of these with the condition that the three integers
are relatively prime.

Definition 3.1 A primitive pythagorean triple is a triplet of positive integers (x, y, z) that satisfy
gcd(x, y, z) = 1 and
x2 + y 2 = z 2

The goal is to derive a formula for a primitive pythagorean triple in two variables so that we may
easily discern the nature of the solutions. For the derivation we will work in the complex integers,
or those of the form x + yi where i is the imaginary unit and x and y range over the integers.
The set of these numbers is conventionally denoted Z[i] and called Gaussian integers in honor of
the mathematician Carl Friedrich Gauss. They form a ring, which is simply a set equipped with
addition and multiplication operations (for example the regular integers Z form a ring).
Note that if y = 0 we have a regular integer. Therefore the ordinary integers are a subset of
the Gaussian integers, or Z ⊂ Z[i]. This is advantageous because if we can prove a result for Z[i]
we have automatically proved it for Z. In general proving a result for other rings instead of the
integers themselves can be much easier. This technique was pioneered by Gauss and we will rely
on it for the proof of Lemma 3.1 and Theorem 6.1.

Figure 3: The Gaussian integers form a square lattice in the complex plane [3]

4

Meijke Balay-Mickelson

Fermat’s Last Theorem

Having some basic knowledge of complex numbers should ensure that Gaussian integers are not
too alien. In fact, they behave much like the integers, Z. For example, if you add, subtract, or
multiply Gaussian integers you get another Gaussian integer, however this is not always the case
with division. Knowing the units of Z[i] are {±i, ±1} say the associates of a Gaussian integer α∗ are
of the form uα where u is a unit, just as the associates of a regular integer k are ±k. It is then easy
to extend the notion of divisibility and primality to the Gaussian integers. Just as we say a | b iff
there exists c such that ac = b, we have
α | β ⇐⇒ ∃γ : αγ = β
where γ is unique up to associates. Accordingly a Gaussian prime is a Gaussian integer divisible
by only units and its associates.
An important property of Z[i] is that it is a unique factorization domain, meaning each number
can be written as a unique product of prime elements (again up to associates), just as the fundamental theorem of arithmetic asserts for Z. Essentially the Gaussian integers differ from regular
integers only in the notion of magnitude. Whereas in Z the magnitude of k is |k|, in Z[i] we have
to use a norm function N : Z[i] 7→ Z, however this is unimportant right now. In summary, for our
purposes we can assume Z[i] behaves identically to the familiar integers. Now we’re prepared for
the lemma:
Lemma 3.1 If (x, y, z) is a primitive pythagorean triple then
(x, y, z) = (a2 − b2 , 2ab, a2 + b2 )
where a and b are opposite parity, co-prime integers
Proof: First consider x2 + y 2 = z 2 modulo 4† . We have for all integers x and some integer k

(2k)2 = 4k 2 ≡ 0
if x is even
2
x =
(mod 4)
(2k + 1)2 = 4k 2 + 4k + 1 ≡ 1
if x is odd
Having both x2 and y 2 even would violate gcd(x, y) = 1 and having both odd would imply z 2 ≡ 2
(mod 4) which is impossible. Then exactly one of x2 , y 2 must be odd so that x2 +y 2 ≡ 0+1 = 1 ≡ z 2 .
Hence z 2 and z are odd. Now we’ll move to Z[i] as planned by rewriting x2 + y 2 as a product of
Gaussian integers:
(x + iy)(x − iy) = z 2

(1)

2

Suppose a Gaussian prime π divides x + iy, then π | z and so both sides of (1) are divisible by
π 2 . However if π | x − yi then π also divides (x + yi) + (x − yi) = 2x. But gcd(2x, z) = 1.
Contradiction. Therefore x + yi and x − yi are co-prime and both must be perfect squares in order
for their product to be a square. Keeping in mind associates, we can express either in the form uα2
where u ∈ {±i, ±1}, so let α = a + ib with a, b ∈ Z and write
x + iy = u(a + ib)2 = u(a2 + 2abi − b2 )
∗ In

(2)

this paper, Greek letters will denote integers of complex rings and Latin letters, ordinary integers.
modular arithmetic is unfamiliar, you need only know that a ≡ b (mod c) means a has remainder b upon
division by c.
† If

5

Meijke Balay-Mickelson

Fermat’s Last Theorem

Our solutions are found by examining the different cases of u. x is the real part of (2) and y is the
imaginary part.
u
x
y

−1

1
2

a −b
2ab

2

2

−i

i
2

b −a

−2ab
2

−2ab

2ab
2

b −a

a2 − b2

Note that in all cases z = a2 +b2 . Since we’re interested in only positive solutions we can assume
a > b > 0 and select (x, y, z) = (a2 − b2 , 2ab, a2 + b2 ). Since z 2 = (a2 + b2 )2 ≡ 1 (mod 4) clearly
a and b must have opposite parity. Finally, a and b are co-prime because if they had a common
divisor d, then d | (a − b)(a + b) = x and d | 2ab = y contradicting the fact that gcd(x, y) = 1
At this point our analysis of ‘FLT for squares,’ if you will, becomes easy because we have a
general form for the solutions. In particular, we know
Theorem 3.2 There are infinitely many non-trivial integer solutions to
x2 + y 2 = z 2

Proof: This is equivalent to saying there are infinitely many primitive pythagorean triples.
Using the notation from Lemma 3.1 fix b arbitrarily, say b = 2. Then there are infinitely many
choices for a (in this case all the odd primes) and obviously each choice of a will determine a unique
triple (x, y, z).

Since it turned out that the equation in (?) for n = 2 has infinitely many solutions, as does n = 1,
we may have na¨ıvely expected the trend to continue for higher exponents. Of course we know that
nothing could be further from the truth!

6

Meijke Balay-Mickelson

4

Fermat’s Last Theorem

FLT for n = 4

With this, we’re prepared to prove Fermat’s Last Theorem for n = 4, which states that
x4 + y 4 = z 4

(3)

has no non-trivial solutions over the integers (xyz 6= 0). This is the only case that Fermat published
a proof of, and as Andrew Wiles wittily remarked the proof does fit in the margin. It relies on the
principle of infinite descent, which was likely the strategy Fermat had in mind for his “marvelous
demonstration,” however it fails to generalize to other prime exponents.
Infinite descent is the process of showing that the existence of a solution implies the existence
of another smaller solution, which implies the existence of another smaller solution, ad infinitum.
This is impossible if the solutions are in positive integers because there is no strictly decreasing
infinite sequence of positive integers. More concisely, proof by infinite descent can be described as
the following:
Definition 4.1 If the assumption that a positive integer k0 has a property P implies that there
exists another positive integer k1 < k0 that also has property P, then no positive integer has the
property P.
Now note that if x4 + y 4 = z 2 has no solutions, then clearly neither does (3) because one could
instead write x4 + y 4 = (z 2 )2 . Therefore it is sufficient to prove
Theorem 4.1
x4 + y 4 = z 2

(4)

has no non-trivial integer solutions.
Proof: Seeking a contradiction, suppose there is a non-trivial solution (x, y, z). We can assume
(x2 , y 2 , z) is a primitive pythagorean triple because we can simply divide through by any common
divisor. Without loss of generality let x be even so that by Lemma 3.2 there exist co-prime, opposite
parity a and b with a > b > 0 such that
x2

=

2ab

2

= a2 − b2

z

= a2 + b2

y

On the other hand, (b, y, a) is also a primitive pythagorean triple so again by Lemma 3.2 there exist
another pair of opposite parity integers p and q with p > q > 0, gcd(p, q) = 1 which satisfy
b

=

2pq

y

= p2 − q 2

a

= p2 + q 2

7

Meijke Balay-Mickelson

Fermat’s Last Theorem

Substituting, we find
x2 = 2ab = 4pq(p2 + q 2 )
First notice (x/2)2 = pq(p2 + q 2 ) is the square of an integer. Similar to the proof of Lemma 3.1
we show that p, q, and p2 + q 2 are relatively prime implying they are all squares. Now suppose
a prime s divides pq. Then either s | p or s | q since gcd(p, q) = 1. Therefore s - p2 + q 2 =⇒
gcd(pq, p2 + q 2 ) = 1. Thus p, q, and p2 + q 2 all must be squares in order for their product to be a
square, and we can let p = X 2 , q = Y 2 , p2 + q 2 = Z 2 . for some X, Y, Z ∈ Z. But then
X4 + Y 4 = Z2
which is another solution to (5). In addition Z 2 = p2 + q 2 = a < a2 + b2 = z < z 2 , so using
the terminology of Definition 4.1 assuming k0 = z 2 has the property of being part of a non-trivial
solution of (4) leads to Z 2 = k1 < k0 = z 2 that also has that property. Hence we have infinite
descent and we’re done.

Theorem 4.1 immediately implies the following corollary:
Corollary 4.1
xn + y n = z n
has no non-trivial solutions for all n > 2 if 4 | n
Proof: If n = 4k (k = 1, 2, 3, ...) then we can write (?) as
xk

4

+ yk

4

= zk

4

As k ranges over the positive integers Theorem 4.1 ensures that the equation will never have a
non-trivial solution. Hence we have proven FLT for all exponents that are multiples of 4.

Now it’s clear that if FLT is proven for any particular n, it is also proven for all positive
multiples of n. Therefore it is sufficient to prove FLT for all the prime numbers. That is precisely
why mathematicians like Lam´e and Legendre proved prime cases like n = 7 and n = 5, instead of
n = 6 which follows from n = 3.

8

Meijke Balay-Mickelson

5

Fermat’s Last Theorem

Eisenstein Integers

The proof of cubic exponents is slightly more complicated than the one for fourth powers, but
will use the same kind of logic as the previous case. The original proof by Euler which works only
in Z is very lengthy, but thanks to Gauss’ technique of working in other rings we can explore a more
elegant proof using Eisenstein integers, another set of complex numbers. The Eisenstein integers
are named after Gotthold Eisenstein. They are very similar
√ to Z[i] except instead of i they are
based upon the primitive cube root of unity, ω = 12 (−1 + i 3), called such because it is the first
complex solution to z 3 = 1 (one may verify).

Definition 5.1 The Eisenstein integers are the set
Z[ω] = {x + ωy : x, y ∈ Z}
where ω is the primitive cube root of unity.

Z[ω] is also a unique factorization domain [5], so
associates, divisibility, and primality are defined identiFigure 4: The Eisenstein integers form cally to Z[i], except the units of Z[ω] are
a triangular lattice in C [3]
{±1, ±ω, ±ω 2 }. One thing to note about ω. It is a solution to z 3 − 1 = 0 which can be factorized as
(z − 1)(1 + z + z 2 ) = 0
Since z 6= 1 we get
1 + ω + ω2 = 0

(♥)

a nice property which will be useful in computation. From
this property and the fact that ω 3 = 1 it’s clear that a
major advantage Z[ω] will be the interesting new possible factorizations. For example, similar to
how we factored x2 + y 2 = (x + iy)(x − iy) in Z[i], we can now factor x3 + y 3 :
x3 + y 3 = (x + y)(x2 − xy + y 2 ) = (x + y)(xω + yω 2 )(xω 2 + yω)

9


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