Catalan's Conjecture - A learning exercise for the bored mind.

I was going through a video by Numberphile where they were talking about the Catalan’s conjecture.
This is another conjecture which is simple to state; however extremely difficult to prove; just like Collatz conjecture, about which I already have a blog in place.
At the very outset of the blog; let me bring it to your knowledge that this blog is just for learning and understanding and contains “very little”-to-“null” original work on the subject. However, this blog will be exhaustive and will contain a lot of references that will help a newbie (like myself) get into the depths of the conjecture and fully understand concepts used in its proof.
Let us now define the conjecture;
This statement was conjectured by Eugene Catalan (1814–1894) and was sent to the editor of Journal fur die Reine und Angewandte Mathematik$^{[1]-\text{Introduction}}$.
The conjecture is as follows,
$2^3$ and $3^2$are two powers of natural numbers whose values are consecutive (i.e., 8 and 9); the conjecture is for a mathematical statement such as,
$$\boxed{x^a-y^b=1} \tag*{(1)}$$
The only solution of $(1)$ is for $a$, $b$ $>$ $1$ and $x$, $y$ $>$ $0$ is $x=3$, $a=2$, $y=2$, $b=3$.
Catalan’s conjecture was proven true by Preda Mihăilescu in 2002; the proof involves the theory of cyclotomic fields and Galois Modules$^{[2]-\text{History}}$.
So, as you see now, the breadth of the subject blew up! On second were talking about squares and cubes and now we are talking of Cyclotomic fields and Galois Modules!
Nevertheless, I will try and cover each topic in brief and quench our mathematical thirst!
Let us consider $(1)$ (with $a=p$ and $b=q$; for the sake of eliminating any confusion between $a$ and an english “a”), unless otherwise stated, $x$ and $y$ can be negative integers as well. Now, we re-write $(1)$ as follows,
$$(x-1)\frac{x^p-1}{x-1}=y^q \tag*{(2)}$$
The GCD of the two factors on the left hand side of the equation (after considering $x^p = ((x-1)+1)^p$) is either $p$ or $1$ (How did this happen?).

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Some concepts before we move ahead:

The Wieferich pairs
$^{[3]-\text{Wieferich pair}}$In mathematics,
a Wieferich pair is a pair of prime number $p$ and $q$ that satisfy,
$$p^{q-1} \equiv 1( \mod p^2)$$
Let us write $(2)$ as, $$(x-1)\frac{x^p-1}{x-1}=pb^q$$ for, $p \nmid b$
This suggests a traditional approach of factorizing the left hand side
in $\mathbb{Z}[\zeta]$, the ring of integers in the $p$th cyclotomic
field $\mathbb{Q}(\zeta)(\zeta=e^{2\pi i/p})$

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Ring$^{[4]-\text{Ring-Wolfram}}$:
A ring in the mathematical sense is a set $S$ together with two binary
operators $+$ and $.$ (Addition and multiplication), satisfying the
following conditions:
1. Additive associativity: For all $a,b,c \in S$, $(a+b)+c=(a+b)+c$,
2. Additive commutativity: For all $a,b\in S$, $a+b=b+a$,
3. Additive identity: There exists an element $0$ in $S$ such that for all a in $S$, $0+a=a+0=a$.
4. Additive inverse: For every a in $S$ there exists $-a \in S$ such that $a+(-a)=(-a)+a=0$,
5. Left and right distributivity: For all $a,b,c \in S$, $a.(b+c)=(a.b)+(a.c)$ and $(b+c).a=(b.a)+(c.a)$,
6. Multiplicative associativity: For all $a,b,c \in S$, $(a.b).c=a.(b.c)$ ($a$ ring satisfying this property is sometimes
explicitly termed an associative ring). Conditions 1-5 are always
required. Though non-associative rings exist, virtually all texts also
require condition 6.
7. Multiplicative commutativity: For all $a,b \in S$, $a.b=b.a$ ($a$ ring satisfying this property is termed a commutative ring),
8. Multiplicative identity: There exists an element $1 \in S$ such that for all $a \neq 0 \in S$, $1.a=a.1=a$ (a ring satisfying this
property is termed a unit ring, or sometimes a “ring with identity”),
9. Multiplicative inverse: For each $a \neq 0 \in S$, there exists an element $a^{-1} \in S$ such that $\forall a \neq 0 \in S, a.a^{-1}=a^{-1}.a=1$, where $1$ is the identity element.

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A brief history of the past developments on this conjecture is a must to be read and understood; the significance comes due to the period of 150 years for which it remained an open problem$^{[5]-\text{Catalan's Conjecture: Are 8 and 9 the Only Consecutive Powers?}}$,

• Only after six years after Catalan formally defined the conjecture, a result was proposed by French mathematician, Victor Lebesgue. He stated that, for the equation, $x^p-y^2=1$; where $p$ is a prime; has no solutions for positive values of $x$ and $y$. A proof of the same will be discussed in brief in the later part of the blog.
• After Lebesgue’s work, all development solely consisted of small exponents, and then Naggel showed in 1921 that the difference between a third power and an other perfect power never is equal to 1.
• In 1932, Selberg proved that, $x^4-y^n=1$ has no solution in positive integers when $n>1$. A stronger result to this was proved by Ko Cho in 1965, that stated that the equation $x^2-y^q=1$ has no solutions for positive integers when $q \geq 5$.
• Cassels made some observations for $x^p-y^q=1$ where $p$ and $q$ are odd-primes. He proved that is this equality holds for positive integers $x$ and $y$, then $p$ divides $y$ and $q$ divides $x$. For the case $p=2$, this had already been shown by Naggel
• Inkeri defined the concept of a Wieferich pair [the definition and explanation of the same is given above] in the concept of Catalan equation as follows:
  If the Catalan’s equation $(1)$ holds, then either $(p,q)$ is a Wieferich Pair, or $q$ divides $h_p$, the class number of the cyclotomic field $\mathbb{Q}(\zeta_p)$, or $p$ divides $h_q$, the class number of the cyclotomic field $\mathbb{Q}(\zeta_q)$

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Cyclotomic Field$^{[5]-\text{Cyclotomic field}}$:
In number theory, a cyclotomic field is a number field obtained by adjoining a complex primitive root of unity to $\mathbb{Q}$, the field of rational numbers. The $n$-th cyclotomic field $\mathbb{Q}(\zeta_n)$ (where $n > 2$) is obtained by adjoining a primitive $n$-th root of
Primitive root of unity$^{[6]-\text{Root of unity}}$
In mathematics, a root of unity, occasionally called a de Moivre number, is any complex number that gives $1$ when raised to some positive integer power $n$.

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• Some time later Mihailescu proved that the Catalan equation has no solutions if $p$ and $q$ are odd and $q$ does not divide $p-1$. By this result the Catalan conjecture became a theorem

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More on Cyclotomic Fields:
Let $p$ be an odd-prime number. Let $\Phi_p$ be the $p$-th cyclotomic polynomial in $\mathbb{Q}[X]$ i.e., $\Phi_p = \frac{X^p-1}{X-1}$. Consider the field extension $\mathbb{Q}[X]/(\Phi_p) \cong \mathbb{Q}(\zeta)$ of $\mathbb{Q}$, where $\zeta$ denotes a primitive $p$-th root of unity. This is a field extension of degree $p-1$ and it is reducible in $\mathbb{Q}[X]$. We denote $\mathbb{Q}(\zeta)$ by $k$ from now on.
This field extension is Galois with Galois Group,
$$G=\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q}) \cong (\mathbb{Z}/p\mathbb{Z})^{*},$$
Since the map,
$$(\mathbb{Z}/p\mathbb{Z})^{*} {{\sim} \atop {\to}}\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$$
$$a (\mod p) \mapsto (\sigma_a : \zeta \mapsto \zeta^a)$$
is an isomorphism
The automorphism $\sigma_{p-1}$ acts in all embeddings as complex conjugation. Therefore, we call $\sigma_{p-1}$ complex conjugation.
The fixed field of complex conjugation is $\mathbb{Q}(\zeta+\zeta^{-1})$, which is called the maximal real subfield of $\mathbb{Q}(\zeta)$. We denote $\mathbb{Q}(\zeta + \zeta^{-1})$ by $K^{+}$. The field extension $\mathbb{Q}(\zeta+\zeta^{-1})$ of $\mathbb{Q}$ has degree $\frac{p-1}{2}$ and it is Galois with Galois theory.
$$G^{+}=\text{Gal}(\mathbb{Q}(\zeta+\zeta^{-1})/\mathbb{Q}) \cong (\mathbb{Z}/p\mathbb{Z})^{*}/(\pm 1)$$

Another important concept that is

Mihailescu’s proof

To be contd…in the next blogpost!