The Gaussian $\mathbb{Z}[i]$ and Eisenstein $\mathbb{Z}[\omega]$ integers have been used to solve some diophantine equations. I have never seen any examples of the golden integers $\mathbb{Z}[\varphi]$ used in number theory though. If anyone happens to know some equations we can apply this in and how it's done I would greatly appreciate it!
Number Theory – Golden Number Theory
algebraic-number-theorydiophantine equationsnumber theory
Related Solutions
Quite a lot of algebraic number theory was invented through trying to prove Fermat's last theorem and other Diophantine problems.
For example, if I asked you to solve the equation $x^2 - y^2 = 5$ in integers it is very simple, you can factorise $(x+y)(x-y) = 5$ and solve the problem by linking to divisors of $5$, in order to get the solutions $x = \pm 3, y=\pm 2$.
Now say I ask you to solve the equation $y^3 = x^2 + 2$ in integers. This is not so easy if we work entirely in $\mathbb{Z}$ and use elementary methods. However, if we shift focus into a bigger ring of numbers, $\mathbb{Z}[\sqrt{-2}] = \{x+y\sqrt{-2}\,|\,x,y\in\mathbb{Z}\}$ then the problem again turns into a multiplicative problem:
$(x + \sqrt{-2})(x- \sqrt{-2}) = y^3$
so that solving the original Diophantine is really the same as solving a "product" style equation in $\mathbb{Z}[\sqrt{-2}]$.
The point of (basic) algebraic number theory is to study rings like this. How do the elements in these rings factorise?
In our problem above it turns out that the ring $\mathbb{Z}[\sqrt{-2}]$ has properties that are strangely close to properties of $\mathbb{Z}$. In fact the elements in this ring "factorise uniquely" into irreducible elements (the analogue of prime numbers in $\mathbb{Z}$).
The phrase "factorise uniquely" does not have quite the meaning you might think, we have to allow for multiplication of units (things that "divide $1$"). It is the ordering of $\mathbb{Z}$ allows us to consider unique factorisations into "positive primes".
There is also a notion of coprimality. This allows us to solve our problem since for odd $x$ it can be shown that $x\pm\sqrt{-2}$ are coprime in $\mathbb{Z}[\sqrt{-2}]$. But their product is a cube so (as in $\mathbb{Z}$), we must have that $x + \sqrt{-2} = (a+b\sqrt{-2})^3$ for some $a+b\sqrt{-2}\in \mathbb{Z}[\sqrt{-2}]$. Comparing coefficients lets you find the possibilities for $a,b$, hence for $x$.
The ideas of Lame and Kummer were to study FLT in the same way by considering the factorisation (for $\zeta$ a primitive $p$-th root of unity):
$z^p = x^p + y^p = (x + y)(x + \zeta y) ... (x + \zeta^{p-1} y)$
forming yet another product equation, now in the ring $\mathbb{Z}[\zeta]$.
Now this is not the entire story, since some of the rings we study in algebraic number theory do not have unique factorisation. For example the ring $\mathbb{Z}[\sqrt{-5}]$ does not since:
$6 = 2\times 3 = (1+\sqrt{-5})(1 - \sqrt{-5})$
gives two totally different factorisations of $6$. Actually the ring $\mathbb{Z}[\zeta]$ does not have unique factorisation for $p=23$, so that FLT could not be solved entirely by the above method.
The thing that stopped factorisation being unique was the fact that the ring wasn't big enough to factorise everything further into the same things.
Fortunately we can restore unique factorisation without having to extend! Through the genius of Kummer and Dedekind, they realised that by considering the "multiples" of an element as an object in its own right, we can reform factorisation in a way that becomes unique upto ordering.
In modern language these objects are called ideals of a ring. There is a notion of a prime ideal, capturing the notion of prime number. The different factorisations of $6$ above can be explained as reordering of the prime ideals in the factorisation of the ideal generated by $6$. These prime ideals are NOT generated by one element, so they dont correspond to "multiples" of something in $\mathbb{Z}[\sqrt{-5}]$, they correspond more to "multiples" of something that doesn't exist in the ring, but would exist after making an extension.
Kummer was able to prove a huge number of cases of FLT by using the ideal theory. This is outlined in many books.
Focus in algebraic number theory now turns to studying these algebraic constructions. We see that in a given "nice" ring, certain prime numbers may factorise, whereas others don't.
For example, in $\mathbb{Z}[i]$ we find that a prime $p$ factorises further if and only if $p=2$ or $p \equiv 1$ mod $4$. The factorisation of $2$ is different to the others in that $2 = (1+i)^2$ is not "square-free". All the others factorise into two different factors. We say that $2$ ramifies, primes $p\equiv 1$ mod $4$ split and primes $p\equiv 3$ mod $4$ are inert.
This congruence relationship describing the factorisation of primes is in some sense really explained by the values of the Legendre symbol $\left(\frac{-1}{p}\right)$, which is also explaining sums of two squares! Working in similar rings gives you the entire quadratic reciprocity law.
The goal of class field theory is to explain the splitting of primes in ANY extension of "number fields" to get similar characterisations in terms of congruences. In fact I just told a lie, we cannot yet do this for ANY extension, class field theory does it for abelian extensions (ones with abelian Galois group) but never-the-less it is quite a strong theory that has many applications (for example it solves the question of which primes can be written as $x^2 + ny^2$.
In the case of abelian extensions of $\mathbb{Q}$ we find that there are simple congruence conditions mod some integer $N$ that completely describe splitting behaviour of primes!
Another side of class field theory is the Cebotarev density theorem, which states essentially that most splitting types happen infinitely often. This is a huge generalisation of Dirichlet's theorem on primes in arithmetic progressions...in fact it provides an infinite amount of Dirichlet theorems, one for each abelian extension.
These days the (mostly unsolved) Langlands program is supposed to be filling in the gaps for non-abelian extensions but this is very difficult to understand and is not yet completely understood. When this is fully understood it will prove to be the holy grail of number theory, it will characterise in a huge way the splitting of primes.
Anyway, I hope this somewhat rushed introduction will whet your appetite. The book I first started with was Stewart/Tall - Algebraic number theory and fermat's last theorem. This is a good book to start you off. Also Lang - Algebraic number theory, Cox - Primes of the form $x^2 + ny^2$ and Childress - Class field theory are good ones to start with for class field theory.
Best Answer
Using the fact that $\mathbb{Z}[\varphi]$ is a unique factorization domain in which we can decompose $$ x^5+y^5=(x+y)(x^2+y^2-\varphi xy)(x^2+y^2-\bar\varphi xy),\quad\qquad{\rm(1)} $$ we can give a proof of Fermat's Last Theorem for the case of exponent 5. Here, I am using a bar over a number to denote conjugation, so $\varphi=(1+\sqrt{5})/2$ and $\bar\varphi=(1-\sqrt{5})/2$.
That is, for exponent 5, FLT holds in the ring $\mathbb{Z}[\varphi]$ and, in particular, it holds in the integers.
Before going any further, let's note a few facts about factorization in $\mathbb{Z}[\varphi]$. As is well known, it is norm-Euclidean, so is a unique factorization domain. We have the prime factorizations $5=(\sqrt{5})^2$ and $11=q\bar q$, where I am setting $q=4-\sqrt{5}$ (for the remainder of this post). The identity $\varphi\bar\varphi=-1$ shows that $\varphi$ is a unit. In fact, it is a fundamental unit, so that every unit in $\mathbb{Z}[\varphi]$ is of the form $\pm\varphi^r$ for integer $r$. It will also be useful to use mod-q arithmetic (with $q$ as above). Then, $\varphi=(1+\sqrt{5})/2=8$ (mod q). Therefore every element of the quotent $\mathbb{Z}[\varphi]/(q)$ is equal to a rational integer mod q. As 11 = 0 mod q, this gives $\mathbb{Z}[\varphi]/(q)\cong\mathbb{Z}/(11)$. So, mod-q arithmetic in $\mathbb{Z}[\varphi]$ is exactly the same as mod-11 arithmetic in the integers. In particular, every 5'th power is equal to one of $0,1,-1$ mod q. Applying this to the equation $x^5+y^5=z^5$ shows that at least one of $x,y,z$ must have a factor of q. By dividing through by their highest common factor, we reduce to the case where $x,y,z$ are coprime, so exactly one is a multiple of q. Rearranging as $(-z)^5+y^5=(-x)^5$ if necessary, we can always bring the multiple of q to the right hand side. This reduces the problem to the following.
Let's prove this by showing that, if we have one solution, then we can find another solution for which $xyz$ has strictly fewer distinct prime factors. Applied to a minimal solution, this would give a contradiction. This is essentially the method of descent used by Fermat himself for the case of exponent 4.
So, suppose we have one solution. Writing $c_0=x+y$, $c_1=x^2+y^2-\varphi xy$ and $c_2=x^2+y^2-\bar\varphi xy$, (1) gives the decomposition $uz^5=c_0c_1c_2$. Also, $$ c_0^2-\bar\varphi c_1-\varphi c_2=0.\qquad\qquad{\rm(2)} $$ We would like to show that the factors $c_0,c_1,c_2$ are 5'th powers, which will be easier if they are coprime. Using the fact that $x,y$ are coprime to $z$, the identities $$ \begin{align} &c_0^2-c_1=\sqrt{5}\varphi xy,\\ &c_0^2-c_2=-\sqrt{5}\bar\varphi xy,\\ &c_1-c_2=-\sqrt{5}xy \end{align} $$ show that the highest common factor of $c_0^2,c_1,c_2$ is either 1 or $\sqrt{5}$. Consider the case where $\sqrt{5}$ divides $z$. Then it will also divide at least one of $c_i$, and the identities above show that it divides each $c_i$. In particular, 5 divides $c_0^2$, so the identities above show that $\sqrt{5}$ divides each of $c_1,c_2$ exactly once.
In the case where $z$ is not a multiple of $\sqrt{5}$, let us set $\tilde c_0=c_0^2,\tilde c_1=c_1,\tilde c_2=c_2$ and, in the case where $\sqrt{5}$ divides $z$, set $\tilde c_0=c_0^2/\sqrt{5},\tilde c_1=c_1/\sqrt{5},\tilde c_2=c_2/\sqrt{5}$. These are coprime and $$ \tilde c_0\tilde c_1^2\tilde c_2^2 = u^2\left(z^{2}/\sqrt{5}^{m}\right)^5 $$ where $m=0$ if $\sqrt{5}$ does not divide $z$ and $m=1$ if it does. As they are coprime, each prime factor of $z$ divides exactly one of the $\tilde c_i$, and its exponent is a multiple of 5. So, considering prime factorizations, each $\tilde c_i$ is equal to a unit multiplied by a fifth power $w_i^5$. So, (2) gives $$ u_0w_0^5+u_1w_1^5+u_2w_2^5=0 $$ for units $u_i$. Without loss of generality, we assume that $q$ divides $w_0$ and, dividing through by $-u_1$ if necessary, we suppose that $u_1=-1$. Then, $u_2=\pm1$ mod q. However, being a unit, we have $u_2=\pm(\varphi)^r=\pm 8^r$ (mod q), and, looking at this mod 11, 5 must divide $r$. So, $u_2$ is a fifth power and, by absorbing $-u_2^{1/5}$ into $w_2$, we can take $u_2=-1$. So we have arrived at $$ w_1^5+w_2^5=u_0w_0^5. $$ Also, all prime factors of $w_0w_1w_2$ are factors of $z$. So, except in the case where $x,y$ are units, we have a solution with strictly fewer prime factors, and we are done.
So suppose that we have a solution to Theorem 2. Iteratively applying the procedure above will keep generating new solutions and, as the number of prime factors of $xyz$ cannot decrease indefinitely, we must eventually settle on the case where $x,y$ are units, so that $x/y=\pm\varphi^r$. Exchanging $x,y$ if necessary, we suppose that $r > 0$. Then, $q^5$ is a factor of $1\pm\varphi^{5r}$. Using the identity $\varphi^5=-1+\varphi^4q$, and applying the binomial identity, it can be seen that $rq$ must be a multiple of $q^{5}$, so $r$ is a multiple of $11^4$. In particular, $\vert x/y\vert=\vert\varphi\vert^r$ will be very large (note, $\vert\varphi\vert^{11^4} > 10^{3000}$). Then, the definitions above for $c_0^2,c_1,c_2$ are dominated by the $x^2$ terms, so the ratios $\tilde c_i/\tilde c_j$ are close to one. Going through these details bounds the ratios $w_i/w_j$ and, in particular, none of them will be as large as $\vert\varphi\vert^{11^4}$. This means that we cannot have $x,y$ and $w_1,w_2$ all units. So, continuing the induction will generate solutions with ever fewer prime factors, giving the required contradiction.
This method of approaching FLT for exponent 5 was something I came up with after seeing the exponent 3 case in lectures years ago. It is a bit tedious having to separately deal with the case where $x,y$ are units. Maybe that can be tidied up. Essentially, the reason why this method works is because $\mathbb{Z}[\varphi]$ consists precisely of the real algebraic integers of the cyclotomic field $\mathbb{Q}(\zeta_5)$.