Golden Number Theory

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!

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.

Theorem: There are no solutions to x^5+y^5=z^5 for nonzero x,y,z in \mathbb{Z}[\varphi].

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.

Theorem 2: There are no solutions to x^5+y^5=uz^5 for nonzero coprime x,y,z\in\mathbb{Z}[\varphi] with u\in\mathbb{Z}[\varphi] a unit and q dividing z.

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).

Attribution
Source : Link , Question Author : quanta , Answer Author : George Lowther

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