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!

  • 2
    $\begingroup$ one place you might check are the archives of fibonacci quarterly. $\endgroup$ – Harry Stern Jan 23 '11 at 2:08
  • 1
    $\begingroup$ Not a solution to any diophantine equations, but something related to the Golden Ratio: jstor.org/stable/3029218. $\endgroup$ – Arturo Magidin Jan 23 '11 at 2:10
  • 1
    $\begingroup$ @Willie: In fact, according to Wikipedia, $\mathbb{Z}[\varphi]$ does have class number $1$. $\endgroup$ – Arturo Magidin Jan 23 '11 at 2:19
  • 9
    $\begingroup$ @quanta: Well, $\mathbb{Z}[\varphi]$ is a UFD in which you can decompose $x^5+y^5=(x+y)(x^2+y^2-\varphi xy)(x^2+y^2-\bar\varphi xy)$, giving one way to prove FLT for exponent 5. $\endgroup$ – George Lowther Jan 23 '11 at 4:12
  • 1
    $\begingroup$ The interesting paper of Bergman "A number system with an irrational base" proposed by Arturo may be seen here. $\endgroup$ – Raymond Manzoni Jun 20 '12 at 14:53

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 $\mathbb{Q}(\zeta_5)$.

  • $\begingroup$ Thanks for this! Would anyone be able to explain how the highest common factor of c_0^2,c_1,c_2 is found to be 1 or sqrt(5)? $\endgroup$ – quanta Jan 24 '11 at 0:32
  • 1
    $\begingroup$ Because it divides $\sqrt{5}xy$. And, as it divides a power of $z$ (so is coprime to $x,y$), it must divide $\sqrt{5}$. As this is prime, $\sqrt{5}$ and 1 are the only possibilities. $\endgroup$ – George Lowther Jan 24 '11 at 0:36
  • 11
    $\begingroup$ btw, you can prove FLT for exponents 3,4,7 in a very similar way. For exponent 3, use $x^3+y^3=(x+y)(x+\omega y)(x+\omega^2y)$ in $\mathbb{Z}[\omega]$. For 4, which is the easiest, use $x^4-y^4=(x-y)(x+y)(x^2+y^2)$ in the integers. For 7, which gets rather messy, use $x^7+y^7=(x+y)(x^3+y^3-\alpha x^2y-\bar\alpha xy^2)(x^3+y^3-\bar\alpha x^2y-\alpha xy^2)$ in $\mathbb{Z}[\alpha]$ with $\alpha=(1+\sqrt{-7})/2$. $\endgroup$ – George Lowther Jan 24 '11 at 1:08
  • 4
    $\begingroup$ @George Lowther: +1, Very nice post. $\endgroup$ – Eric Naslund Jan 24 '11 at 2:28
  • 1
    $\begingroup$ Because $w_0=0$, $u_1=-1$, $w_1^5=\pm1$ and $w_2^5=\pm1$ mod $q$. As I mentioned earlier up, mod $q$ arithmetic is the same as mod 11 in the integeres, so 5th powers are equal to 0,1 or -1 mod $q$. $\endgroup$ – George Lowther Feb 13 '11 at 3:50

While not nearly as impressive as George's amazing answer, it's worth noting that $\mathbb{Q}[\phi]$ (though not quite $\mathbb{Z}[\phi]$, the elements are in $\frac{1}{2}\mathbb{Z}[\phi]$) shows up in the icosians (a subgroup of order 120 of the group of unit quaternions) and the theory of the icosahedral group and the 600-cell (and even, tangentially, $E_8$); check out the Wikipedia page on the icosians for more details. (It's, loosely, a higher-dimensional version of the description of the vertices of the icosahedron as the corners of the golden rectangle $(0, \pm 1, \pm \phi)$ and the other two rectangles given by cyclic permutations of the coordinates)

  • 2
    $\begingroup$ +1 The tangential relation to $E_8$ and such comes from the fact that when we build the division algebra of the usual Hamiltonian quaternions, but with center restricted to $\mathbb{Q}(\phi)$ instead of all the reals, we get the algebra known as Icosians. Its non-trivial Hasse invariants occur only at the infinite places. Hence the maximal orders of this algebra have a very small discriminant, and therefore they are destined to give rise to dense lattices. $\endgroup$ – Jyrki Lahtonen Jun 20 '12 at 11:11

Formally the Riemann zeta-function can be expressed as

$$ \zeta(z)=\prod_{k=0}^{\infty}\;\;\prod_{p\in \mathbb{P}}\bigg\{\left( 1 -\varphi^{-1}\;p^{-5^{k}z} +p^{-2\cdot5^{k}z}\right)\left( 1 +\varphi\;p^{-5^{k}z} +p^{-2\cdot5^{k}z}\right)\bigg\} \;for\;z>1$$

where $ \varphi=\frac{1+\sqrt{5}}{2} $ is the Golden Ratio. This follows from the fact that the zeta function can be expressed as

$$ \zeta(z)=\prod_{p\in \mathbb{P}}\;\;\sum_{k=0}^{\infty}\frac{1}{p^{k\;z}}$$

and after some manipulations its easy to get the above representation.

To justify this, take the series $$ f(x)=1+x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}+x^{7}+x^{8}+x^{9}+x^{10}+x^{11}+x^{12}+\cdots $$ we can write this as $$ f(x)=1+x+x^{2}+x^{3}+x^{4}+x^{5}(1+x+x^{2}+x^{3}+x^{4})+x^{10}(1+x+x^{2}+x^{3}+x^{4})+\cdots $$ or equivalently $$ f(x)=(1+x+x^{2}+x^{3}+x^{4})(1+x^{5}+x^{10}+x^{15}+x^{20}+x^{25}+x^{30}+x^{35}\cdots) $$ and again $$ f(x)=(1+x+x^{2}+x^{3}+x^{4})(1+x^{5}+x^{10}+x^{15}+x^{20}+x^{25}(1+x^{5}+x^{10}+x^{15}+x^{20})+x^{25}(1+x^{5}+x^{10}+x^{15}+x^{20})+\cdots) $$ or $$ f(x)=(1+x+x^{2}+x^{3}+x^{4})(1+x^{5}+x^{10}+x^{15}+x^{20})(1+x^{25}+x^{50}+x^{75}+x^{100}+\cdots) $$
so the general pattern is $$ f(x)=\prod_{k=0}^{\infty}(1+x^{1\cdot 5^{k}}+x^{2\cdot 5^{k}}+x^{3\cdot 5^{k}}+x^{4\cdot 5^{k}}) $$ now make $y=x^{5^{k}}$, one has that $$ 1+y+y^{2}+y^{3}+y^{4}=\left(y^{2}-\frac{\sqrt{5}-1}{2}y+1\right)\left(y^{2}+\frac{\sqrt{5}+1}{2}y+1\right) $$ where $\varphi=\frac{\sqrt{5}+1}{2}$ and $\frac{1}{\varphi}=\frac{\sqrt{5}-1}{2}$ so $$ 1+y+y^{2}+y^{3}+y^{4}=\left(y^{2}-\frac{1}{\varphi}y+1\right)\left(y^{2}+\varphi y+1\right) $$ now remember that $$ \zeta(s)=\prod_{p \in \mathbb{P}}\left(1+\frac{1}{p^{s}}+\frac{1}{p^{2s}}+\frac{1}{p^{3s}}+\frac{1}{p^{4s}}+\cdots \right) $$

  • $\begingroup$ Could you please give me a reference to this formula? $\endgroup$ – dot dot Jul 23 '12 at 17:35
  • $\begingroup$ Where is it valid ? I assume Re (s) > 1. $\endgroup$ – mick Sep 10 '12 at 15:46
  • $\begingroup$ Oh i read 2,5 instead of 2.5 haha. Makes sense now. Thanks for the clarification. One more comment : i think it might be intresting to generalize this so that it holds for $Re(z)>0$. Who knows you might be able to solve the RH then ;) $\endgroup$ – mick Sep 18 '12 at 18:32

You would probably solve the Mordell equation $y^2=x^3+5$ by working in that field.


I wonder why Binet's formula to calculate Fibonacci number wasn't already posted. Here it is: $$ F\left(n\right) = {{\varphi^n-(-\varphi)^{-n}} \over {\sqrt 5}}. $$ Since you asked for $\mathbb Z(\varphi)$, reformulate it to $$ 5 F\left(n\right)^2 +2 = \varphi^{2n}+\varphi^{-2n}. $$ to get a nice identity for your Fibonacci integers.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.