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!
 A: You would probably solve the Mordell equation $y^2=x^3+5$ by working in that field. 
A: 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)
A: 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
      $$
      \zeta(s)=\prod_{p \in \mathbb{P}}\left(1+\frac{1}{p^{s}}+\frac{1}{p^{2s}}+\frac{1}{p^{3s}}+\frac{1}{p^{4s}}+\cdots \right)
      $$
A: 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.
