<s>[INST] How to show that $\mathbb{Z}\left[\frac{1 + \sqrt{5}}{2}\right]$ is finitely generated? We can say that  $\mathbb{Z}\left[\cfrac{1 + \sqrt{5}}{2}\right]$ is finitely generated if minimal polynomial of $\cfrac{1 + \sqrt{5}}{2}$ is in $\mathbb{Z}[X]$. After some calculations it can be shown that $f(X) = X^2 - X - 1 \in \mathbb{Z}[X]$ is the minimal polynomial of $\cfrac{1 + \sqrt{5}}{2}$. I think that $\mathbb{Z}\left[\cfrac{1 + \sqrt{5}}{2}\right]$ is generated by $\left \{1, \cfrac{1 + \sqrt{5}}{2}\ \right \}$ so that any element of it can be written in the form $a + b \cfrac{1 + \sqrt 5}{2}$ or simply $\cfrac{c}{2} + \cfrac{d \sqrt{5}}{2}$ with $c, d \in \mathbb{Z}$.  
By this claim, if I take some $f\in \mathbb{Z}\left[\cfrac{1 + \sqrt{5}}{2}\right]$ then I should be able to write it in the form above. Now let $f = a_n \left(\cfrac{1 + \sqrt{5}}{2} \right)^n + a_{n - 1} \left(\cfrac{1 + \sqrt{5}}{2} \right)^{n - 1} + \dots + a_1 \left(\cfrac{1 + \sqrt{5}}{2} \right) + a_0$, each $a_i \in \mathbb{Z}$. The $k^{th}$ term in the partial sum above is equal to $a_k.\cfrac{n!.5^{\frac{k}{2}}}{k!(n-k)!2^k}$ . I can not see that how these terms cancel out each other and in the end we have something like $\cfrac{c}{2}+\cfrac{d\sqrt{5}}{2}$. How do we show this?
 [/INST] You might want to check that
$$\left(\frac{1+\sqrt{5}}2\right)^2=\frac{3+\sqrt5}2,$$
$$\left(\frac{1+\sqrt{5}}2\right)^3=2+\sqrt5.$$
$$\left(\frac{1+\sqrt{5}}2\right)^4=\frac{7+3\sqrt5}2$$
etc.
Indeed you can prove (by induction maybe) that
$$\left(\frac{1+\sqrt{5}}2\right)^n=\frac{a_n+b_n\sqrt5}2$$
where $a_n$ and $b_n$ are integers, moreover they are either both odd
integers or both even integers.
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