Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer. How do we prove that $\dfrac{\phi^{400}+1}{\phi^{200}}$ is an integer, where $\phi$ is the golden ratio?
This appeared in an answer to a question I asked previously, but I do not see how to prove this..
 A: We have $\phi^2=\phi+1$. We can use this to iterate powers of $\phi$. We have $\phi^3=2\phi+1$, $\phi^4=3\phi+2$, etc.
We can iterate this, and finally obtain
$$
\frac{\phi^{400}+1}{\phi^{200}}=627376215338105766356982006981782561278127.
$$
This is a squarefree composite number.
A: $$\color{blue}{(\phi^2+\phi^{-2})}\in \mathbb N,$$
$$(\phi^2+\phi^{-2})^2=\color{blue}{(\phi^4+\phi^{-4})}+2\in \mathbb N,$$
$$(\phi^2+\phi^{-2})^3=\color{blue}{(\phi^6+\phi^{-6})}+3(\phi^2+\phi^{-2})\in \mathbb N,$$
$$(\phi^2+\phi^{-2})^4=\color{blue}{(\phi^8+\phi^{-8})}+4(\phi^4+\phi^{-4})+6\in \mathbb N,$$
$$(\phi^2+\phi^{-2})^5=\color{blue}{(\phi^{10}+\phi^{-10})}+5(\phi^6+\phi^{-6})+10(\phi^2+\phi^{-2})\in \mathbb N,$$$$...$$
$$(\phi^2+\phi^{-2})^{100}=\color{blue}{(\phi^{200}+\phi^{-200})}+100(\phi^{196}+\phi^{-196})+4950(\phi^{192}+\phi^{-192})+...\in \mathbb N.$$
A: We can prove by induction that 
if $x+\dfrac1x$ is an integer,  $x^n+\dfrac1{x^n}$ will be an integer
as $$\left(x^n+\frac1{x^n}\right)\left(x+\frac1x\right)=x^{n+1}+\frac1{x^{n+1}}+x^{n-1}+\frac1{x^{n-1}}$$
$$\iff x^{n+1}+\frac1{x^{n+1}}=\left(x^n+\frac1{x^n}\right)\left(x+\frac1x\right)-\left(x^{n-1}+\frac1{x^{n-1}}\right)$$
The base cases being
$n=1\implies x^2+\dfrac1{x^2}=\left(x+\dfrac1x\right)^2-2$ and
$x^3+\dfrac1{x^3}=\left(x+\dfrac1x\right)^3-3\left(x+\dfrac1x\right)$
or $n=2\implies x^3+\dfrac1{x^3}=\left(x^2+\dfrac1{x^2}\right)\left(x+\dfrac1x\right)-\left(x^1+\dfrac1{x^1}\right)$
As Golden Ratio$(\phi)$ satisfies $x^2-x-1=0$
we have $x^2-1=x\implies x-\dfrac1x=1\implies x^2+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2=1^2+2$
Here $n=100$
A: In general if $x+\dfrac1x$ is an integer then for all $n$, $x^n+\dfrac{1}{x^n}$ is an integer.
Both, $x$ and $\dfrac 1x$ are roots of the same equation $X^2-aX+1=0$ 
where $a$ is an integer. It follows that any equation deduced from it is also an equation of both  $x$ and $\dfrac 1x$.
We have 
$$X^2=aX-1$$ $$X^3=aX^2-X=a(aX-1)-X=(a^2-1)X-a$$
 $$X^4=(a^2-1)X^2-aX=(a^2-1)(aX-1)-aX=(a^3-2a)X-(a^2-1)$$ 
For $X^n$ one has by iteration$$X^n=f_n(a)X+g_n(a)$$ where $f_n(a)$ and $g_n(a)$ are integers.
Since also $$\left(\frac{1}{X}\right)^n=f_n(a)\left(\frac{1}{X}\right)+g_n(a)$$ we conclude that
$$x^n+\frac{1}{x^n}=f_n(a)(x+\frac1x)+2g_n(a)=af_n(a)+2g_n(a)\in\mathbb Z$$
(Note that this mode allows us to calculate the integer values of  $x^n+\dfrac{1}{x^n}$).
A: More generally,
$\dfrac{\phi^{2n}+1}{\phi^{n}}$ is an integer for $n$ even.
Indeed, let $n=2m$ and $\alpha=\phi^2$. Then
$$
\dfrac{\phi^{2n}+1}{\phi^{n}}
= \phi^{n}+\dfrac{1}{\phi^{n}}
= \alpha^{m}+\dfrac{1}{\alpha^{m}}
=: y_m
$$
Since $\alpha$ and $\dfrac{1}{\alpha}$ are the roots of $x^2=3x-1$, we have
$ y_{k+2} = 3y_{k+1}-y_{k}$ for all $k \in \mathbb N$.
Since $y_0=2$ and $y_1=3$ are integers, so is $y_k$ for every $k \in \mathbb N$.
In fact, $y_k=L_{2k}$, the $2k$-th Lucas number.
Yet more generally,
$\dfrac{\phi^{2n}+(-1)^n}{\phi^{n}}$ is an integer for all $n \in \mathbb N$. In fact, it is the  $n$-th Lucas number.
A: Using the golden ratio $\phi=\frac{1+\sqrt{5}}{2}$ and silver ratio $\psi=\frac{1-\sqrt{5}}{2}$, we get:
$$\dfrac{\phi^{400}+1}{\phi^{200}}=\phi^{200}+\frac{1}{\phi^{200}}=\phi^{200}+\psi^{200}.$$
The Fibonacci formula:
$$\begin{align}F_n&=\frac{\phi^n-\psi^n}{\sqrt{5}} \Rightarrow \\
\phi^{100}-\psi^{100}&=\sqrt{5}\cdot F_{100} \Rightarrow \\
\phi^{200}+\psi^{200}&=5F_{100}^2+2.\end{align}$$
Note: $F_{100}=354224848179261915075$.
