Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer.

Solution 1:

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 $n=100$

Solution 2:

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.

Solution 3:

$$\color{blue}{(\phi^2+\phi^{-2})}\in \mathbb N,$$ $$(\phi^2+\phi^{-2})^2=\color{blue}{(\phi^4+\phi^{-4})}+2\in \mathbb N,$$ $$(\phi^2+\phi^{-2})^3=\color{blue}{(\phi^6+\phi^{-6})}+3(\phi^2+\phi^{-2})\in \mathbb N,$$ $$(\phi^2+\phi^{-2})^4=\color{blue}{(\phi^8+\phi^{-8})}+4(\phi^4+\phi^{-4})+6\in \mathbb N,$$ $$(\phi^2+\phi^{-2})^5=\color{blue}{(\phi^{10}+\phi^{-10})}+5(\phi^6+\phi^{-6})+10(\phi^2+\phi^{-2})\in \mathbb N,$$$$...$$ $$(\phi^2+\phi^{-2})^{100}=\color{blue}{(\phi^{200}+\phi^{-200})}+100(\phi^{196}+\phi^{-196})+4950(\phi^{192}+\phi^{-192})+...\in \mathbb N.$$

Solution 4:

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 $x^n+\dfrac{1}{x^n}$).