#subcesse

[pupyzu18]

prisub || subcesse || royalsub

prisub (pri - sub) : a gender connected to being a sub prince; this gender is connected to being a sub, sub aesthetics, prince aesthetics, and non-binary boyhood.

subcesse (sub - cess): a gender connected to being a sub princess; this gender is connected to being a sub, sub aesthetics, princess aesthetics, and non-binary girlhood.

royalsub (royal - sub) : a gender connected to being sub royalty; this gender is connected to being a sub, sub aesthetics, royal aesthetics, and being non-binary.

flags/terms created by me ; gendersub

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[mathematicianadda]

Prove that $x_n$ and $n$ are coprimes

I'm here again with a problem of Italian National Math Olympiads 2007. Given the following subcession:

Given the following subcession

$$\left\{\begin{matrix} x_1=2\\ x_{n+1}=2x_n^2-1 \end{matrix}\right.$$

Prove that $\gcd(x_n,n)=1 \ \ \ \forall n> 1$

My attempt

I thought that maybe, having the close formula for $x_n$ could simplify the problem. I noticed the the recursion formula is analogous to the cosine duplication formula:

$$\cos(2x)=2\cos^2(x)-1$$

So basically at each iteration step we are duplicating the cosine and:

$$x_n=\cos(2^{n-1}\arccos(x_1))=\cos(2^{n-1}\arccos(2))$$

Calculating $\arccos(2)$ is equivalent to this equation:

$$\cos(x)=2$$ $$\frac{e^{ix}+e^{-ix}}{2}=2$$

Substituting $e^{ix}=t$:

$$t+\frac 1t=4$$ $$t^2-4t+1=0$$ $$t=2\pm \sqrt{3}$$ $$e^{ix}=2\pm \sqrt{3}$$ $$x=\frac{\ln(2\pm \sqrt{3})}{i}$$

So:

$$x_n=\cos\left(2^{n-1}\frac{\ln(2\pm \sqrt{3})}{i}\right) $$

By Euler's formula:

$$x_n=\frac{e^{i2^{n-1}\frac{\ln(2\pm \sqrt{3})}{i}}+e^{-i2^{n-1}\frac{\ln(2\pm \sqrt{3})}{i}}}{2}$$ $$x_n=\frac{(2\pm \sqrt{3})^{2^{n-1}}+(2\pm \sqrt{3})^{-2^{n-1}}}{2}$$

The signs can be determined by checking some values. In the end: $$x_n=\frac{(2+\sqrt{3})^{2^{n-1}}+(2+ \sqrt{3})^{-2^{n-1}}}{2}$$ $$x_n=\frac{(\sqrt{3}+2)^{2^{n-1}}+(2-\sqrt{3})^{2^{n-1}}}{2}$$

So now we have to proof that if $p|n$ then $p \nmid \frac{(\sqrt{3}+2)^{2^{n-1}}+(2-\sqrt{3})^{2^{n-1}}}{2} $ with $p\in \Bbb{P} $. If $p=2$ the proof is trivial because clearly $x_n \equiv 1 \pmod{2} \ \ \ \ \forall n\geq 2$ . So we can limitate us to study the simplified expression:

$$(\sqrt{3}+2)^{2^{n-1}}+(2-\sqrt{3})^{2^{n-1}}$$

Then I don't know how to continue :(

I tried using Newton binomial we obtain:

$$(\sqrt{3}+2)^{2^{n-1}}+(2-\sqrt{3})^{2^{n-1}}= \sum_{i=0}^{2^{n-1}} {2^{n-1}\choose i} (\sqrt{3})^i 2^{2^{n-1}-i}+\sum_{i=0}^{2^{n-1}} (-1)^i{2^{n-1}\choose i} (\sqrt{3})^i 2^{2^{n-1}-i}$$

Notice that if $i\equiv 1 \pmod{2}$ the terms get simplified so:

$$\sum_{i=0}^{2^{n-2}} {2^{n-1}\choose 2i} 3^{i} 2^{2^{n-1}-2i+1} $$

But then I can't see the pattern :(

Can you help me, I would like to know how to solve this problem and if possible how to continue my solution.

Thank you for your time

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger http://bit.ly/2Z74wR4