How to prove that following conjecture is true?
Definition
Let $P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)$, where $m$ and $x$ are nonnegative integers.
Conjecture
Let $F_n(b)=b^{2^n}+1$ such that $n \ge 2$ and $b$ is even number .
Let $S_i=P_b(S_{i-1})$ with $S_0=P_{b}(6)$, thus
If $F_n(b)$ is prime, then $S_{2^n-1} \equiv 2\pmod{F_n(b)}$.
I have checked this conjecture for all even $b \in [2,2000] $ with $n \in [2,10]$ .
PARI/GP implementation of test :
CTGF(b,n)=
{
my(s=Mod(2*polchebyshev(b,1,3),b^2^n+1));
for(i=1,2^n-1,s=2*polchebyshev(b,1,s/2));
s==2
}
Any hint would be greatly appreciated .
First of all, we prove by induction that $$S_i=\alpha^{b^{i+1}}+\beta^{b^{i+1}}\tag1$$ where $\alpha=3-2\sqrt 2,\beta=3+2\sqrt 2$ with $\alpha\beta=1$.
Proof for $(1)$ :
$$\begin{align}S_0&=P_{b}(6)\\&=2^{-b}\cdot \left(\left(6-4\sqrt{2}\right)^{b}+\left(6+4\sqrt{2}\right)^{b}\right)\\&=2^{-b}\cdot\left(2^b(3-2\sqrt 2)^b+2^b(3+2\sqrt 2)^b\right)\\&=\alpha^b+\beta^b\end{align}$$
Suppose that $(1)$ holds for $i$. Using the fact that $$(\alpha^m+\beta^m)^2-4=(\beta^m-\alpha^m)^2$$ we get $$\small\begin{align}S_{i+1}&=P_{b}(S_i)\\&=2^{-b}\cdot \left(\left(\alpha^{b^{i+1}}+\beta^{b^{i+1}}-\sqrt{(\alpha^{b^{i+1}}+\beta^{b^{i+1}})^2-4}\right)^{b}+\left(\alpha^{b^{i+1}}+\beta^{b^{i+1}}+\sqrt{(\alpha^{b^{i+1}}+\beta^{b^{i+1}})^2-4}\right)^{b}\right)\\&=2^{-b}\cdot\left(\left(\alpha^{b^{i+1}}+\beta^{b^{i+1}}-\sqrt{(\beta^{b^{i+1}}-\alpha^{b^{i+1}})^2}\right)^b+\left(\alpha^{b^{i+1}}+\beta^{b^{i+1}}+\sqrt{(\beta^{b^{i+1}}-\alpha^{b^{i+1}})^2}\right)^b\right)\\&=2^{-b}\cdot\left(\left(2\alpha^{b^{i+1}}\right)^b+\left(2\beta^{b^{i+1}}\right)^b\right)\\&=\alpha^{b^{i+2}}+\beta^{b^{i+2}}\qquad\blacksquare\end{align}$$
Let $N:=F_n(b)=b^{2^n}+1$. Then, from $(1)$, $$S_{2^n-1}=\alpha^{b^{2^n}}+\beta^{b^{2^n}}=\alpha^{N-1}+\beta^{N-1}$$
Since $\alpha\beta=1$, $$\begin{align}S_{2^n-1}&=\alpha^{N-1}+\beta^{N-1}\\&=\alpha\beta(\alpha^{N-1}+\beta^{N-1})\\&=\beta\cdot\alpha^N+\alpha\cdot\beta^N\\&=3(\alpha^N+\beta^N)-2\sqrt 2\ (\beta^N-\alpha^N)\tag2\end{align}$$
So, in the following, we find $\alpha^N+\beta^N\pmod N$ and $\sqrt 2\ (\beta^N-\alpha^N)\pmod N$.
Using the binomial theorem, $$\begin{align}\alpha^N+\beta^N&=(3-2\sqrt 2)^N+(3+2\sqrt 2)^N\\&=\sum_{i=0}^{N}\binom Ni3^i\cdot ((-2\sqrt 2)^{N-i}+(2\sqrt 2)^{N-i})\\&=\sum_{j=1}^{(N+1)/2}\binom{N}{2j-1}3^{2j-1}\cdot 2(2\sqrt 2)^{N-(2j-1)}\end{align}$$
Since $\binom{N}{2j-1}\equiv 0\pmod N$ for $1\le j\le (N-1)/2$, we get $$\alpha^N+\beta^N\equiv \binom{N}{N}3^{N}\cdot 2(2\sqrt 2)^{0}\equiv 2\cdot 3^N\pmod N$$
Now, by Fermat's little theorem, $$\alpha^N+\beta^N\equiv 2\cdot 3^N\equiv 2\cdot 3\equiv 6\pmod N\tag3$$
Similarly, $$\begin{align}\sqrt 2\ (\beta^N-\alpha^N)&=\sqrt 2\ ((3+2\sqrt 2)^N-(3-2\sqrt 2)^N)\\&=\sqrt 2\ \sum_{i=0}^{N}\binom Ni3^i\cdot((2\sqrt 2)^{N-i}-(-2\sqrt 2)^{N-i})\\&=\sqrt 2\ \sum_{j=0}^{(N-1)/2}\binom{N}{2j}3^{2j}\cdot 2(2\sqrt 2)^{N-2j}\\&\equiv \sqrt 2\ \binom{N}{0}3^{0}\cdot 2(2\sqrt 2)^{N}\quad\pmod N\\&\equiv 2^{N+1}\cdot 2^{(N+1)/2}\quad\pmod N\\&\equiv 4\cdot 2^{(N+1)/2}\quad\pmod N\tag4\end{align}$$
By the way, since $b$ is even with $n\ge 2$, $$N=b^{2^n}+1\equiv 1\pmod 8$$ from which $$2^{(N-1)/2}\equiv \left(\dfrac{2}{N}\right) \equiv (-1)^{(N^2-1)/8}\equiv 1\pmod N$$ follows where $\left(\dfrac{q}{p}\right)$ denotes the Legendre symbol.
So, from $(4)$, $$\sqrt 2\ (\beta^N-\alpha^N)\equiv 4\cdot 2^{(N+1)/2}\equiv 4\cdot 2\equiv 8\pmod N\tag5$$
Therefore, finally, from $(2)(3)$ and $(5)$, $$S_{2^n-1}\equiv 3(\alpha^N+\beta^N)-2\sqrt 2\ (\beta^N-\alpha^N)\equiv 3\cdot 6-2\cdot 8\equiv 2\pmod{F_n(b)}$$ as conjectured.