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 $E_n(b)=\Large\frac{b^{2^n}+1}{2}$ such that $n>1$, $b$ is odd number greater than one .
Let $S_i=P_b(S_{i-1})$ with $S_0=P_{b}(6)$, thus
If $E_n(b)$ is prime, then $S_{2^n-1} \equiv 6\pmod{E_n(b)}$.
I have checked this conjecture for all odd $b \in [3,1999] $ with $n \in [2,10]$ .
Searching for counterexample (PARI/GP) :
CEGF(g1,g2)=
{
forstep(b=g1,g2,[2],
for(n=2,10,my(s=Mod(2*polchebyshev(b,1,3),(b^2^n+1)/2));
for(i=1,2^n-1, s=2*polchebyshev(b,1,s/2));
if(!s==6 && isprime((b^2^n+1)/2),print("b=",b,"n=",n))))
}
PARI/GP implementation of test .
GFOdd(b,n)=
{
my(s=Mod(2*polchebyshev(b,1,3),(b^2^n+1)/2));
for(i=1,2^n-1,s=2*polchebyshev(b,1,s/2));
s==6
}
Any hint would be greatly appreciated .
First of all, we prove by induction that $$S_i=p^{2b^{i+1}}+q^{2b^{i+1}}\tag1$$ where $p=\sqrt 2-1,q=\sqrt 2+1$ with $pq=1$.
Proof for $(1)$ : $$S_0=P_{b}(6)=2^{-b}\cdot \left(\left(6-4\sqrt{2}\right)^{b}+\left(6+4\sqrt{2}\right)^{b}\right)=(3-2\sqrt 2)^b+(3+2\sqrt 2)^b=p^{2b}+q^{2b}$$ Supposing that $(1)$ holds for $i$ gives $$\begin{align}S_{i+1}&=P_b(S_i)\\&=2^{-b}\cdot \left(\left(S_i-\sqrt{S_i^2-4}\right)^{b}+\left(S_i+\sqrt{S_i^2-4}\right)^{b}\right)\\&=2^{-b}\cdot \left(\left(p^{2b^{i+1}}+q^{2b^{i+1}}-\sqrt{\left(q^{2b^{i+1}}-p^{2b^{i+1}}\right)^2}\right)^{b}+\left(p^{2b^{i+1}}+q^{2b^{i+1}}+\sqrt{\left(q^{2b^{i+1}}-p^{2b^{i+1}}\right)^2}\right)^{b}\right)\\&=2^{-b}\cdot \left(\left(p^{2b^{i+1}}+q^{2b^{i+1}}-\left(q^{2b^{i+1}}-p^{2b^{i+1}}\right)\right)^{b}+\left(p^{2b^{i+1}}+q^{2b^{i+1}}+\left(q^{2b^{i+1}}-p^{2b^{i+1}}\right)\right)^{b}\right)\\&=2^{-b}\cdot \left(\left(2p^{2b^{i+1}}\right)^{b}+\left(2q^{2b^{i+1}}\right)^{b}\right)\\&=p^{2b^{i+2}}+q^{2b^{i+2}}\quad\blacksquare\end{align}$$
Let $N:=2^n-1,M:=E_n(b)=(b^{N+1}+1)/2$. From $(1)$, we have $$\begin{align}S_{2^n-1}&=S_N\\&=p^{2b^{N+1}}+q^{2b^{N+1}}\\&=p^{2(2M-1)}+q^{2(2M-1)}\\&=p^{4M-2}+q^{4M-2}\\&=(pq)^2(p^{4M-2}+q^{4M-2})\\&=3(p^{4M}+q^{4M})-2\sqrt 2\ (q^{4M}-p^{4M})\end{align}$$
Now using the binomial theorem and Fermat's little theorem, $$\begin{align}p^{4M}+q^{4M}&=(17-12\sqrt 2)^M+(17+12\sqrt 2)^M\\&=\sum_{i=0}^{M}\binom{M}{i}17^i((-12\sqrt 2)^{M-i}+(12\sqrt 2)^{M-i})\\&=\sum_{j=1}^{(M+1)/2}\binom{M}{2j-1}17^{2j-1}\cdot 2(12\sqrt 2)^{M-(2j-1)}\\&\equiv\binom{M}{M}17^{M}\cdot 2(12\sqrt 2)^{0}\quad\pmod M\\&\equiv 17\cdot 2\qquad\pmod M\\&\equiv 34\qquad\pmod M\end{align}$$ Similarly, $$\begin{align}2\sqrt 2\ (q^{4M}-p^{4M})&=2\sqrt 2\ ((17+12\sqrt 2)^M-(17-12\sqrt 2)^M)\\&=2\sqrt 2\ \sum_{i=0}^{M}\binom{M}{i}17^i((12\sqrt 2)^{M-i}-(-12\sqrt 2)^{M-i})\\&=2\sqrt 2\ \sum_{j=0}^{(M-1)/2}\binom{M}{2j}17^{2j}\cdot 2(12\sqrt 2)^{M-2j}\\&=\sum_{j=0}^{(M-1)/2}\binom{M}{2j}17^{2j}\cdot 4\cdot 12^{M-2j}\cdot 2^{(M-2j+1)/2}\\&\equiv \binom{M}{0}17^{0}\cdot 4\cdot 12^{M}\cdot 2^{(M+1)/2}\qquad\pmod M\\&\equiv 4\cdot 12\cdot 2\qquad\pmod M\\&\equiv 96\qquad\pmod M\end{align}$$ since $2^{(M-1)/2}\equiv (-1)^{(M^2-1)/8}\equiv 1\pmod M$ (this is because $M\equiv 1\pmod 8$ from $b^2\equiv 1,9\pmod{16}$)
It follows from these that $$\begin{align}S_{2^n-1}&=3(p^{4M}+q^{4M})-2\sqrt 2\ (q^{4M}-p^{4M})\\&\equiv 3\cdot 34-96\qquad\pmod M\\&\equiv 6\qquad\pmod{E_n(b)}\end{align}$$ as conjectured.