How to prove that following hypothesis 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 .
Hypothesis
Let $F_n(6)=6^{2^n}+1$ with $n\ge 2$ .
Let $S_i=P_{6}(S_{i-1})$ with $S_0=P_{3}(P_{3}(43))$
thus ,
$F_n(6)$ is prime iff $S_{2^n-2} \equiv 0 \pmod{F_n(6)}$
PARI/GP implementation of test :
GeneFer6(n)=
{
S=2*polchebyshev(3,1,polchebyshev(3,1,43/2));
F=6^2^n+1;
ctr=1;
while(ctr<=2^n-2,
S=Mod(2*polchebyshev(6,1,S/2),F);
ctr+=1);
S==0
}
Any hint would be greatly appreciated .
This is a partial answer.
This answer proves the following :
$$\text{If $F_n(6)$ is prime, then $S_{2^n-2}\equiv 0\pmod{F_n(6)}$}$$
First of all, we prove by induction that $$S_i=\alpha^{9\cdot 6^{i}}+\beta^{9\cdot 6^{i}}\tag1$$ where $\alpha=(43-\sqrt{1845})/2,\beta=(43+\sqrt{1845})/2$ with $\alpha\beta=1$.
Since $$P_{3}(43)=2^{-3}\cdot \left(\left(43-\sqrt{1845}\right)^{3}+\left(43+\sqrt{1845}\right)^{3}\right)=\alpha^3+\beta^3$$ we get $$\begin{align}S_0&=P_3(P_3(43))\\&=2^{-3}\cdot \left(\left(\alpha^3+\beta^3-\sqrt{(\alpha^3+\beta^3)^2-4}\right)^{3}+\left(\alpha^3+\beta^3+\sqrt{(\alpha^3+\beta^3)^2-4}\right)^{3}\right)\\&=2^{-3}\cdot \left(\left(\alpha^3+\beta^3-\sqrt{(\beta^3-\alpha^3)^2}\right)^{3}+\left(\alpha^3+\beta^3+\sqrt{(\beta^3-\alpha^3)^2}\right)^{3}\right)\\&=2^{-3}\cdot \left(\left(\alpha^3+\beta^3-(\beta^3-\alpha^3)\right)^{3}+\left(\alpha^3+\beta^3+(\beta^3-\alpha^3)\right)^{3}\right)\\&=2^{-3}\cdot\left(\left(2\alpha^3\right)^3+\left(2\beta^3\right)^3\right)\\&=\alpha^{9}+\beta^{9}\end{align}$$
Supposing that $(1)$ holds for $i$ gives $$\small\begin{align}S_{i+1}&=P_6(S_i)\\&=2^{-6}\cdot \left(\left(\alpha^{9\cdot 6^{i}}+\beta^{9\cdot 6^{i}}-\sqrt{(\alpha^{9\cdot 6^{i}}+\beta^{9\cdot 6^{i}})^2-4}\right)^{6}+\left(\alpha^{9\cdot 6^{i}}+\beta^{9\cdot 6^{i}}+\sqrt{(\alpha^{9\cdot 6^{i}}+\beta^{9\cdot 6^{i}})^2-4}\right)^{6}\right)\\&=2^{-6}\cdot \left(\left(\alpha^{9\cdot 6^{i}}+\beta^{9\cdot 6^{i}}-\sqrt{(\beta^{9\cdot 6^{i}}-\alpha^{9\cdot 6^{i}})^2}\right)^{6}+\left(\alpha^{9\cdot 6^{i}}+\beta^{9\cdot 6^{i}}+\sqrt{(\beta^{9\cdot 6^{i}}-\alpha^{9\cdot 6^{i}})^2}\right)^{6}\right)\\&=2^{-6}\cdot \left(\left(\alpha^{9\cdot 6^{i}}+\beta^{9\cdot 6^{i}}-(\beta^{9\cdot 6^{i}}-\alpha^{9\cdot 6^{i}})\right)^{6}+\left(\alpha^{9\cdot 6^{i}}+\beta^{9\cdot 6^{i}}+(\beta^{9\cdot 6^{i}}-\alpha^{9\cdot 6^{i}})\right)^{6}\right)\\&=2^{-6}\cdot\left(\left(2\alpha^{9\cdot 6^{i}}\right)^6+\left(2\beta^{9\cdot 6^{i}}\right)^6\right)\\&=\alpha^{9\cdot 6^{i+1}}+\beta^{9\cdot 6^{i+1}}\qquad\blacksquare\end{align}$$
Let $N:=F_n(6)=6^{2^n}+1$. Then, from $(1)$, $$\begin{align}S_{2^n-2}&=\alpha^{9\cdot 6^{2^n-2}}+\beta^{9\cdot 6^{2^n-2}}\\&=\alpha^{(N-1)/4}+\beta^{(N-1)/4}\\&=\left(\frac{3\sqrt 5-\sqrt{41}}{2}\right)^{(N-1)/2}+\left(\frac{3\sqrt 5+\sqrt{41}}{2}\right)^{(N-1)/2}\end{align}$$ Squaring the both sides, $$S_{2^n-2}^2=\left(\frac{3\sqrt 5-\sqrt{41}}{2}\right)^{N-1}+\left(\frac{3\sqrt 5+\sqrt{41}}{2}\right)^{N-1}+2$$ Multiplying the both sides by $2^{N-1}$, $$2^{N-1}\cdot S_{2^n-2}^2=(3\sqrt 5-\sqrt{41})^{N-1}+(3\sqrt 5+\sqrt{41})^{N-1}+2^N$$ Multiplying the both sides by $4=(3\sqrt 5-\sqrt{41})(3\sqrt 5+\sqrt{41})$, $$\small\begin{align}&2^{N+1}\cdot S_{2^n-2}^2\\&=(3\sqrt 5+\sqrt{41})(3\sqrt 5-\sqrt{41})^{N}+(3\sqrt 5-\sqrt{41})(3\sqrt 5+\sqrt{41})^N+2^{N+2}\\&=3\sqrt 5\ ((3\sqrt 5-\sqrt{41})^{N}+(3\sqrt 5+\sqrt{41})^{N})-\sqrt{41}\ ((3\sqrt 5+\sqrt{41})^{N}-(3\sqrt 5-\sqrt{41})^{N})+2^{N+2}\end{align}$$ Using the binomial theorem and Fermat's little theorem, $$\begin{align}&\sqrt 5\ ((3\sqrt 5-\sqrt{41})^N+(3\sqrt 5+\sqrt{41})^N)\\&=\sqrt 5\ \sum_{i=0}^{N}\binom Ni(3\sqrt 5)^i\left((-\sqrt{41})^{N-i}+(\sqrt{41})^{N-i}\right)\\&=\sum_{j=1}^{(N+1)/2}\binom{N}{2j-1}\cdot 3^{2j-1}\cdot (\sqrt 5)^{2j}\cdot 2(\sqrt{41})^{N-(2j-1)}\\&\equiv \binom{N}{N}\cdot 3^{N}\cdot (\sqrt 5)^{N+1}\cdot 2(\sqrt{41})^{0}\quad\pmod N\\&\equiv 2\cdot 3^N\cdot 5^{(N+1)/2}\quad\pmod N\\&\equiv 2\cdot 3\cdot 5^{(N+1)/2}\quad\pmod N\tag2\end{align}$$ Now since $N\equiv 2\pmod 5$, we get $\left(\frac N5\right)=-1$, so $$5^{(N-1)/2}\equiv\left(\dfrac 5N\right)\equiv \dfrac{(-1)^{\frac{5-1}{2}\cdot\frac{N-1}{2}}}{\left(\frac N5\right)}\equiv\frac{(-1)^{N-1}}{-1}\equiv -1\pmod N$$ where $\left(\dfrac{q}{p}\right)$ denotes the Legendre symbol.
So from $(2)$, $$\small\sqrt 5\ ((3\sqrt 5-\sqrt{41})^N+(3\sqrt 5+\sqrt{41})^N)\equiv 2\cdot 3\cdot 5^{(N+1)/2}\equiv 2\cdot 3\cdot (-5)\equiv -30\pmod N\tag3$$
Similarly, $$\begin{align}&\sqrt{41}\ ((3\sqrt 5+\sqrt{41})^{N}-(3\sqrt 5-\sqrt{41})^{N})\\&=\sqrt{41}\ \sum_{i=0}^{N}\binom Ni(3\sqrt 5)^i\left((\sqrt{41})^{N-i}-(-\sqrt{41})^{N-i}\right)\\&=\sum_{j=0}^{(N-1)/2}\binom{N}{2j}(3\sqrt 5)^{2j}\cdot 2(\sqrt{41})^{N-2j+1}\\&\equiv \binom{N}{0}(3\sqrt 5)^{0}\cdot 2(\sqrt{41})^{N+1}\quad\pmod N\\&\equiv 2\cdot 41^{(N+1)/2}\quad\pmod N\tag4\end{align}$$ Now we have $F_2(6)=6^{2^2}+1\equiv 26\pmod{41}$ and it can be proven by induction that for $k\ge 1$, $$F_{4k-1}(6)=6^{2^{4k-1}}+1\equiv 11\pmod{41},\quad F_{4k}(6)=6^{2^{4k}}+1\equiv 19\pmod{41}$$ $$F_{4k+1}(6)=6^{2^{4k+1}}+1\equiv 38\pmod{41},\quad F_{4k+2}(6)=6^{2^{4k+2}}+1\equiv 17\pmod{41}$$ from which $\left(\frac{N}{41}\right)=-1$ follows.
It follows from this that $$41^{(N-1)/2}\equiv \left(\dfrac{41}{N}\right)\equiv\dfrac{(-1)^{\frac{41-1}{2}\cdot\frac{N-1}{2}}}{\left(\frac{N}{41}\right)}\equiv \frac{(-1)^{10(N-1)}}{-1}\equiv -1\pmod N$$ so from $(4)$, $$\small\sqrt{41}\ ((3\sqrt 5+\sqrt{41})^{N}-(3\sqrt 5-\sqrt{41})^{N})\equiv 2\cdot 41^{(N+1)/2}\equiv 2\cdot (-41)\equiv -82\pmod N\tag5$$
Therefore, from $(3)(5)$, $$\small\begin{align}&2^{N+1}\cdot S_{2^n-2}^2\\&=3\sqrt 5\ ((3\sqrt 5-\sqrt{41})^{N}+(3\sqrt 5+\sqrt{41})^{N})-\sqrt{41}\ ((3\sqrt 5+\sqrt{41})^{N}-(3\sqrt 5-\sqrt{41})^{N})+2^{N+2}\\&\equiv 3\cdot (-30)-(-82)+8\quad\pmod N\\&\equiv 0\quad\pmod N\end{align}$$ from which $$S_{2^n-2}\equiv 0\pmod{F_n(6)}$$ follows since $2^{N+1}\equiv 4\pmod N$ is coprime to $N$.