How to prove that a claim given below 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 .
Theorem (Inkeri)
Let $F_n=2^{2^n}+1$ with $n\ge 2$ .
Let $S_i=P_{2}(S_{i-1})$ with $S_0=8$
then ,
$F_n$ is prime iff $S_{2^n-2} \equiv 0 \pmod{F_n}$
Claim
Let $N=k\cdot b^n+1$ such that $n>2$ , $0<k<b^n$ ,
$k \equiv 1,7 \pmod{30}$ , $b$ is even , $3 \not \mid b$ , $5 \not \mid b$ and $n \equiv 0 \pmod{4} $ .
Let $S_i=P_b(S_{i-1})$ with $S_0=P_{kb/2}(P_{b/2}(8))$
then ,
$N$ is prime iff $S_{n-2} \equiv 0 \pmod{N}$
For $b=2$ I managed to prove a claim , at least I think so . You can see my proof in this paper (Theorem 3.4.) . But I have no idea how to prove this claim for $b>2$ .
Any hint will be greatly appreciated .
This is a partial answer.
This answer proves the following :
$$\text{if $N$ is prime, then $S_{n-2}\equiv 0\pmod N$}\tag0$$
Proof :
First of all, we prove by induction that $$S_i=p^{kb^{i+2}/4}+q^{kb^{i+2}/4}\tag1$$ where $p=4-\sqrt{15}, q=4+\sqrt{15}$ with $pq=1$.
Proof for $(1)$ :
$$S_0=P_{kb/2}(P_{b/2}(8))$$$$\small=2^{-kb/2}\left(2^{-b/2}\cdot\left((2p)^{b/2}+(2q)^{b/2}\right)-(q^{b/2}-p^{b/2})\right)^{kb/2}+2^{-kb/2}\left(2^{-b/2}\cdot\left((2p)^{b/2}+(2q)^{b/2}\right)+(q^{b/2}-p^{b/2})\right)^{kb/2}$$ $$=\left(p^{b/2}\right)^{kb/2}+\left(q^{b/2}\right)^{kb/2}=p^{kb^2/4}+q^{kb^2/4}$$
Supposing that $(1)$ holds for $i$ gives $$S_{i+1}=P_b(S_i)=2^{-b}\left(p^{kb^{i+2}/4}+q^{kb^{i+2}/4}-\sqrt{\left(p^{kb^{i+2}/4}+q^{kb^{i+2}/4}\right)^2-4}\right)^{b}+2^{-b}\left(p^{kb^{i+2}/4}+q^{kb^{i+2}/4}+\sqrt{\left(p^{kb^{i+2}/4}+q^{kb^{i+2}/4}\right)^2-4}\right)^{b}$$ $$=2^{-b}\left(p^{kb^{i+2}/4}+q^{kb^{i+2}/4}-\left(q^{kb^{i+2}/4}-p^{kb^{i+2}/4}\right)\right)^{b}+2^{-b}\left(p^{kb^{i+2}/4}+q^{kb^{i+2}/4}+\left(q^{kb^{i+2}/4}-p^{kb^{i+2}/4}\right)\right)^{b}$$ $$=p^{kb^{i+3}/4}+q^{kb^{i+3}/4}\qquad\blacksquare$$
Let $N=k\cdot b^n+1$. Then, from $(1)$, we have $$S_{n-2}=p^{kb^{n}/4}+q^{kb^{n}/4}=p^{(N-1)/4}+q^{(N-1)/4}$$
Now, in order to prove that $S_{n-2}\equiv 0\pmod N$, it is sufficient to prove that $S_{n-2}^2\equiv 0\pmod N$, i.e. $$p^{(N-1)/2}+q^{(N-1)/2}\equiv -2\pmod N$$
We use $$p^{(N+3)/2}+q^{(N+3)/2}=(p+q)(p^{(N+1)/2}+q^{(N+1)/2})-(p^{(N-1)/2}+q^{(N-1)/2})\tag2$$
Using that $$\sqrt{4\pm\sqrt{15}}=\frac{\sqrt{10}\pm\sqrt 6}{2}$$ we have $$\begin{align}&2^{N+1}(p^{(N+1)/2}+q^{(N+1)/2})\\&=\left(\sqrt{10}-\sqrt 6\right)^{N+1}+\left(\sqrt{10}+\sqrt 6\right)^{N+1}\\&=\sum_{i=0}^{N+1}\binom{N+1}{i}(\sqrt{10})^i((-\sqrt 6)^{N+1-i}+(\sqrt 6)^{N+1-i})\\&=\sum_{j=0}^{(N+1)/2}\binom{N+1}{2j}(\sqrt{10})^{2j}\cdot 2(\sqrt{6})^{N+1-2j}\\&\equiv 2\cdot 6^{(N+1)/2}+10^{(N+1)/2}\cdot 2\qquad\pmod N\\&\equiv 2\cdot (-6)+(-10)\cdot 2\qquad\pmod N\\&\equiv -32\qquad\pmod N\end{align}$$ (because $6^{(N-1)/2}\equiv -1\equiv 10^{(N-1)/2}\pmod N$) from which we have $$p^{(N+1)/2}+q^{(N+1)/2}\equiv -8\qquad\pmod N$$ since $2^{N+1}\equiv 4\pmod N$ is coprime to $N$.
Also, $$\small\begin{align}&2^{N+3}(p^{(N+3)/2}+q^{(N+3)/2})\\&=\left(\sqrt{10}-\sqrt 6\right)^{N+3}+\left(\sqrt{10}+\sqrt 6\right)^{N+3}\\&=\sum_{i=0}^{N+3}\binom{N+3}{i}(\sqrt{10})^i((-\sqrt 6)^{N+3-i}+(\sqrt 6)^{N+3-i})\\&=\sum_{j=0}^{(N+3)/2}\binom{N+3}{2j}(\sqrt{10})^{2j}\cdot 2(\sqrt 6)^{N+3-2j}\\&\equiv 2\cdot 6^{(N+3)/2}+\frac{(N+3)(N+2)}{2}\cdot 10\cdot 2\cdot 6^{(N+1)/2}+\frac{(N+3)(N+2)}{2}\cdot 10^{(N+1)/2}\cdot 2\cdot 6+10^{(N+3)/2}\cdot 2\quad\pmod N\\&\equiv 2\cdot 6^{(N+3)/2}+6\cdot 10\cdot 6^{(N+1)/2}+6\cdot 10^{(N+1)/2}\cdot 6+10^{(N+3)/2}\cdot 2\qquad\pmod N\\&\equiv 2(-6^2)+60(-6)+6\cdot (-10)\cdot 6+(-10^2)\cdot 2\qquad\pmod N\\&\equiv -992\qquad\pmod N\end{align}$$ from which we have $$p^{(N+3)/2}+q^{(N+3)/2}\equiv -62\qquad\pmod N$$
Hence, from $(2)$, $$\begin{align}p^{(N-1)/2}+q^{(N-1)/2}&=8(p^{(N+1)/2}+q^{(N+1)/2})-(p^{(N+3)/2}+q^{(N+3)/2})\\&\equiv 8(-8)-(-62)\qquad\pmod N\\&\equiv -2\qquad\pmod N\end{align}$$
Thus, $$\begin{align}S_{n-2}^2&=(p^{(N-1)/4}+q^{(N-1)/4})^2\\&=p^{(N-1)/2}+q^{(N-1)/2}+2\\&\equiv 0\qquad\pmod N\end{align}$$ from which $S_{n-2}\equiv 0\pmod{N}$ follows.