Can you provide a proof for the following claim:

Let $P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4})^m+(x+\sqrt{x^2-4}\right)^m\right)$ .

Let $N= 12k \cdot 5^{n} - 1 $ where $n\ge3$ , $12k <5^n$ and

$\begin{cases} k \equiv 1 \pmod{14} \text{ with } n \equiv 0,3,4 \pmod{6} \\ k \equiv 3 \pmod{14} \text{ with } n \equiv 1,4,5 \pmod{6} \\ k \equiv 5 \pmod{14} \text{ with } n \equiv 2,3,5 \pmod{6} \\ k \equiv 9 \pmod{14} \text{ with } n \equiv 0,2,5 \pmod{6} \\ k \equiv 11 \pmod{14} \text{ with } n \equiv 1,2,4 \pmod{6} \\ k \equiv 13 \pmod{14} \text{ with } n \equiv 0,1,3 \pmod{6} \end{cases}$

Let $S_i=S_{i-1}^5-5S_{i-1}^3+5S_{i-1}$ with $S_0=P_{75k}(5)$, then $N$ is a prime iff $S_{n-2} \equiv 0 \pmod{N}$ .

You can run this test here. I have verified this claim for $k \in [1,111]$ with $n \in [3,1000]$ .

Can you provide a proof for the following claim:

Let $P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4})^m+(x+\sqrt{x^2-4}\right)^m\right)$ .

Let $N= 12k \cdot 5^{n} - 1 $ where $n\ge3$ , $12k <5^n$ and

$\begin{cases} k \equiv 1 \pmod{7} \text{ with } n \equiv 0,3,4 \pmod{6} \\ k \equiv 2 \pmod{7} \text{ with } n \equiv 0,2,5 \pmod{6} \\ k \equiv 3 \pmod{7} \text{ with } n \equiv 1,4,5 \pmod{6} \\ k \equiv 4 \pmod{7} \text{ with } n \equiv 1,2,4 \pmod{6} \\ k \equiv 5 \pmod{7} \text{ with } n \equiv 2,3,5 \pmod{6} \\ k \equiv 6 \pmod{7} \text{ with } n \equiv 0,1,3 \pmod{6} \end{cases}$

Let $S_i=S_{i-1}^5-5S_{i-1}^3+5S_{i-1}$ with $S_0=P_{75k}(5)$, then $N$ is a prime iff $S_{n-2} \equiv 0 \pmod{N}$ .

You can run this test here. I have verified this claim for $k \in [1,111]$ with $n \in [3,1000]$ .