Can you prove or disprove the following claim:
Let $P_m(x)=2^{-m}\cdot\left(\left(x-\sqrt{x^2-4}\right)^m+\left(x+\sqrt{x^2-4}\right)^m\right)$ . Let $N= 12k \cdot 5^{n} + 1 $ where $k$ is an odd natural number , $n\ge3$ and $12k <5^n$ . Let $S_i=S_{i-1}^5-5S_{i-1}^3+5S_{i-1}$ with $S_0=P_{75k}(4)$, then: $$N \text{ is a prime if and only if } S_{n-2} \equiv 0 \pmod{N}$$ .
You can run this test here.
I have verified this claim for $k \in [1,1000]$ with $n \in [3,800]$ .
I was searching for counterexample using the following PARI/GP code:
CE12k5n1(k1,k2,n1,n2)={
i=0;
forstep(k=k1,k2,[2],
if(i==1,break,
for(n=n1,n2,
if(12*k<5^n,
N=12*k*5^n+1;
S=Mod(2*polchebyshev(75*k,1,2),N);
ctr=1;
while(ctr<=n-2,
S=Mod(2*polchebyshev(5,1,S/2),N);
ctr+=1);
if(S==0 && !ispseudoprime(N),print("k="k";""n="n);i=i++;break)))))
}
This is a partial answer.
This answer proves that if $N$ is a prime, then $S_{n-2}\equiv 0\pmod N$.
Proof :
Let us first prove by induction that $$S_i=(2-\sqrt 3)^{5^i75k}+(2+\sqrt 3)^{5^i75k}\tag1$$
$(1)$ holds for $i=0$ since $$\begin{align}S_0&=P_{75k}(4) \\\\&=2^{-75k}\cdot\left(\left(4-2\sqrt{3}\right)^{75k}+\left(4+2\sqrt{3}\right)^{75k}\right) \\\\&=(2-\sqrt 3)^{75k}+(2+\sqrt 3)^{75k}\end{align}$$
Supposing that $(1)$ holds for $i$ and letting $x=(2-\sqrt 3)^{5^i75k},y=(2+\sqrt 3)^{5^i75k}$ where $xy=1$ give $$\begin{align}S_{i+1}&=S_{i}^5-5S_{i}^3+5S_{i} \\\\&=(x+y)^5-5(x+y)^3+5(x+y) \\\\&=(x^5 + 5 x^3 + 10 x + 10y + 5 y^3 + y^5)-5(x^3 + 3 x + 3y + y^3)+5(x+y) \\\\&=x^5+y^5 \\\\&=(2-\sqrt 3)^{5^{i+1}75k}+(2+\sqrt 3)^{5^{i+1}75k}\qquad\square\end{align}$$
From $(1)$, we get $$\begin{align}S_{n-2}&=(2-\sqrt 3)^{5^{n-2}75k}+(2+\sqrt 3)^{5^{n-2}75k} \\\\&=(2-\sqrt 3)^{5^n3k}+(2+\sqrt 3)^{5^n3k} \\\\&=(2-\sqrt 3)^{(N-1)/4}+(2+\sqrt 3)^{(N-1)/4}\end{align}$$ Using $2\pm\sqrt 3=\bigg(\frac{\sqrt 6\pm\sqrt 2}{2}\bigg)^2$, we get
$$\begin{align}&2^{N+1}S_{n-2}^2-2^{N+2} \\\\&=(\sqrt 6+\sqrt 2)(\sqrt 6-\sqrt 2)^{N}+(\sqrt 6-\sqrt 2)(\sqrt 6+\sqrt 2)^{N} \\\\&=\sqrt 6\bigg((\sqrt 6+\sqrt 2)^{N}+(\sqrt 6-\sqrt 2)^{N}\bigg) \\&\qquad\qquad -\sqrt 2\bigg((\sqrt 6+\sqrt 2)^{N}-(\sqrt 6-\sqrt 2)^{N}\bigg) \\\\&=\sqrt 6\sum_{i=0}^{N}\binom Ni(\sqrt 6)^{N-i}\bigg((\sqrt 2)^i+(-\sqrt 2)^i\bigg) \\&\qquad\qquad-\sqrt 2\sum_{i=0}^{N}\binom Ni(\sqrt 6)^{N-i}\bigg((\sqrt 2)^i-(-\sqrt 2)^i\bigg) \\\\&=\sum_{j=0}^{(N-1)/2}\binom{N}{2j}6^{(N-2j+1)/2}2^{j+1}-\sum_{j=1}^{(N+1)/2}\binom{N}{2j-1}6^{(N-2j+1)/2}2^{j+1}\end{align}$$
Since $\binom{N}{k}\equiv 0\pmod N$ for $1\le k\le N-1$, we have $$\begin{align}&2^{N+1}S_{n-2}^2-2^{N+2} \\\\&\equiv 6^{(N+1)/2}2^{1}-2^{(N+3)/2}\pmod N \\\\&\equiv 12\cdot 2^{(N-1)/2}\cdot 3^{(N-1)/2}-4\cdot 2^{(N-1)/2}\pmod N \\\\&\equiv 12\cdot (-1)^{(N^2-1)/8}\cdot \frac{(-1)^{(N-1)/2}}{\bigg(\frac N3\bigg)}-4\cdot (-1)^{(N^2-1)/8}\pmod N\end{align}$$ where $\left(\frac{q}{p}\right)$ denotes the Legendre symbol.
Since $k$ is odd, writing $k=2m+1$ gives $$N=12(2m+1)5^n+1\equiv 4\cdot 5^n+1\equiv 5\pmod 8$$ from which we have $$(-1)^{(N^2-1)/8}=-1$$
So, we get $$2^{N+1}S_{n-2}^2-2^{N+2}\equiv 12\cdot (-1)\cdot \frac{1}{1}-4\cdot (-1)\equiv -8\pmod N$$
It follows from $2^{N-1}\equiv 1\pmod N$ that $$S_{n-2}\equiv 0\pmod N$$