How to prove that this conjecture is true ?
Definition : $\text{Let}~ P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)~ , \text{where}~ m ~\text{and}~ x ~\text{are nonnegative integers} .$
Conjecture : $$\text{Let} ~N=k\cdot 6^n-1~ \text{such that} ~ n>2 , k>0 ,$$
$$k \equiv 2,5 \pmod{7}~ \text{and}~ k<6^n .$$ $$\text{Let}~ S_i=P_6(S_{i-1})~ \text{with}~ S_0=P_{3k}(P_3(5)) ,~ \text{thus} $$ $$N ~\text{is prime iff} ~ S_{n-2} \equiv 0 \pmod{N}$$
Pari GP implementation of the test :
Tk6n1(k,n)=
{
my(s=Mod(2*polchebyshev(3*k,1,polchebyshev(3,1,5/2)),k*6^n-1));
for(i=1,n-2, s=2*polchebyshev(6,1,s/2));
s==0
}
Searching for counterexample (Pari GP)
Necessity
CEk6n1(k,lb,ub)=
{
for(n=lb,ub,s=2*polchebyshev(3*k,1,polchebyshev(3,1,5/2));
N=k*6^n-1;
for(i=1,n-2,s=Mod(2*polchebyshev(6,1,s/2),N));
if(!s==0 && isprime(N),print(n)))
}
Sufficiency
CEk6n1(k,lb,ub)=
{
for(n=lb,ub,s=2*polchebyshev(3*k,1,polchebyshev(3,1,5/2));
N=k*6^n-1;
for(i=1,n-2,s=Mod(2*polchebyshev(6,1,s/2),N));
if(s==0 && !isprime(N),print(n)))
}
Any help would be much appreciated .
EDIT:
One can formulate following conjecture also .
Conjecture : $$\text{Let} ~N=k\cdot 6^n-1~ \text{such that} ~ n>2 , k>0 ,$$
$$k \equiv 3,4 \pmod{5}~ \text{and}~ k<6^n .$$ $$\text{Let}~ S_i=P_6(S_{i-1})~ \text{with}~ S_0=P_{3k}(P_3(3)) ,~ \text{thus} $$ $$N ~\text{is prime iff} ~ S_{n-2} \equiv 0 \pmod{N}$$
This is a partial answer.
Edit : I'm going to prove that $$\text{if $N$ is prime, then $S_{n-2}\equiv 0\pmod N$}$$ for both conjectures.
(For the first conjecture)
First of all, $$P_3(5)=2^{-3}\cdot \left(\left(5-\sqrt{21}\right)^3+\left(5+\sqrt{21}\right)^3\right)=110$$ So, $$\begin{align}S_0=P_{3k}(P_3(5))=P_{3k}(110)&=2^{-3k}\cdot \left(\left(110-\sqrt{110^2-4}\right)^{3k}+\left(110+\sqrt{110^2-4}\right)^{3k}\right)\\&=\left(\frac{110-\sqrt{110^2-4}}{2}\right)^{3k}+\left(\frac{110+\sqrt{110^2-4}}{2}\right)^{3k}\\&=\left(55-12\sqrt{21}\right)^{3k}+\left(55+12\sqrt{21}\right)^{3k}\\&=(a^2)^{3k}+(b^2)^{3k}\\&=a^{6k}+b^{6k}\end{align}$$ where $a=2\sqrt 7-3\sqrt 3,b=2\sqrt 7+3\sqrt 3$ with $ab=1$.
From this, we can prove by induction that $$S_i=a^{6^{i+1}k}+b^{6^{i+1}k}.$$ Thus, $$S_{n-2}=a^{\frac{N+1}{6}}+b^{\frac{N+1}{6}}=\left(\frac{\sqrt 7}{2}-\frac{\sqrt 3}{2}\right)^{\frac{N+1}{2}}+\left(\frac{\sqrt 7}{2}+\frac{\sqrt 3}{2}\right)^{\frac{N+1}{2}}=2^{-\frac{N+1}{2}}\left((\sqrt 7-\sqrt 3)^{\frac{N+1}{2}}+(\sqrt 7+\sqrt 3)^{\frac{N+1}{2}}\right).$$ By the way, for $N$ prime, $$\begin{align}(\sqrt 7-\sqrt 3)^{N+1}+(\sqrt 7+\sqrt 3)^{N+1}&=\sum_{i=0}^{N+1}\binom{N+1}{i}\left(\sqrt 7\right)^{i}((-\sqrt 3)^{N+1-i}+(\sqrt 3)^{N+1-i})\\&=\sum_{j=0}^{(N+1)/2}\binom{N+1}{2j}\left(\sqrt 7\right)^{2j}\cdot 2(\sqrt 3)^{N+1-2j}\\&=\sum_{j=0}^{(N+1)/2}\binom{N+1}{2j}7^{j}\cdot 2\cdot 3^{\frac{N+1}{2}-j}\\&\equiv 2\cdot 3^{\frac{N+1}{2}}+7^{\frac{N+1}{2}}\cdot 2\pmod{N}\\&\equiv 2\cdot 3+(-7)\cdot 2\pmod N\\&\equiv -8\pmod N\end{align}$$ This is because $N\equiv 2\pmod 3$ and $N\equiv \pm 2\cdot (-1)^n-1\equiv 1,4\pmod 7$ implies that $$3^{(N-1)/2}\equiv 1\pmod N,\quad 7^{(N-1)/2}\equiv -1\pmod N.$$
From this, since $2^{N-1}\equiv 1\pmod N$, $$\begin{align}2^{N+1}S_{n-2}^2&=(\sqrt 7-\sqrt 3)^{N+1}+(\sqrt 7+\sqrt 3)^{N+1}+2\cdot 4^{\frac{N+1}{2}}\\&\equiv -8+2\cdot 2^{N-1}\cdot 4\pmod N\\&\equiv 0\pmod N\end{align}$$
Thus, $S_{n-2}\equiv 0\pmod N$.
(For the second conjecture)
$$P_3(3)=2^{-3}\cdot\left(\left(3-\sqrt{5}\right)^3+\left(3+\sqrt{5}\right)^3\right)=18$$
$$S_0=P_{3k}(P_3(3))=2^{-3k}\cdot\left(\left(18-\sqrt{18^2-4}\right)^{3k}+\left(18+\sqrt{18^2-4}\right)^{3k}\right)$$ $$=(9-4\sqrt{5})^{3k}+(9+4\sqrt 5)^{3k}=c^{6k}+d^{6k}$$ where $c=\sqrt 5-2,d=\sqrt 5+2$ with $cd=1$.
We can prove by induction that $$S_i=c^{6^{i+1}k}+d^{6^{i+1}k}$$
Thus, $$S_{n-2}=c^{\frac{N+1}{6}}+d^{\frac{N+1}{6}}=\left(\frac{\sqrt 5}{2}-\frac 12\right)^{\frac{N+1}{2}}+\left(\frac{\sqrt 5}{2}+\frac 12\right)^{\frac{N+1}{2}}=2^{-\frac{N+1}{2}}\left(\left(\sqrt 5-1\right)^{\frac{N+1}{2}}+\left(\sqrt 5+1\right)^{\frac{N+1}{2}}\right).$$
By the way, for $N$ prime, $$\begin{align}\left(\sqrt 5-1\right)^{N+1}+\left(\sqrt 5+1\right)^{N+1}&=\sum_{i=0}^{N+1}\binom{N+1}{i}(\sqrt 5)^{i}\left((-1)^{N+1-i}+1^{N+1-i}\right)\\&=\sum_{j=0}^{(N+1)/2}\binom{N+1}{2j}(\sqrt 5)^{2j}\cdot 2\\&=\sum_{j=0}^{(N+1)/2}\binom{N+1}{2j}5^j\cdot 2\\&\equiv 2+5^{\frac{N+1}{2}}\cdot 2\pmod N\\&\equiv 2+(-5)\cdot 2\pmod N\\&\equiv -8\pmod N\end{align}$$
This is because $N\equiv 2,3\pmod 5$ implies that $$5^{\frac{N-1}{2}}\equiv -1\pmod N.$$
From this, since $2^{N-1}\equiv 1\pmod N$, $$\begin{align}2^{N+1}S_{n-2}^2&=(\sqrt 5-1)^{N+1}+(\sqrt 5+1)^{N+1}+2\cdot 4^{\frac{N+1}{2}}\\&\equiv -8+2\cdot 2^{N-1}\cdot 4\pmod N\\&\equiv 0\pmod N\end{align}$$
Thus, $S_{n-2}\equiv 0\pmod N$.