5
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Can you provide proof or counterexample for the claim given below?

Inspired by Lucas-Lehmer primality test I have formulated the following claim:

Let $P_m(x)=2^{-m}\cdot((x-\sqrt{x^2-4})^m+(x+\sqrt{x^2-4})^m)$ . Let $N= 4 \cdot 3^{n}-1 $ where $n\ge3$ . Let $S_i=S_{i-1}^3-3 S_{i-1}$ with $S_0=P_9(6)$ . Then $N$ is prime if and only if $S_{n-2} \equiv 0 \pmod{N}$ .

You can run this test here .

Numbers $n$ such that $4 \cdot 3^n-1$ is prime can be found here .

I was searching for counterexample using the following PARI/GP code:

CE431(n1,n2)=
{
for(n=n1,n2,
N=4*3^n-1;
S=2*polchebyshev(9,1,3);
ctr=1;
while(ctr<=n-2,
S=Mod(2*polchebyshev(3,1,S/2),N);
ctr+=1);
if(S==0 && !ispseudoprime(N),print("n="n)))
}
$\endgroup$
2
+100
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This is a partial answer.

This answer proves that if $N$ is prime, then $S_{n-2}\equiv 0\pmod N$.

Proof :

First of all, let us prove by induction on $i$ that $$S_i=s^{2\cdot 3^{i+2}}+t^{2\cdot 3^{i+2}}\tag1$$ where $s=\sqrt 2-1,t=\sqrt 2+1$ with $st=1$.

We see that $(1)$ holds for $i=0$ since $$S_0=P_9(6)=(3-2\sqrt 2)^9+(3+2\sqrt 2)^9=s^{2\cdot 3^2}+t^{2\cdot 3^2}$$

Supposing that $(1)$ holds for $i$ gives $$\begin{align}S_{i+1}&=S_i^3-3S_i \\\\&=(s^{2\cdot 3^{i+2}}+t^{2\cdot 3^{i+2}})^3-3(s^{2\cdot 3^{i+2}}+t^{2\cdot 3^{i+2}}) \\\\&=s^{2\cdot 3^{i+3}}+t^{2\cdot 3^{i+3}}\qquad\square\end{align}$$

Using $(1)$ and $N=4\cdot 3^n-1$, we get $$S_{n-2}=s^{2\cdot 3^{n}}+t^{2\cdot 3^{n}}=s^{(N+1)/2}+t^{(N+1)/2}$$

So, we have, by the binomial theorem, $$\begin{align}&S_{n-2}^2-2 \\\\&=s\cdot s^{N}+t\cdot t^{N} \\\\&=(\sqrt 2-1)(\sqrt 2-1)^N+(\sqrt 2+1)(\sqrt 2+1)^N \\\\&=\sqrt 2\sum_{i=0}^{N}\binom Ni(\sqrt 2)^i((-1)^{N-i}+1^{N-i}) \\&\qquad\quad +\sum_{i=0}^{N}\binom Ni(\sqrt 2)^i(1^{N-i}-(-1)^{N-i}) \\\\&=\sum_{j=1}^{(N+1)/2}\binom N{2j-1}2^{j+1}+\sum_{j=0}^{(N-1)/2}\binom{N}{2j}2^{j+1}\end{align}$$

Using that $\binom{N}{m}\equiv 0\pmod N$ for $1\le m\le N-1$, we get $$S_{n-2}^2-2\equiv 2^{\frac{N+1}{2}+1}+2=4\cdot 2^{\frac{N-1}{2}}+2\pmod N\tag 2$$

Here, since $$4\cdot 3^{2k}-1\equiv 4\cdot 9^k-1\equiv 3\pmod 8$$ and $$4\cdot 3^{2k+1}-1\equiv 12\cdot 9^k-1\equiv 3\pmod 8$$ we see that $$N\equiv 3\pmod 8$$ from which we have $$2^{\frac{N-1}{2}}\equiv \left(\frac 2N\right)=(-1)^{(N^2-1)/8}=-1\pmod N\tag3$$ where $\left(\frac{q}{p}\right)$ denotes the Legendre symbol.

From $(2)(3)$, we have $$S_{n-2}^2-2\equiv 4\cdot (-1)+2\pmod N$$ from which $$S_{n-2}\equiv 0\pmod N$$ follows.$\quad\blacksquare$

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