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:

if(S==0 && !ispseudoprime(N),print("n="n)))

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.$\quad\blacksquare$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.