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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 $S_0=P_9(4)$ . 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(9,1,2);
 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)))
 }
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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 $$S_i=s^{3^{i+2}}+t^{3^{i+2}}\tag1$$ where $s=2-\sqrt 3,t=2+\sqrt 3$ with $st=1$.

We see that $(1)$ holds for $i=0$ since $$S_0=P_9(4)=2^{-9}((4-2\sqrt 3)^9+(4+2\sqrt 3)^9)=s^{3^2}+t^{3^2}$$

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

Using $(1)$ and $N=4\cdot 3^n+1$, we have $$S_{n-2}=s^{3^{n}}+t^{3^{n}}=s^{(N-1)/4}+t^{(N-1)/4}$$ Noting that $$\sqrt{2\pm\sqrt 3}=\frac{\sqrt 6\pm\sqrt 2}{2}$$ we have $$S_{n-2}^2-2=s^{(N-1)/2}+t^{(N-1)/2}=\left(\frac{\sqrt 6-\sqrt 2}{2}\right)^{N-1}+\left(\frac{\sqrt 6+\sqrt 2}{2}\right)^{N-1}$$ Multiplying this by $2^{N+1}=2^{N-1}(\sqrt 6-\sqrt 2)(\sqrt 6+\sqrt 2)$, we get, by the binomial theorem, $$\begin{align}&2^{N+1}(S_{n-2}^2-2) \\\\&=(\sqrt 6+\sqrt 2)(\sqrt 6-\sqrt 2)^N+(\sqrt 6-\sqrt 2)(\sqrt 6+\sqrt 2)^N \\\\&=\sqrt 6\ ((\sqrt 6-\sqrt 2)^N+(\sqrt 6+\sqrt 2)^N) \\&\qquad\qquad -\sqrt 2\ ((\sqrt 6+\sqrt 2)^N-(\sqrt 6-\sqrt 2)^N) \\\\&=\sqrt 6\sum_{i=0}^{N}\binom Ni(\sqrt 6)^{N-i}((-\sqrt 2)^i+(\sqrt 2)^i) \\&\qquad\qquad -\sqrt 2\sum_{i=0}^{N}\binom Ni(\sqrt 6)^{N-i}((\sqrt 2)^{i}-(-\sqrt 2)^{i}) \\\\&=\sum_{j=0}^{(N-1)/2}\binom{N}{2j}\cdot 6^{(N-2j+1)/2}\cdot 2^{j+1} \\&\qquad\qquad -\sum_{j=1}^{(N+1)/2}\binom{N}{2j-1}\cdot 6^{(N-2j+1)/2}\cdot 2^{j+1}\end{align}$$

Here, using that $\binom Nm\equiv 0\pmod N$ for $1\le m\le N-1$, we get $$2^{N+1}(S_{n-2}^2-2)\equiv 2\cdot 2^{(N+1)/2}\cdot 3^{(N+1)/2}-2^{(N+3)/2}\tag2$$

Since $$4\cdot 3^{2k}+1\equiv 4\cdot 1+1\equiv 5\pmod 8$$ and $$4\cdot 3^{2k+1}+1\equiv 12\cdot 1+1\equiv 5\pmod 8$$ we see that $$N\equiv 5\pmod 8$$ from which we have $$2^{(N-1)/2}\equiv \left(\frac{2}{N}\right)=(-1)^{(N^2-1)/8}=-1\pmod N\tag3$$where $\left(\frac{q}{p}\right)$ denotes the Legendre symbol.

Using $2^{N-1}\equiv 1\pmod N$ and $$3^{(N-1)/2}\equiv\left(\frac{3}{N}\right)=\frac{(-1)^{(N-1)/2}}{\left(\frac N3\right)}=\frac{1}{1}=1\pmod N$$ we get, from $(2)(3)$, $$4(S_{n-2}^2-2)\equiv 2\cdot (-2)\cdot 3-(-4)\pmod N$$ from which $$S_{n-2}\equiv 0\pmod N$$ follows.$\quad\blacksquare$

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