8
$\begingroup$

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

Inspired by Lucas-Lehmer-Riesel 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= k \cdot 3^{n}-1 $ where $k$ is a positive even natural number , $ k<2^n$ and $n\ge3$ . Let $a$ be a natural number greater than two such that $\left(\frac{a-2}{N}\right)=1$ and $\left(\frac{a+2}{N}\right)=-1$ where $\left(\frac{}{}\right)$ denotes Jacobi symbol. Let $S_i=S_{i-1}^3-3 S_{i-1}$ with $S_0$ equal to the modular $P_{9k/2}(a)\phantom{5} \text{mod} \phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} \equiv -2 \pmod{N}$ .

You can run this test here .

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

CEk31(k1,k2,n1,n2)=
{
forstep(k=k1,k2,[2],
for(n=n1,n2,
if(k<2^n,
N=k*3^n-1;
a=3;
while(!(kronecker(a-2,N)==1 && kronecker(a+2,N)==-1),a++);
S=Mod(2*polchebyshev(9*k/2,1,a/2),N);
ctr=1;
while(ctr<=n-2,
S=Mod(2*polchebyshev(3,1,S/2),N);
ctr+=1);
if(S==N-2 && !ispseudoprime(N),print("k="k,"n="n)))))
}
$\endgroup$
3
+100
$\begingroup$

This is a partial answer.

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

Proof :

Let us define $s,t$ as $$s:=\frac{a-\sqrt{a^2-4}}{2},\qquad t:=\frac{a+\sqrt{a^2-4}}{2}$$ where $st=1$.

Here, let us prove by induction on $i$ that $$S_i\equiv s^{3^{i+2}k/2}+t^{3^{i+2}k/2}\pmod N\tag1$$

We see that $(1)$ holds for $i=0$ since $$S_0\equiv P_{9k/2}(a)=s^{9k/2}+t^{9k/2}\pmod N$$ Supposing that $(1)$ holds for $i$ gives $$\begin{align}S_{i+1}& \equiv S_i^3-3S_i \\\\&\equiv (s^{3^{i+2}k/2}+t^{3^{i+2}k/2})^3-3(s^{3^{i+2}k/2}+t^{3^{i+2}k/2})\\\\&\equiv s^{3^{i+3}k/2}+t^{3^{i+3}k/2}\pmod N\quad\square\end{align}$$

Now, using $(1)$ and $k\cdot 3^n=N+1$, we get $$S_{n-2}\equiv s^{3^nk/2}+t^{3^nk/2}\equiv s^{(N+1)/2}+t^{(N+1)/2}\pmod N$$ Using that $$\sqrt{\frac{a\pm\sqrt{a^2-4}}{2}}=\frac{\sqrt{a+2}\pm\sqrt{a-2}}{2}$$ we have $$\begin{align}2^{N+1}S_{n-2}&\equiv (\sqrt{a+2}-\sqrt{a-2})^{N+1}+(\sqrt{a+2}+\sqrt{a-2})^{N+1} \\\\&\equiv (\sqrt{a+2}-\sqrt{a-2})(\sqrt{a+2}-\sqrt{a-2})^{N} \\&\qquad\quad +(\sqrt{a+2}+\sqrt{a-2})(\sqrt{a+2}+\sqrt{a-2})^{N} \\\\&\equiv \sqrt{a+2}\sum_{i=0}^{N}\binom Ni(\sqrt{a+2})^i((-\sqrt{a-2})^{N-i}+(\sqrt{a-2})^{N-i}) \\&\qquad \quad +\sqrt{a-2}\sum_{i=0}^{N}\binom Ni(\sqrt{a+2})^i((\sqrt{a-2})^{N-i}-(-\sqrt{a-2})^{N-i}) \\\\&\equiv \sum_{j=1}^{(N+1)/2}\binom{N}{2j-1}(a+2)^{j}\cdot 2(a-2)^{(N-(2j-1))/2} \\& \qquad \quad +\sum_{j=0}^{(N-1)/2}\binom{N}{2j}(a+2)^{j}\cdot 2(a-2)^{(N-2j+1)/2}\pmod N\end{align}$$

Using that $\binom Nm\equiv 0\pmod N$ for $1\le m\le N-1$, we get $$4\cdot 2^{N-1}S_{n-2}\equiv 2(a+2)(a+2)^{(N-1)/2}+2(a-2)(a-2)^{(N-1)/2}\pmod N$$

Since $$2^{N-1}\equiv 1,\qquad (a-2)^{(N-1)/2}\equiv 1,\qquad (a+2)^{(N-1)/2}\equiv -1\pmod N$$ we have $$4S_{n-2}\equiv -2(a+2)+2(a-2)\equiv -8\pmod N$$ from which $$S_{n-2}\equiv -2\pmod N$$ follows.$\quad\blacksquare$

$\endgroup$

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.