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Here is my observation :

Let $N$ = $(3^p-1)/2$ when $p$ is a prime number $p > 3$

Let $S_i=S_{i-1}^3-3 S_{i-1}$ with $S_0=52$ . Then $N$ is prime if and only if $S_{p-1} \equiv S_{0}\pmod{N}$.

I choose 52 because this is one of the "seeds" for the test of Lucas-Lehmer and it seems it works with this "seed" (you can find the seeds for Lucas-Lehmer test here)

For example with $p$ = $7$ I found with Pari GP

 52, 1093
 548, 1093
 682, 1093
 969, 1093
 1033, 1093
 594, 1093
 52, 1093

And $1093$ is a prime number

I checked until $p=1100$ and I didn't find counterexample.

Is there a way to explain this ? I don't know how to start for proving it. If you found a counterexample please tell me.

kijinSeija
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  • $$x^3-3x-52=(x-4)(x^2+4x+13).$$ Not sure if that helps at all. – Thomas Andrews Dec 10 '21 at 21:18
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    I don't think your question makes sense. Please look at the proof of the LLT for Mersenne numbers before inventing new ones. – reuns Dec 10 '21 at 22:47
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    @reuns did you find a counterexample to be so self-confident ? This is just an observation – kijinSeija Dec 11 '21 at 01:35
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    I interpret that you are asking for a primality test, not an explanation for why by chance this sequence is not $S_{p-1}=S_0\bmod N$ for small $N$ composite, a question that would make sense only if you'd know that we always have $S_{p-1}=S_0\bmod N$ for $N$ prime. To this latter question take a look at my answer, if $N$ is not prime then several steps are unlikely to hold, so by chance for small $p$ they don't (try with a concrete $N$ composite and see which steps fail). – reuns Dec 11 '21 at 16:25
  • I'm intersted by the proof of course I will upvote your answer anf thanks for your time. But I try to understand why it works with 52 for S0. I didn't check indeed with composite number for p. Only prime until 2000 now and still no counterexample with Pari Gp. – kijinSeija Dec 11 '21 at 16:52

1 Answers1

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Your sequence is $$S_i = (2+\sqrt3)^{3^{i+1}}+(2-\sqrt3)^{3^{i+1}}$$ Proof: check that $S_0=52$ and $S_{i+1}=S_i^3-3S_i$.

So $S_{p-1}=S_0\bmod N$ iff $$(2+\sqrt3)^{3^p} +(2-\sqrt3)^{3^p}=(2+\sqrt3)^3+(2-\sqrt3)^3\bmod N$$

If $N$ is prime then

  • $N=1\bmod 3$ so $1+\sqrt{-3}^N= (1+\sqrt{-3})^N=(2\zeta_3)^N=2 \zeta_3 =1+\sqrt{-3}\bmod N$

  • $N=1\bmod 4$ so $\sqrt{3}^N = (-i\sqrt{-3})^N= (-i)^N \sqrt{-3}^N= -i \sqrt{-3}=\sqrt3\bmod N$

  • $(2+\sqrt3)^N = 2^N+(\sqrt3)^N=2+\sqrt3\bmod N$

    Similarly $(2-\sqrt3)^N=2-\sqrt3\bmod N$

  • And hence $$S_{p-1}=(2+\sqrt3)^{2N+1}+(2-\sqrt3)^{2N+1}=(2+\sqrt3)^3+ (2-\sqrt3)^3=S_0\bmod N$$

That's a good start. But for the other direction, I don't see why $S_{p-1}=S_0\bmod N$ would imply that $N$ is not composite.

reuns
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