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This question is successor of Primality test for specific class of generalized Fermat numbers .

Can you provide a proof or a 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 $F_n(b)= b^{2^n}+1 $ where $b$ is an even natural number and $n\ge2$ . Let $a$ be a natural number greater than two such that $\left(\frac{a-2}{F_n(b)}\right)=-1$ and $\left(\frac{a+2}{F_n(b)}\right)=-1$ where $\left(\frac{}{}\right)$ denotes Jacobi symbol. Let $S_i=P_b(S_{i-1})$ with $S_0$ equal to the modular $P_{b/2}(P_{b/2}(a))\phantom{5} \text{mod} \phantom{5} F_n(b)$. Then $F_n(b)$ is prime if and only if $S_{2^n-2} \equiv 0 \pmod{F_n(b)}$ .

You can run this test here. A list of generalized Fermat primes sorted by base $b$ can be found here. I have tested this claim for many random values of $b$ and $n$ and there were no counterexamples.

A command line program that implements this test can be found here.

Android app that implements this test can be found here .

Python script that implements this test can be found here.

Mathematica notebook that implements this test can be found here.

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I offer 100 € for a proof of this claim. Proof must be published in Journal of Number Theory.

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  • $\begingroup$ There are no counterexamples for $b \in [2,10000]$ with $n \in [2,10]$ . $\endgroup$ – Peđa Terzić Nov 16 at 14:19
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This is a partial answer.

This answer proves that if $F_n(b)$ is prime, then $S_{2^n-2} \equiv 0 \pmod{F_n(b)}$.

Proof :

Let $N:=F_n(b)=b^{2^n}+1$. It can be proven by induction that $$S_i\equiv 2^{-b^{i+2}/4}(p^{b^{i+2}/4}+q^{b^{i+2}/4})\pmod N\tag1$$ where $p=a-\sqrt{a^2-4},q=a+\sqrt{a^2-4}$.

From $(1)$, we get, using $\sqrt{a\pm\sqrt{a^2-4}}=\frac 1{\sqrt 2}(\sqrt{a+2}\pm\sqrt{a-2})$,

$$\begin{align}&2^{N+1}\cdot S_{2^n-2}^2-2^{N+2} \\\\&\equiv \left(\sqrt{a+2}+\sqrt{a-2}\right)\left(\sqrt{a+2}-\sqrt{a-2}\right)^{N} \\&\qquad\qquad +\left(\sqrt{a+2}-\sqrt{a-2}\right)\left(\sqrt{a+2}+\sqrt{a-2}\right)^{N}\pmod N \\\\&\equiv \sqrt{a+2}\left(\left(\sqrt{a+2}-\sqrt{a-2}\right)^{N} +\left(\sqrt{a+2}+\sqrt{a-2}\right)^{N}\right) \\&\qquad\qquad-\sqrt{a-2}\left(\left(\sqrt{a+2}+\sqrt{a-2}\right)^{N} -\left(\sqrt{a+2}-\sqrt{a-2}\right)^{N}\right)\pmod N \\\\&\equiv \sqrt{a+2}\sum_{k=0}^{N}\binom Nk(\sqrt{a+2})^{N-k}((-\sqrt{a-2})^k+(\sqrt{a-2})^k) \\&\qquad\qquad -\sqrt{a-2}\sum_{k=0}^{N}\binom Nk(\sqrt{a+2})^{N-k}((\sqrt{a-2})^k-(-\sqrt{a-2})^k)\pmod N \\\\&\equiv \sum_{j=0}^{(N-1)/2}\binom{N}{2j}(a+2)^{(N-2j+1)/2}\cdot 2(a-2)^j \\&\qquad\qquad -\sum_{j=1}^{(N+1)/2}\binom{N}{2j-1}(a+2)^{(N-2j+1)/2}\cdot 2(a-2)^j\pmod N \\\\&\equiv 2(a+2)\cdot\left(\frac{a+2}{N}\right)-2(a-2)\cdot\left(\frac{a-2}{N}\right)\pmod N \\\\&\equiv 2(a+2)\cdot (-1)-2(a-2)\cdot (-1)\pmod N \\\\&\equiv -8\pmod N \end{align}$$ So, we get $$2^{N+1}\cdot S_{2^n-2}^2-2^{N+2}\equiv -8\pmod N$$ It follows from $2^{(N-1)/2}\equiv 1\pmod N$ that $$S_{2^n-2}\equiv 0\pmod{F_n(b)}$$

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