Conjectured primality tests for specific classes of $kcdot b^n pm 1$












5












$begingroup$


Can you provide proofs or counterexamples for the claims given below?



Inspired by Lucas-Lehmer-Riesel primality test I have formulated the following two claims:



First claim




Let $P_m(x)=2^{-m}cdot((x-sqrt{x^2-4})^m+(x+sqrt{x^2-4})^m)$ .
Let $M= k cdot b^{n}-1 $ where $k$ is positive natural number , $ k<2^n$ , $b$ is an even positive natural number and $nge3$ . Let $a$ be a natural number greater than two such that $left(frac{a-2}{M}right)=1$ and $left(frac{a+2}{M}right)=-1$ where $left(frac{}{}right)$ denotes Jacobi symbol. Let $S_i=P_b(S_{i-1})$ with $S_0$ equal to the modular $P_{kb/2}(P_{b/2}(a))phantom{5} text{mod} phantom{5} M$. Then $M$ is prime if and only if $S_{n-2} equiv 0 pmod{M}$ .




You can run this test here .



Second claim




Let $P_m(x)=2^{-m}cdot((x-sqrt{x^2-4})^m+(x+sqrt{x^2-4})^m)$ .
Let $N= k cdot b^{n}+1 $ where $k$ is positive natural number , $ k<2^n$ , $b$ is an even positive natural number and $nge3$ . 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 $S_i=P_b(S_{i-1})$ with $S_0$ equal to the modular $P_{kb/2}(P_{b/2}(a))phantom{5} text{mod} phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} equiv 0 pmod{N}$ .




You can run this test here .



I have tested these claims for many random values of $k$, $b$ and $n$ and there were no countereamples.



REMARK



It is possible to reformulate these claims into more compact form:




Let $P_m(x)=2^{-m}cdot((x-sqrt{x^2-4})^m+(x+sqrt{x^2-4})^m)$ .
Let $N= k cdot b^{n}pm 1 $ where $k$ is positive natural number , $ k<2^n$ , $b$ is an even positive natural number and $nge3$ . Let $a$ be a natural number greater than two such that $left(frac{2-a}{N}right)=left(frac{a+2}{N}right)=-1$ where $left(frac{}{}right)$ denotes Jacobi symbol. Let $S_i=P_b(S_{i-1})$ with $S_0$ equal to the modular $P_{kb/2}(P_{b/2}(a))phantom{5} text{mod} phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} equiv 0 pmod{N}$ .




GUI application that implements these tests can be found here .



A command line program that implements these tests can be found here .










share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    You are unlikely to get much interest in this question. There are infinitely many possible conjectures in the world. What makes one interesting is its inherent beauty, its relation to existing work, and partial results in support of it. You have not demonstrated any of these, and yet you're asking strangers to spend hours of their time investigating your conjectures. 100 internet points is unlikely to make much difference. I recommend instead expanding the question substantially, giving partial proofs in support of your conjecture, and details about the "many random" choices you tested.
    $endgroup$
    – vadim123
    Feb 4 at 19:47
















5












$begingroup$


Can you provide proofs or counterexamples for the claims given below?



Inspired by Lucas-Lehmer-Riesel primality test I have formulated the following two claims:



First claim




Let $P_m(x)=2^{-m}cdot((x-sqrt{x^2-4})^m+(x+sqrt{x^2-4})^m)$ .
Let $M= k cdot b^{n}-1 $ where $k$ is positive natural number , $ k<2^n$ , $b$ is an even positive natural number and $nge3$ . Let $a$ be a natural number greater than two such that $left(frac{a-2}{M}right)=1$ and $left(frac{a+2}{M}right)=-1$ where $left(frac{}{}right)$ denotes Jacobi symbol. Let $S_i=P_b(S_{i-1})$ with $S_0$ equal to the modular $P_{kb/2}(P_{b/2}(a))phantom{5} text{mod} phantom{5} M$. Then $M$ is prime if and only if $S_{n-2} equiv 0 pmod{M}$ .




You can run this test here .



Second claim




Let $P_m(x)=2^{-m}cdot((x-sqrt{x^2-4})^m+(x+sqrt{x^2-4})^m)$ .
Let $N= k cdot b^{n}+1 $ where $k$ is positive natural number , $ k<2^n$ , $b$ is an even positive natural number and $nge3$ . 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 $S_i=P_b(S_{i-1})$ with $S_0$ equal to the modular $P_{kb/2}(P_{b/2}(a))phantom{5} text{mod} phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} equiv 0 pmod{N}$ .




You can run this test here .



I have tested these claims for many random values of $k$, $b$ and $n$ and there were no countereamples.



REMARK



It is possible to reformulate these claims into more compact form:




Let $P_m(x)=2^{-m}cdot((x-sqrt{x^2-4})^m+(x+sqrt{x^2-4})^m)$ .
Let $N= k cdot b^{n}pm 1 $ where $k$ is positive natural number , $ k<2^n$ , $b$ is an even positive natural number and $nge3$ . Let $a$ be a natural number greater than two such that $left(frac{2-a}{N}right)=left(frac{a+2}{N}right)=-1$ where $left(frac{}{}right)$ denotes Jacobi symbol. Let $S_i=P_b(S_{i-1})$ with $S_0$ equal to the modular $P_{kb/2}(P_{b/2}(a))phantom{5} text{mod} phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} equiv 0 pmod{N}$ .




GUI application that implements these tests can be found here .



A command line program that implements these tests can be found here .










share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    You are unlikely to get much interest in this question. There are infinitely many possible conjectures in the world. What makes one interesting is its inherent beauty, its relation to existing work, and partial results in support of it. You have not demonstrated any of these, and yet you're asking strangers to spend hours of their time investigating your conjectures. 100 internet points is unlikely to make much difference. I recommend instead expanding the question substantially, giving partial proofs in support of your conjecture, and details about the "many random" choices you tested.
    $endgroup$
    – vadim123
    Feb 4 at 19:47














5












5








5


1



$begingroup$


Can you provide proofs or counterexamples for the claims given below?



Inspired by Lucas-Lehmer-Riesel primality test I have formulated the following two claims:



First claim




Let $P_m(x)=2^{-m}cdot((x-sqrt{x^2-4})^m+(x+sqrt{x^2-4})^m)$ .
Let $M= k cdot b^{n}-1 $ where $k$ is positive natural number , $ k<2^n$ , $b$ is an even positive natural number and $nge3$ . Let $a$ be a natural number greater than two such that $left(frac{a-2}{M}right)=1$ and $left(frac{a+2}{M}right)=-1$ where $left(frac{}{}right)$ denotes Jacobi symbol. Let $S_i=P_b(S_{i-1})$ with $S_0$ equal to the modular $P_{kb/2}(P_{b/2}(a))phantom{5} text{mod} phantom{5} M$. Then $M$ is prime if and only if $S_{n-2} equiv 0 pmod{M}$ .




You can run this test here .



Second claim




Let $P_m(x)=2^{-m}cdot((x-sqrt{x^2-4})^m+(x+sqrt{x^2-4})^m)$ .
Let $N= k cdot b^{n}+1 $ where $k$ is positive natural number , $ k<2^n$ , $b$ is an even positive natural number and $nge3$ . 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 $S_i=P_b(S_{i-1})$ with $S_0$ equal to the modular $P_{kb/2}(P_{b/2}(a))phantom{5} text{mod} phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} equiv 0 pmod{N}$ .




You can run this test here .



I have tested these claims for many random values of $k$, $b$ and $n$ and there were no countereamples.



REMARK



It is possible to reformulate these claims into more compact form:




Let $P_m(x)=2^{-m}cdot((x-sqrt{x^2-4})^m+(x+sqrt{x^2-4})^m)$ .
Let $N= k cdot b^{n}pm 1 $ where $k$ is positive natural number , $ k<2^n$ , $b$ is an even positive natural number and $nge3$ . Let $a$ be a natural number greater than two such that $left(frac{2-a}{N}right)=left(frac{a+2}{N}right)=-1$ where $left(frac{}{}right)$ denotes Jacobi symbol. Let $S_i=P_b(S_{i-1})$ with $S_0$ equal to the modular $P_{kb/2}(P_{b/2}(a))phantom{5} text{mod} phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} equiv 0 pmod{N}$ .




GUI application that implements these tests can be found here .



A command line program that implements these tests can be found here .










share|cite|improve this question











$endgroup$




Can you provide proofs or counterexamples for the claims given below?



Inspired by Lucas-Lehmer-Riesel primality test I have formulated the following two claims:



First claim




Let $P_m(x)=2^{-m}cdot((x-sqrt{x^2-4})^m+(x+sqrt{x^2-4})^m)$ .
Let $M= k cdot b^{n}-1 $ where $k$ is positive natural number , $ k<2^n$ , $b$ is an even positive natural number and $nge3$ . Let $a$ be a natural number greater than two such that $left(frac{a-2}{M}right)=1$ and $left(frac{a+2}{M}right)=-1$ where $left(frac{}{}right)$ denotes Jacobi symbol. Let $S_i=P_b(S_{i-1})$ with $S_0$ equal to the modular $P_{kb/2}(P_{b/2}(a))phantom{5} text{mod} phantom{5} M$. Then $M$ is prime if and only if $S_{n-2} equiv 0 pmod{M}$ .




You can run this test here .



Second claim




Let $P_m(x)=2^{-m}cdot((x-sqrt{x^2-4})^m+(x+sqrt{x^2-4})^m)$ .
Let $N= k cdot b^{n}+1 $ where $k$ is positive natural number , $ k<2^n$ , $b$ is an even positive natural number and $nge3$ . 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 $S_i=P_b(S_{i-1})$ with $S_0$ equal to the modular $P_{kb/2}(P_{b/2}(a))phantom{5} text{mod} phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} equiv 0 pmod{N}$ .




You can run this test here .



I have tested these claims for many random values of $k$, $b$ and $n$ and there were no countereamples.



REMARK



It is possible to reformulate these claims into more compact form:




Let $P_m(x)=2^{-m}cdot((x-sqrt{x^2-4})^m+(x+sqrt{x^2-4})^m)$ .
Let $N= k cdot b^{n}pm 1 $ where $k$ is positive natural number , $ k<2^n$ , $b$ is an even positive natural number and $nge3$ . Let $a$ be a natural number greater than two such that $left(frac{2-a}{N}right)=left(frac{a+2}{N}right)=-1$ where $left(frac{}{}right)$ denotes Jacobi symbol. Let $S_i=P_b(S_{i-1})$ with $S_0$ equal to the modular $P_{kb/2}(P_{b/2}(a))phantom{5} text{mod} phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} equiv 0 pmod{N}$ .




GUI application that implements these tests can be found here .



A command line program that implements these tests can be found here .







elementary-number-theory prime-numbers examples-counterexamples primality-test lucas-lehmer-test






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edited Mar 27 at 12:48







Peđa Terzić

















asked Feb 2 at 19:38









Peđa TerzićPeđa Terzić

7,67922674




7,67922674








  • 6




    $begingroup$
    You are unlikely to get much interest in this question. There are infinitely many possible conjectures in the world. What makes one interesting is its inherent beauty, its relation to existing work, and partial results in support of it. You have not demonstrated any of these, and yet you're asking strangers to spend hours of their time investigating your conjectures. 100 internet points is unlikely to make much difference. I recommend instead expanding the question substantially, giving partial proofs in support of your conjecture, and details about the "many random" choices you tested.
    $endgroup$
    – vadim123
    Feb 4 at 19:47














  • 6




    $begingroup$
    You are unlikely to get much interest in this question. There are infinitely many possible conjectures in the world. What makes one interesting is its inherent beauty, its relation to existing work, and partial results in support of it. You have not demonstrated any of these, and yet you're asking strangers to spend hours of their time investigating your conjectures. 100 internet points is unlikely to make much difference. I recommend instead expanding the question substantially, giving partial proofs in support of your conjecture, and details about the "many random" choices you tested.
    $endgroup$
    – vadim123
    Feb 4 at 19:47








6




6




$begingroup$
You are unlikely to get much interest in this question. There are infinitely many possible conjectures in the world. What makes one interesting is its inherent beauty, its relation to existing work, and partial results in support of it. You have not demonstrated any of these, and yet you're asking strangers to spend hours of their time investigating your conjectures. 100 internet points is unlikely to make much difference. I recommend instead expanding the question substantially, giving partial proofs in support of your conjecture, and details about the "many random" choices you tested.
$endgroup$
– vadim123
Feb 4 at 19:47




$begingroup$
You are unlikely to get much interest in this question. There are infinitely many possible conjectures in the world. What makes one interesting is its inherent beauty, its relation to existing work, and partial results in support of it. You have not demonstrated any of these, and yet you're asking strangers to spend hours of their time investigating your conjectures. 100 internet points is unlikely to make much difference. I recommend instead expanding the question substantially, giving partial proofs in support of your conjecture, and details about the "many random" choices you tested.
$endgroup$
– vadim123
Feb 4 at 19:47










1 Answer
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$begingroup$

I've tried some examples, checking with simple Pari/gp code, and I was unable to find any counter-example. However, I tried few. Someone should run some complete check up to some maximum value for different cases.



Anyway, this is the following of what you already published some years ago. Still very interesting in my opinion. I like the way you define the Seed (S0). However, some explanation would help: Something like this "how the test works" for LLR (https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer%E2%80%93Riesel_test#How_the_test_works) would help.



I hope that someone with Math skills (that I do not have...) will have a look and will provide hints for a proof of these conjectures.



Maybe you should show for b=2 that P_2(x) is x^2-2 , linking this to LLT and LLR more clearly.



Regards,



Tony






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    $begingroup$

    I've tried some examples, checking with simple Pari/gp code, and I was unable to find any counter-example. However, I tried few. Someone should run some complete check up to some maximum value for different cases.



    Anyway, this is the following of what you already published some years ago. Still very interesting in my opinion. I like the way you define the Seed (S0). However, some explanation would help: Something like this "how the test works" for LLR (https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer%E2%80%93Riesel_test#How_the_test_works) would help.



    I hope that someone with Math skills (that I do not have...) will have a look and will provide hints for a proof of these conjectures.



    Maybe you should show for b=2 that P_2(x) is x^2-2 , linking this to LLT and LLR more clearly.



    Regards,



    Tony






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      I've tried some examples, checking with simple Pari/gp code, and I was unable to find any counter-example. However, I tried few. Someone should run some complete check up to some maximum value for different cases.



      Anyway, this is the following of what you already published some years ago. Still very interesting in my opinion. I like the way you define the Seed (S0). However, some explanation would help: Something like this "how the test works" for LLR (https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer%E2%80%93Riesel_test#How_the_test_works) would help.



      I hope that someone with Math skills (that I do not have...) will have a look and will provide hints for a proof of these conjectures.



      Maybe you should show for b=2 that P_2(x) is x^2-2 , linking this to LLT and LLR more clearly.



      Regards,



      Tony






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        I've tried some examples, checking with simple Pari/gp code, and I was unable to find any counter-example. However, I tried few. Someone should run some complete check up to some maximum value for different cases.



        Anyway, this is the following of what you already published some years ago. Still very interesting in my opinion. I like the way you define the Seed (S0). However, some explanation would help: Something like this "how the test works" for LLR (https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer%E2%80%93Riesel_test#How_the_test_works) would help.



        I hope that someone with Math skills (that I do not have...) will have a look and will provide hints for a proof of these conjectures.



        Maybe you should show for b=2 that P_2(x) is x^2-2 , linking this to LLT and LLR more clearly.



        Regards,



        Tony






        share|cite|improve this answer









        $endgroup$



        I've tried some examples, checking with simple Pari/gp code, and I was unable to find any counter-example. However, I tried few. Someone should run some complete check up to some maximum value for different cases.



        Anyway, this is the following of what you already published some years ago. Still very interesting in my opinion. I like the way you define the Seed (S0). However, some explanation would help: Something like this "how the test works" for LLR (https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer%E2%80%93Riesel_test#How_the_test_works) would help.



        I hope that someone with Math skills (that I do not have...) will have a look and will provide hints for a proof of these conjectures.



        Maybe you should show for b=2 that P_2(x) is x^2-2 , linking this to LLT and LLR more clearly.



        Regards,



        Tony







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 14 at 17:08









        Tony ReixTony Reix

        777




        777






























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