Are there further primes of the form $varphi(n)^{varphi(varphi(n))}+1$?












9












$begingroup$


For positive integers $n$ , define $$f(n):=varphi(n)^{varphi(varphi(n))}+1$$ where $varphi(n)$ denotes the totient function.



According to my calculation, for the following positive integers $n$ , $f(n)$ is a prime number : $$[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 18, 97, 119, 153, 194, 195, 208, 224, 23
8, 260, 280, 288, 306, 312, 336, 360, 390, 420]$$ and upto $n=10^4$, no further prime occurs. For $n>6$ , we have $varphi(varphi(n))>1$ and $varphi(n)>1$ hence $varphi(varphi(n))$ must be a power of $2$. The number is then a generalized Fermat-number.




Do further primes $f(n)$ exist ?







number-theory elementary-number-theory prime-numbers totient-function







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    @paw88789 To get a prime , this must be the case.
    $endgroup$
    – Peter
    Jul 26 '18 at 16:22






  • 2




    $begingroup$
    @Joffan if $q$ is an odd prime factor of $n$ , then $a^n+1$ is divisible by $a^{n/q}+1$
    $endgroup$
    – Peter
    Jul 26 '18 at 16:24






  • 1




    $begingroup$
    Have you looked it up in the OEIS? I just did. No relevant results.
    $endgroup$
    – Robert Soupe
    Jul 26 '18 at 18:19






  • 1




    $begingroup$
    Then I tried Select[Range[1000], PrimeQ[EulerPhi[#]^EulerPhi[EulerPhi[#]] + 1] &] in Wolfram Alpha. I probably need Mathematica if I want to confirm your assertion up to $10^4$.
    $endgroup$
    – Robert Soupe
    Jul 26 '18 at 18:22






  • 1




    $begingroup$
    @RobertSoupe : Pari/GP is a free software which is able to deal with that problem nicely. Note btw, that many $n$ have the same value $varphi(n)$ and thus the same $text{isprime}(f(n)$ -result, so time consumption could be much reduced when you avoid multiple computation of the $f(n)$ at different $n$ leading to the same result.
    $endgroup$
    – Gottfried Helms
    Jul 26 '18 at 20:58
















9












$begingroup$


For positive integers $n$ , define $$f(n):=varphi(n)^{varphi(varphi(n))}+1$$ where $varphi(n)$ denotes the totient function.



According to my calculation, for the following positive integers $n$ , $f(n)$ is a prime number : $$[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 18, 97, 119, 153, 194, 195, 208, 224, 23
8, 260, 280, 288, 306, 312, 336, 360, 390, 420]$$ and upto $n=10^4$, no further prime occurs. For $n>6$ , we have $varphi(varphi(n))>1$ and $varphi(n)>1$ hence $varphi(varphi(n))$ must be a power of $2$. The number is then a generalized Fermat-number.




Do further primes $f(n)$ exist ?







number-theory elementary-number-theory prime-numbers totient-function







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    @paw88789 To get a prime , this must be the case.
    $endgroup$
    – Peter
    Jul 26 '18 at 16:22






  • 2




    $begingroup$
    @Joffan if $q$ is an odd prime factor of $n$ , then $a^n+1$ is divisible by $a^{n/q}+1$
    $endgroup$
    – Peter
    Jul 26 '18 at 16:24






  • 1




    $begingroup$
    Have you looked it up in the OEIS? I just did. No relevant results.
    $endgroup$
    – Robert Soupe
    Jul 26 '18 at 18:19






  • 1




    $begingroup$
    Then I tried Select[Range[1000], PrimeQ[EulerPhi[#]^EulerPhi[EulerPhi[#]] + 1] &] in Wolfram Alpha. I probably need Mathematica if I want to confirm your assertion up to $10^4$.
    $endgroup$
    – Robert Soupe
    Jul 26 '18 at 18:22






  • 1




    $begingroup$
    @RobertSoupe : Pari/GP is a free software which is able to deal with that problem nicely. Note btw, that many $n$ have the same value $varphi(n)$ and thus the same $text{isprime}(f(n)$ -result, so time consumption could be much reduced when you avoid multiple computation of the $f(n)$ at different $n$ leading to the same result.
    $endgroup$
    – Gottfried Helms
    Jul 26 '18 at 20:58














9












9








9


3



$begingroup$


For positive integers $n$ , define $$f(n):=varphi(n)^{varphi(varphi(n))}+1$$ where $varphi(n)$ denotes the totient function.



According to my calculation, for the following positive integers $n$ , $f(n)$ is a prime number : $$[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 18, 97, 119, 153, 194, 195, 208, 224, 23
8, 260, 280, 288, 306, 312, 336, 360, 390, 420]$$ and upto $n=10^4$, no further prime occurs. For $n>6$ , we have $varphi(varphi(n))>1$ and $varphi(n)>1$ hence $varphi(varphi(n))$ must be a power of $2$. The number is then a generalized Fermat-number.




Do further primes $f(n)$ exist ?







number-theory elementary-number-theory prime-numbers totient-function







share|cite|improve this question









$endgroup$




For positive integers $n$ , define $$f(n):=varphi(n)^{varphi(varphi(n))}+1$$ where $varphi(n)$ denotes the totient function.



According to my calculation, for the following positive integers $n$ , $f(n)$ is a prime number : $$[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 18, 97, 119, 153, 194, 195, 208, 224, 23
8, 260, 280, 288, 306, 312, 336, 360, 390, 420]$$ and upto $n=10^4$, no further prime occurs. For $n>6$ , we have $varphi(varphi(n))>1$ and $varphi(n)>1$ hence $varphi(varphi(n))$ must be a power of $2$. The number is then a generalized Fermat-number.




Do further primes $f(n)$ exist ?







number-theory elementary-number-theory prime-numbers totient-function




number-theory elementary-number-theory prime-numbers totient-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jul 26 '18 at 16:06









PeterPeter

48.8k1240137




48.8k1240137








  • 1




    $begingroup$
    @paw88789 To get a prime , this must be the case.
    $endgroup$
    – Peter
    Jul 26 '18 at 16:22






  • 2




    $begingroup$
    @Joffan if $q$ is an odd prime factor of $n$ , then $a^n+1$ is divisible by $a^{n/q}+1$
    $endgroup$
    – Peter
    Jul 26 '18 at 16:24






  • 1




    $begingroup$
    Have you looked it up in the OEIS? I just did. No relevant results.
    $endgroup$
    – Robert Soupe
    Jul 26 '18 at 18:19






  • 1




    $begingroup$
    Then I tried Select[Range[1000], PrimeQ[EulerPhi[#]^EulerPhi[EulerPhi[#]] + 1] &] in Wolfram Alpha. I probably need Mathematica if I want to confirm your assertion up to $10^4$.
    $endgroup$
    – Robert Soupe
    Jul 26 '18 at 18:22






  • 1




    $begingroup$
    @RobertSoupe : Pari/GP is a free software which is able to deal with that problem nicely. Note btw, that many $n$ have the same value $varphi(n)$ and thus the same $text{isprime}(f(n)$ -result, so time consumption could be much reduced when you avoid multiple computation of the $f(n)$ at different $n$ leading to the same result.
    $endgroup$
    – Gottfried Helms
    Jul 26 '18 at 20:58














  • 1




    $begingroup$
    @paw88789 To get a prime , this must be the case.
    $endgroup$
    – Peter
    Jul 26 '18 at 16:22






  • 2




    $begingroup$
    @Joffan if $q$ is an odd prime factor of $n$ , then $a^n+1$ is divisible by $a^{n/q}+1$
    $endgroup$
    – Peter
    Jul 26 '18 at 16:24






  • 1




    $begingroup$
    Have you looked it up in the OEIS? I just did. No relevant results.
    $endgroup$
    – Robert Soupe
    Jul 26 '18 at 18:19






  • 1




    $begingroup$
    Then I tried Select[Range[1000], PrimeQ[EulerPhi[#]^EulerPhi[EulerPhi[#]] + 1] &] in Wolfram Alpha. I probably need Mathematica if I want to confirm your assertion up to $10^4$.
    $endgroup$
    – Robert Soupe
    Jul 26 '18 at 18:22






  • 1




    $begingroup$
    @RobertSoupe : Pari/GP is a free software which is able to deal with that problem nicely. Note btw, that many $n$ have the same value $varphi(n)$ and thus the same $text{isprime}(f(n)$ -result, so time consumption could be much reduced when you avoid multiple computation of the $f(n)$ at different $n$ leading to the same result.
    $endgroup$
    – Gottfried Helms
    Jul 26 '18 at 20:58








1




1




$begingroup$
@paw88789 To get a prime , this must be the case.
$endgroup$
– Peter
Jul 26 '18 at 16:22




$begingroup$
@paw88789 To get a prime , this must be the case.
$endgroup$
– Peter
Jul 26 '18 at 16:22




2




2




$begingroup$
@Joffan if $q$ is an odd prime factor of $n$ , then $a^n+1$ is divisible by $a^{n/q}+1$
$endgroup$
– Peter
Jul 26 '18 at 16:24




$begingroup$
@Joffan if $q$ is an odd prime factor of $n$ , then $a^n+1$ is divisible by $a^{n/q}+1$
$endgroup$
– Peter
Jul 26 '18 at 16:24




1




1




$begingroup$
Have you looked it up in the OEIS? I just did. No relevant results.
$endgroup$
– Robert Soupe
Jul 26 '18 at 18:19




$begingroup$
Have you looked it up in the OEIS? I just did. No relevant results.
$endgroup$
– Robert Soupe
Jul 26 '18 at 18:19




1




1




$begingroup$
Then I tried Select[Range[1000], PrimeQ[EulerPhi[#]^EulerPhi[EulerPhi[#]] + 1] &] in Wolfram Alpha. I probably need Mathematica if I want to confirm your assertion up to $10^4$.
$endgroup$
– Robert Soupe
Jul 26 '18 at 18:22




$begingroup$
Then I tried Select[Range[1000], PrimeQ[EulerPhi[#]^EulerPhi[EulerPhi[#]] + 1] &] in Wolfram Alpha. I probably need Mathematica if I want to confirm your assertion up to $10^4$.
$endgroup$
– Robert Soupe
Jul 26 '18 at 18:22




1




1




$begingroup$
@RobertSoupe : Pari/GP is a free software which is able to deal with that problem nicely. Note btw, that many $n$ have the same value $varphi(n)$ and thus the same $text{isprime}(f(n)$ -result, so time consumption could be much reduced when you avoid multiple computation of the $f(n)$ at different $n$ leading to the same result.
$endgroup$
– Gottfried Helms
Jul 26 '18 at 20:58




$begingroup$
@RobertSoupe : Pari/GP is a free software which is able to deal with that problem nicely. Note btw, that many $n$ have the same value $varphi(n)$ and thus the same $text{isprime}(f(n)$ -result, so time consumption could be much reduced when you avoid multiple computation of the $f(n)$ at different $n$ leading to the same result.
$endgroup$
– Gottfried Helms
Jul 26 '18 at 20:58










1 Answer
1






active

oldest

votes


















0












$begingroup$

This is mostly just a summary of what I have found that is to big for a comment.



Since $varphi(varphi(n))$ must be a power of $2$, $varphi(n)=2^m p_1 p_2 p_3...p_m$. Where each of the $p_i$ is a distinct Fermat prime. Thus we have
$$varphi(n)^{varphi(varphi(n))}+1=(2^mp_1p_2p_3...p_m)^{2^r}+1.tag{1}$$
We also have that
$$n=2^uq_1q_2...q_sp_1^{e_1}p_2^{e_2}...p_m^{e_m}$$
where the $q_i$ are primes of the form $2^dp_i+1$, and the $e_i$ are each $0,1$ or $2$. If a $q_i$ is present in the factorization, then $e_i$ is at most $1$.



If we want to answer your question, it will probably be easiest to work with the right side of $(1)$.






share|cite|improve this answer









$endgroup$














    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863551%2fare-there-further-primes-of-the-form-varphin-varphi-varphin1%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    This is mostly just a summary of what I have found that is to big for a comment.



    Since $varphi(varphi(n))$ must be a power of $2$, $varphi(n)=2^m p_1 p_2 p_3...p_m$. Where each of the $p_i$ is a distinct Fermat prime. Thus we have
    $$varphi(n)^{varphi(varphi(n))}+1=(2^mp_1p_2p_3...p_m)^{2^r}+1.tag{1}$$
    We also have that
    $$n=2^uq_1q_2...q_sp_1^{e_1}p_2^{e_2}...p_m^{e_m}$$
    where the $q_i$ are primes of the form $2^dp_i+1$, and the $e_i$ are each $0,1$ or $2$. If a $q_i$ is present in the factorization, then $e_i$ is at most $1$.



    If we want to answer your question, it will probably be easiest to work with the right side of $(1)$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      This is mostly just a summary of what I have found that is to big for a comment.



      Since $varphi(varphi(n))$ must be a power of $2$, $varphi(n)=2^m p_1 p_2 p_3...p_m$. Where each of the $p_i$ is a distinct Fermat prime. Thus we have
      $$varphi(n)^{varphi(varphi(n))}+1=(2^mp_1p_2p_3...p_m)^{2^r}+1.tag{1}$$
      We also have that
      $$n=2^uq_1q_2...q_sp_1^{e_1}p_2^{e_2}...p_m^{e_m}$$
      where the $q_i$ are primes of the form $2^dp_i+1$, and the $e_i$ are each $0,1$ or $2$. If a $q_i$ is present in the factorization, then $e_i$ is at most $1$.



      If we want to answer your question, it will probably be easiest to work with the right side of $(1)$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        This is mostly just a summary of what I have found that is to big for a comment.



        Since $varphi(varphi(n))$ must be a power of $2$, $varphi(n)=2^m p_1 p_2 p_3...p_m$. Where each of the $p_i$ is a distinct Fermat prime. Thus we have
        $$varphi(n)^{varphi(varphi(n))}+1=(2^mp_1p_2p_3...p_m)^{2^r}+1.tag{1}$$
        We also have that
        $$n=2^uq_1q_2...q_sp_1^{e_1}p_2^{e_2}...p_m^{e_m}$$
        where the $q_i$ are primes of the form $2^dp_i+1$, and the $e_i$ are each $0,1$ or $2$. If a $q_i$ is present in the factorization, then $e_i$ is at most $1$.



        If we want to answer your question, it will probably be easiest to work with the right side of $(1)$.






        share|cite|improve this answer









        $endgroup$



        This is mostly just a summary of what I have found that is to big for a comment.



        Since $varphi(varphi(n))$ must be a power of $2$, $varphi(n)=2^m p_1 p_2 p_3...p_m$. Where each of the $p_i$ is a distinct Fermat prime. Thus we have
        $$varphi(n)^{varphi(varphi(n))}+1=(2^mp_1p_2p_3...p_m)^{2^r}+1.tag{1}$$
        We also have that
        $$n=2^uq_1q_2...q_sp_1^{e_1}p_2^{e_2}...p_m^{e_m}$$
        where the $q_i$ are primes of the form $2^dp_i+1$, and the $e_i$ are each $0,1$ or $2$. If a $q_i$ is present in the factorization, then $e_i$ is at most $1$.



        If we want to answer your question, it will probably be easiest to work with the right side of $(1)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 21:37









        coDE_RPcoDE_RP

        799




        799






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863551%2fare-there-further-primes-of-the-form-varphin-varphi-varphin1%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Bressuire

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
            Read more

            Gyllenstierna

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more