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









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    1 Answer
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    active

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    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

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