Is $2^{2^k}+1$ always a prime number? [on hold]












0















Is $2^{2^k}+1$ always a prime number ?




If yes how to prove? Remark : $k$ is any natural number.






number-theory elementary-number-theory prime-numbers







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Kshitij Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by José Carlos Santos, Giuseppe Negro, Nosrati, Eevee Trainer, KM101 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Giuseppe Negro, Nosrati, Eevee Trainer, KM101

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 14




    Yes if you are Fermat, no if you are not. Look up "Fermat number".
    – KCd
    2 days ago






  • 7




    $641mid(2^{2^5}+1)$.
    – Lord Shark the Unknown
    2 days ago






  • 2




    It seems that the converse statement is true : For every $nge 5$, the number $$2^{2^n}+1$$ is composite.
    – Peter
    2 days ago






  • 2




    @Peter No one knows that, though it's suspected.
    – Ethan Bolker
    2 days ago






  • 2




    The first unknown case is $$F_{33}=2^{2^{33}}+1$$
    – Peter
    2 days ago
















0















Is $2^{2^k}+1$ always a prime number ?




If yes how to prove? Remark : $k$ is any natural number.






number-theory elementary-number-theory prime-numbers







share|cite|improve this question









New contributor




Kshitij Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











put on hold as off-topic by José Carlos Santos, Giuseppe Negro, Nosrati, Eevee Trainer, KM101 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Giuseppe Negro, Nosrati, Eevee Trainer, KM101

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 14




    Yes if you are Fermat, no if you are not. Look up "Fermat number".
    – KCd
    2 days ago






  • 7




    $641mid(2^{2^5}+1)$.
    – Lord Shark the Unknown
    2 days ago






  • 2




    It seems that the converse statement is true : For every $nge 5$, the number $$2^{2^n}+1$$ is composite.
    – Peter
    2 days ago






  • 2




    @Peter No one knows that, though it's suspected.
    – Ethan Bolker
    2 days ago






  • 2




    The first unknown case is $$F_{33}=2^{2^{33}}+1$$
    – Peter
    2 days ago














0












0








0








Is $2^{2^k}+1$ always a prime number ?




If yes how to prove? Remark : $k$ is any natural number.






number-theory elementary-number-theory prime-numbers







share|cite|improve this question









New contributor




Kshitij Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












Is $2^{2^k}+1$ always a prime number ?




If yes how to prove? Remark : $k$ is any natural number.






number-theory elementary-number-theory prime-numbers




number-theory elementary-number-theory prime-numbers






share|cite|improve this question









New contributor




Kshitij Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Kshitij Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Peter

46.7k1039125




46.7k1039125






New contributor




Kshitij Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









Kshitij Singh

21




21




New contributor




Kshitij Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Kshitij Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Kshitij Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




put on hold as off-topic by José Carlos Santos, Giuseppe Negro, Nosrati, Eevee Trainer, KM101 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Giuseppe Negro, Nosrati, Eevee Trainer, KM101

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by José Carlos Santos, Giuseppe Negro, Nosrati, Eevee Trainer, KM101 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Giuseppe Negro, Nosrati, Eevee Trainer, KM101

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 14




    Yes if you are Fermat, no if you are not. Look up "Fermat number".
    – KCd
    2 days ago






  • 7




    $641mid(2^{2^5}+1)$.
    – Lord Shark the Unknown
    2 days ago






  • 2




    It seems that the converse statement is true : For every $nge 5$, the number $$2^{2^n}+1$$ is composite.
    – Peter
    2 days ago






  • 2




    @Peter No one knows that, though it's suspected.
    – Ethan Bolker
    2 days ago






  • 2




    The first unknown case is $$F_{33}=2^{2^{33}}+1$$
    – Peter
    2 days ago














  • 14




    Yes if you are Fermat, no if you are not. Look up "Fermat number".
    – KCd
    2 days ago






  • 7




    $641mid(2^{2^5}+1)$.
    – Lord Shark the Unknown
    2 days ago






  • 2




    It seems that the converse statement is true : For every $nge 5$, the number $$2^{2^n}+1$$ is composite.
    – Peter
    2 days ago






  • 2




    @Peter No one knows that, though it's suspected.
    – Ethan Bolker
    2 days ago






  • 2




    The first unknown case is $$F_{33}=2^{2^{33}}+1$$
    – Peter
    2 days ago








14




14




Yes if you are Fermat, no if you are not. Look up "Fermat number".
– KCd
2 days ago




Yes if you are Fermat, no if you are not. Look up "Fermat number".
– KCd
2 days ago




7




7




$641mid(2^{2^5}+1)$.
– Lord Shark the Unknown
2 days ago




$641mid(2^{2^5}+1)$.
– Lord Shark the Unknown
2 days ago




2




2




It seems that the converse statement is true : For every $nge 5$, the number $$2^{2^n}+1$$ is composite.
– Peter
2 days ago




It seems that the converse statement is true : For every $nge 5$, the number $$2^{2^n}+1$$ is composite.
– Peter
2 days ago




2




2




@Peter No one knows that, though it's suspected.
– Ethan Bolker
2 days ago




@Peter No one knows that, though it's suspected.
– Ethan Bolker
2 days ago




2




2




The first unknown case is $$F_{33}=2^{2^{33}}+1$$
– Peter
2 days ago




The first unknown case is $$F_{33}=2^{2^{33}}+1$$
– Peter
2 days ago










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