Finding all primes $p,$ such that for all integers $a$, $a^{25} equiv apmod p$. [closed]












-1














Find all primes $p$ such that the following congruence holds for all integers $a$: $quad a^{25}equiv apmod{p}$.



I suspect there is a very simple solution, but I can't find it.






elementary-number-theory







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closed as off-topic by amWhy, jgon, KReiser, Cesareo, Shailesh Jan 5 at 2:54


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." – amWhy, jgon, KReiser, Cesareo, Shailesh

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









  • 1




    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.
    – amWhy
    Jan 4 at 20:55










  • Special case of this Theorem.
    – Bill Dubuque
    Jan 4 at 21:21
















-1














Find all primes $p$ such that the following congruence holds for all integers $a$: $quad a^{25}equiv apmod{p}$.



I suspect there is a very simple solution, but I can't find it.






elementary-number-theory







share|cite|improve this question















closed as off-topic by amWhy, jgon, KReiser, Cesareo, Shailesh Jan 5 at 2:54


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." – amWhy, jgon, KReiser, Cesareo, Shailesh

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









  • 1




    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.
    – amWhy
    Jan 4 at 20:55










  • Special case of this Theorem.
    – Bill Dubuque
    Jan 4 at 21:21














-1












-1








-1







Find all primes $p$ such that the following congruence holds for all integers $a$: $quad a^{25}equiv apmod{p}$.



I suspect there is a very simple solution, but I can't find it.






elementary-number-theory







share|cite|improve this question















Find all primes $p$ such that the following congruence holds for all integers $a$: $quad a^{25}equiv apmod{p}$.



I suspect there is a very simple solution, but I can't find it.






elementary-number-theory




elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 20:58









amWhy

192k28225439




192k28225439










asked Jan 4 at 20:47









John John

394




394




closed as off-topic by amWhy, jgon, KReiser, Cesareo, Shailesh Jan 5 at 2:54


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." – amWhy, jgon, KReiser, Cesareo, Shailesh

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




closed as off-topic by amWhy, jgon, KReiser, Cesareo, Shailesh Jan 5 at 2:54


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." – amWhy, jgon, KReiser, Cesareo, Shailesh

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








  • 1




    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.
    – amWhy
    Jan 4 at 20:55










  • Special case of this Theorem.
    – Bill Dubuque
    Jan 4 at 21:21














  • 1




    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.
    – amWhy
    Jan 4 at 20:55










  • Special case of this Theorem.
    – Bill Dubuque
    Jan 4 at 21:21








1




1




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.
– amWhy
Jan 4 at 20:55




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.
– amWhy
Jan 4 at 20:55












Special case of this Theorem.
– Bill Dubuque
Jan 4 at 21:21




Special case of this Theorem.
– Bill Dubuque
Jan 4 at 21:21










1 Answer
1






active

oldest

votes


















2














Hint:



If $a$ is not divisible by $p$, it is equivalent to $a^{24}equiv 1pmod p$.



By lil' Fermat, this implies that $p-1$ divides $24$.






share|cite|improve this answer

















  • 1




    LOOL Lil' Fermat xD
    – Cristian Baeza
    Jan 4 at 21:01










  • Thanks for the reply. Yes, I now see that it holds for every $p$ where $p-1$ divides $24$. But I don't see why it can't hold for some other $p$ as well.
    – John
    Jan 4 at 21:19






  • 1




    Because it has to hold for every $a$, and in particular for the generator of the cyclic group $(mathbf Z/pmathbf Z)^times$. This generator has order $p-1$.
    – Bernard
    Jan 4 at 21:33




















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Hint:



If $a$ is not divisible by $p$, it is equivalent to $a^{24}equiv 1pmod p$.



By lil' Fermat, this implies that $p-1$ divides $24$.






share|cite|improve this answer

















  • 1




    LOOL Lil' Fermat xD
    – Cristian Baeza
    Jan 4 at 21:01










  • Thanks for the reply. Yes, I now see that it holds for every $p$ where $p-1$ divides $24$. But I don't see why it can't hold for some other $p$ as well.
    – John
    Jan 4 at 21:19






  • 1




    Because it has to hold for every $a$, and in particular for the generator of the cyclic group $(mathbf Z/pmathbf Z)^times$. This generator has order $p-1$.
    – Bernard
    Jan 4 at 21:33


















2














Hint:



If $a$ is not divisible by $p$, it is equivalent to $a^{24}equiv 1pmod p$.



By lil' Fermat, this implies that $p-1$ divides $24$.






share|cite|improve this answer

















  • 1




    LOOL Lil' Fermat xD
    – Cristian Baeza
    Jan 4 at 21:01










  • Thanks for the reply. Yes, I now see that it holds for every $p$ where $p-1$ divides $24$. But I don't see why it can't hold for some other $p$ as well.
    – John
    Jan 4 at 21:19






  • 1




    Because it has to hold for every $a$, and in particular for the generator of the cyclic group $(mathbf Z/pmathbf Z)^times$. This generator has order $p-1$.
    – Bernard
    Jan 4 at 21:33
















2












2








2






Hint:



If $a$ is not divisible by $p$, it is equivalent to $a^{24}equiv 1pmod p$.



By lil' Fermat, this implies that $p-1$ divides $24$.






share|cite|improve this answer












Hint:



If $a$ is not divisible by $p$, it is equivalent to $a^{24}equiv 1pmod p$.



By lil' Fermat, this implies that $p-1$ divides $24$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 20:57









BernardBernard

118k639112




118k639112








  • 1




    LOOL Lil' Fermat xD
    – Cristian Baeza
    Jan 4 at 21:01










  • Thanks for the reply. Yes, I now see that it holds for every $p$ where $p-1$ divides $24$. But I don't see why it can't hold for some other $p$ as well.
    – John
    Jan 4 at 21:19






  • 1




    Because it has to hold for every $a$, and in particular for the generator of the cyclic group $(mathbf Z/pmathbf Z)^times$. This generator has order $p-1$.
    – Bernard
    Jan 4 at 21:33
















  • 1




    LOOL Lil' Fermat xD
    – Cristian Baeza
    Jan 4 at 21:01










  • Thanks for the reply. Yes, I now see that it holds for every $p$ where $p-1$ divides $24$. But I don't see why it can't hold for some other $p$ as well.
    – John
    Jan 4 at 21:19






  • 1




    Because it has to hold for every $a$, and in particular for the generator of the cyclic group $(mathbf Z/pmathbf Z)^times$. This generator has order $p-1$.
    – Bernard
    Jan 4 at 21:33










1




1




LOOL Lil' Fermat xD
– Cristian Baeza
Jan 4 at 21:01




LOOL Lil' Fermat xD
– Cristian Baeza
Jan 4 at 21:01












Thanks for the reply. Yes, I now see that it holds for every $p$ where $p-1$ divides $24$. But I don't see why it can't hold for some other $p$ as well.
– John
Jan 4 at 21:19




Thanks for the reply. Yes, I now see that it holds for every $p$ where $p-1$ divides $24$. But I don't see why it can't hold for some other $p$ as well.
– John
Jan 4 at 21:19




1




1




Because it has to hold for every $a$, and in particular for the generator of the cyclic group $(mathbf Z/pmathbf Z)^times$. This generator has order $p-1$.
– Bernard
Jan 4 at 21:33






Because it has to hold for every $a$, and in particular for the generator of the cyclic group $(mathbf Z/pmathbf Z)^times$. This generator has order $p-1$.
– Bernard
Jan 4 at 21:33





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