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
edited Jan 4 at 20:58
amWhy
192k28225439
asked Jan 4 at 20:47
John John
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.
add a comment |
-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
edited Jan 4 at 20:58
amWhy
192k28225439
asked Jan 4 at 20:47
John John
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.
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
add a comment |
-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
edited Jan 4 at 20:58
amWhy
192k28225439
asked Jan 4 at 20:47
John John
394
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
edited Jan 4 at 20:58
amWhy
192k28225439
asked Jan 4 at 20:47
John John
394
edited Jan 4 at 20:58
amWhy
192k28225439
asked Jan 4 at 20:47
John John
394
edited Jan 4 at 20:58
amWhy
192k28225439
edited Jan 4 at 20:58
amWhy
192k28225439
edited Jan 4 at 20:58
amWhy
192k28225439
192k28225439
asked Jan 4 at 20:47
John John
394
asked Jan 4 at 20:47
John John
394
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
add a comment |
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
add a comment |
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 $24$.
answered Jan 4 at 20:57
BernardBernard
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
add a comment |
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 $24$.
answered Jan 4 at 20:57
BernardBernard
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
add a comment |
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 $24$.
answered Jan 4 at 20:57
BernardBernard
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
add a comment |
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 $24$.
answered Jan 4 at 20:57
BernardBernard
118k639112
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 $24$.
answered Jan 4 at 20:57
BernardBernard
118k639112
answered Jan 4 at 20:57
BernardBernard
118k639112
answered Jan 4 at 20:57
BernardBernard
118k639112
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
add a comment |
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
add a comment |
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