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

-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.

edited yesterday

amWhy

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asked yesterday

John

394

put on hold as off-topic by amWhy, jgon, KReiser, Cesareo, Shailesh yesterday

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
yesterday

Special case of this Theorem.
– Bill Dubuque
yesterday

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.

edited yesterday

amWhy

192k28224439

asked yesterday

John

394

put on hold as off-topic by amWhy, jgon, KReiser, Cesareo, Shailesh yesterday

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
yesterday

Special case of this Theorem.
– Bill Dubuque
yesterday

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.

edited yesterday

amWhy

192k28224439

asked yesterday

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

edited yesterday

amWhy

192k28224439

asked yesterday

John

394

edited yesterday

amWhy

192k28224439

asked yesterday

John

394

edited yesterday

amWhy

192k28224439

edited yesterday

amWhy

192k28224439

edited yesterday

amWhy

192k28224439

asked yesterday

John

394

asked yesterday

John

394

asked yesterday

John

394

put on hold as off-topic by amWhy, jgon, KReiser, Cesareo, Shailesh yesterday

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.

put on hold as off-topic by amWhy, jgon, KReiser, Cesareo, Shailesh yesterday

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
yesterday

Special case of this Theorem.
– Bill Dubuque
yesterday

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
yesterday

Special case of this Theorem.
– Bill Dubuque
yesterday

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
yesterday

Special case of this Theorem.
– Bill Dubuque
yesterday

add a comment |

1 Answer
1

active

oldest

votes

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 yesterday

Bernard

118k639112

1

LOOL Lil' Fermat xD
– Cristian Baeza
yesterday

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
yesterday

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
yesterday

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

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 yesterday

Bernard

118k639112

1

LOOL Lil' Fermat xD
– Cristian Baeza
yesterday

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
yesterday

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
yesterday

add a comment |

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 yesterday

Bernard

118k639112

1

LOOL Lil' Fermat xD
– Cristian Baeza
yesterday

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
yesterday

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
yesterday

add a comment |

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 yesterday

Bernard

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 yesterday

Bernard

118k639112

answered yesterday

Bernard

118k639112

answered yesterday

Bernard

118k639112

answered yesterday

Bernard

118k639112

1

LOOL Lil' Fermat xD
– Cristian Baeza
yesterday

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
yesterday

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
yesterday

add a comment |

1

LOOL Lil' Fermat xD
– Cristian Baeza
yesterday

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
yesterday

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
yesterday

LOOL Lil' Fermat xD
– Cristian Baeza
yesterday

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
yesterday

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
yesterday

add a comment |

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