How many solutions do $x^{p-1} equiv 1 pmod p$ and $x^{p-1} equiv 2 pmod p$ have?
2
$begingroup$
This is my first post so I apologize for any kind of error.
I'm preparing a magistral degree exam in number theory, and I'm performing some exercise.
I'm asking here this question: how can I prove how many solutions there are for $x^{p-1} equiv 1pmod p$ and $x^{p-1} equiv 2 pmod p$?
Edit: $p$ is an odd prime.
elementary-number-theory modular-arithmetic
edited Nov 26 '18 at 9:42
Batominovski
1
asked Nov 26 '18 at 9:07
AlessarAlessar
27115
$endgroup$
add a comment |
2
$begingroup$
This is my first post so I apologize for any kind of error.
I'm preparing a magistral degree exam in number theory, and I'm performing some exercise.
I'm asking here this question: how can I prove how many solutions there are for $x^{p-1} equiv 1pmod p$ and $x^{p-1} equiv 2 pmod p$?
Edit: $p$ is an odd prime.
elementary-number-theory modular-arithmetic
edited Nov 26 '18 at 9:42
Batominovski
1
asked Nov 26 '18 at 9:07
AlessarAlessar
27115
$endgroup$
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Nov 26 '18 at 9:15
$begingroup$
@Alessar: You may take example for $p$ say $p=5$ to get an intuitive understanding.
$endgroup$
– Yadati Kiran
Nov 26 '18 at 9:32
add a comment |
2
2
2
$begingroup$
This is my first post so I apologize for any kind of error.
I'm preparing a magistral degree exam in number theory, and I'm performing some exercise.
I'm asking here this question: how can I prove how many solutions there are for $x^{p-1} equiv 1pmod p$ and $x^{p-1} equiv 2 pmod p$?
Edit: $p$ is an odd prime.
elementary-number-theory modular-arithmetic
edited Nov 26 '18 at 9:42
Batominovski
1
asked Nov 26 '18 at 9:07
AlessarAlessar
27115
$endgroup$
This is my first post so I apologize for any kind of error.
I'm preparing a magistral degree exam in number theory, and I'm performing some exercise.
I'm asking here this question: how can I prove how many solutions there are for $x^{p-1} equiv 1pmod p$ and $x^{p-1} equiv 2 pmod p$?
Edit: $p$ is an odd prime.
elementary-number-theory modular-arithmetic
elementary-number-theory modular-arithmetic
edited Nov 26 '18 at 9:42
Batominovski
1
asked Nov 26 '18 at 9:07
AlessarAlessar
27115
edited Nov 26 '18 at 9:42
Batominovski
1
asked Nov 26 '18 at 9:07
AlessarAlessar
27115
edited Nov 26 '18 at 9:42
Batominovski
1
edited Nov 26 '18 at 9:42
Batominovski
1
edited Nov 26 '18 at 9:42
Batominovski
1
1
asked Nov 26 '18 at 9:07
AlessarAlessar
27115
asked Nov 26 '18 at 9:07
AlessarAlessar
27115
asked Nov 26 '18 at 9:07
AlessarAlessar
27115
27115
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Nov 26 '18 at 9:15
$begingroup$
@Alessar: You may take example for $p$ say $p=5$ to get an intuitive understanding.
$endgroup$
– Yadati Kiran
Nov 26 '18 at 9:32
add a comment |
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Nov 26 '18 at 9:15
$begingroup$
@Alessar: You may take example for $p$ say $p=5$ to get an intuitive understanding.
$endgroup$
– Yadati Kiran
Nov 26 '18 at 9:32
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Nov 26 '18 at 9:15
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Nov 26 '18 at 9:15
$begingroup$
@Alessar: You may take example for $p$ say $p=5$ to get an intuitive understanding.
$endgroup$
– Yadati Kiran
Nov 26 '18 at 9:32
$begingroup$
@Alessar: You may take example for $p$ say $p=5$ to get an intuitive understanding.
$endgroup$
– Yadati Kiran
Nov 26 '18 at 9:32
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
Do you know Fermat‘s little theorem?
Consider the multiplicative group $Bbb Z^times_p$.
answered Nov 26 '18 at 9:12
Lukas KoflerLukas Kofler
1,2632519
$endgroup$
$begingroup$
Yes I know it, this is the same as consider x^{p} equiv x(mod p), but for the second questions I'm really stuck
$endgroup$
– Alessar
Nov 26 '18 at 9:31
2
$begingroup$
If you know Fermat's Little Theorem, then you would know that, if $p$ is a prime natural number and $x$ is an integer such that $pnmid x$, then $x^{p-1}equiv 1pmod{p}$. Thus, unless $p=2$, $x^{p-1}equiv 2pmod{p}$ has no solution.
$endgroup$
– Batominovski
Nov 26 '18 at 9:40
$begingroup$
Thank you so much, it's the first time I study number theory
$endgroup$
– Alessar
Nov 26 '18 at 9:44
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Do you know Fermat‘s little theorem?
Consider the multiplicative group $Bbb Z^times_p$.
answered Nov 26 '18 at 9:12
Lukas KoflerLukas Kofler
1,2632519
$endgroup$
$begingroup$
Yes I know it, this is the same as consider x^{p} equiv x(mod p), but for the second questions I'm really stuck
$endgroup$
– Alessar
Nov 26 '18 at 9:31
2
$begingroup$
If you know Fermat's Little Theorem, then you would know that, if $p$ is a prime natural number and $x$ is an integer such that $pnmid x$, then $x^{p-1}equiv 1pmod{p}$. Thus, unless $p=2$, $x^{p-1}equiv 2pmod{p}$ has no solution.
$endgroup$
– Batominovski
Nov 26 '18 at 9:40
$begingroup$
Thank you so much, it's the first time I study number theory
$endgroup$
– Alessar
Nov 26 '18 at 9:44
add a comment |
2
$begingroup$
Do you know Fermat‘s little theorem?
Consider the multiplicative group $Bbb Z^times_p$.
answered Nov 26 '18 at 9:12
Lukas KoflerLukas Kofler
1,2632519
$endgroup$
$begingroup$
Yes I know it, this is the same as consider x^{p} equiv x(mod p), but for the second questions I'm really stuck
$endgroup$
– Alessar
Nov 26 '18 at 9:31
2
$begingroup$
If you know Fermat's Little Theorem, then you would know that, if $p$ is a prime natural number and $x$ is an integer such that $pnmid x$, then $x^{p-1}equiv 1pmod{p}$. Thus, unless $p=2$, $x^{p-1}equiv 2pmod{p}$ has no solution.
$endgroup$
– Batominovski
Nov 26 '18 at 9:40
$begingroup$
Thank you so much, it's the first time I study number theory
$endgroup$
– Alessar
Nov 26 '18 at 9:44
add a comment |
2
2
2
$begingroup$
Do you know Fermat‘s little theorem?
Consider the multiplicative group $Bbb Z^times_p$.
answered Nov 26 '18 at 9:12
Lukas KoflerLukas Kofler
1,2632519
$endgroup$
Do you know Fermat‘s little theorem?
Consider the multiplicative group $Bbb Z^times_p$.
answered Nov 26 '18 at 9:12
Lukas KoflerLukas Kofler
1,2632519
answered Nov 26 '18 at 9:12
Lukas KoflerLukas Kofler
1,2632519
answered Nov 26 '18 at 9:12
Lukas KoflerLukas Kofler
1,2632519
answered Nov 26 '18 at 9:12
Lukas KoflerLukas Kofler
1,2632519
1,2632519
$begingroup$
Yes I know it, this is the same as consider x^{p} equiv x(mod p), but for the second questions I'm really stuck
$endgroup$
– Alessar
Nov 26 '18 at 9:31
2
$begingroup$
If you know Fermat's Little Theorem, then you would know that, if $p$ is a prime natural number and $x$ is an integer such that $pnmid x$, then $x^{p-1}equiv 1pmod{p}$. Thus, unless $p=2$, $x^{p-1}equiv 2pmod{p}$ has no solution.
$endgroup$
– Batominovski
Nov 26 '18 at 9:40
$begingroup$
Thank you so much, it's the first time I study number theory
$endgroup$
– Alessar
Nov 26 '18 at 9:44
add a comment |
$begingroup$
Yes I know it, this is the same as consider x^{p} equiv x(mod p), but for the second questions I'm really stuck
$endgroup$
– Alessar
Nov 26 '18 at 9:31
2
$begingroup$
If you know Fermat's Little Theorem, then you would know that, if $p$ is a prime natural number and $x$ is an integer such that $pnmid x$, then $x^{p-1}equiv 1pmod{p}$. Thus, unless $p=2$, $x^{p-1}equiv 2pmod{p}$ has no solution.
$endgroup$
– Batominovski
Nov 26 '18 at 9:40
$begingroup$
Thank you so much, it's the first time I study number theory
$endgroup$
– Alessar
Nov 26 '18 at 9:44
$begingroup$
Yes I know it, this is the same as consider x^{p} equiv x(mod p), but for the second questions I'm really stuck
$endgroup$
– Alessar
Nov 26 '18 at 9:31
$begingroup$
Yes I know it, this is the same as consider x^{p} equiv x(mod p), but for the second questions I'm really stuck
$endgroup$
– Alessar
Nov 26 '18 at 9:31
2
2
$begingroup$
If you know Fermat's Little Theorem, then you would know that, if $p$ is a prime natural number and $x$ is an integer such that $pnmid x$, then $x^{p-1}equiv 1pmod{p}$. Thus, unless $p=2$, $x^{p-1}equiv 2pmod{p}$ has no solution.
$endgroup$
– Batominovski
Nov 26 '18 at 9:40
$begingroup$
If you know Fermat's Little Theorem, then you would know that, if $p$ is a prime natural number and $x$ is an integer such that $pnmid x$, then $x^{p-1}equiv 1pmod{p}$. Thus, unless $p=2$, $x^{p-1}equiv 2pmod{p}$ has no solution.
$endgroup$
– Batominovski
Nov 26 '18 at 9:40
$begingroup$
Thank you so much, it's the first time I study number theory
$endgroup$
– Alessar
Nov 26 '18 at 9:44
$begingroup$
Thank you so much, it's the first time I study number theory
$endgroup$
– Alessar
Nov 26 '18 at 9:44
add a comment |
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$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Nov 26 '18 at 9:15
$begingroup$
@Alessar: You may take example for $p$ say $p=5$ to get an intuitive understanding.
$endgroup$
– Yadati Kiran
Nov 26 '18 at 9:32