How many solutions do $x^{p-1} equiv 1 pmod p$ and $x^{p-1} equiv 2 pmod p$ have?












2














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.










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  • 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.
    – José Carlos Santos
    Nov 26 at 9:15










  • @Alessar: You may take example for $p$ say $p=5$ to get an intuitive understanding.
    – Yadati Kiran
    Nov 26 at 9:32


















2














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.










share|cite|improve this question
























  • 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.
    – José Carlos Santos
    Nov 26 at 9:15










  • @Alessar: You may take example for $p$ say $p=5$ to get an intuitive understanding.
    – Yadati Kiran
    Nov 26 at 9:32
















2












2








2







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.










share|cite|improve this question















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






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share|cite|improve this question













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edited Nov 26 at 9:42









Batominovski

33.7k33292




33.7k33292










asked Nov 26 at 9:07









Alessar

19613




19613












  • 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.
    – José Carlos Santos
    Nov 26 at 9:15










  • @Alessar: You may take example for $p$ say $p=5$ to get an intuitive understanding.
    – Yadati Kiran
    Nov 26 at 9:32




















  • 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.
    – José Carlos Santos
    Nov 26 at 9:15










  • @Alessar: You may take example for $p$ say $p=5$ to get an intuitive understanding.
    – Yadati Kiran
    Nov 26 at 9:32


















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.
– José Carlos Santos
Nov 26 at 9:15




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.
– José Carlos Santos
Nov 26 at 9:15












@Alessar: You may take example for $p$ say $p=5$ to get an intuitive understanding.
– Yadati Kiran
Nov 26 at 9:32






@Alessar: You may take example for $p$ say $p=5$ to get an intuitive understanding.
– Yadati Kiran
Nov 26 at 9:32












1 Answer
1






active

oldest

votes


















2














Do you know Fermat‘s little theorem?



Consider the multiplicative group $Bbb Z^times_p$.






share|cite|improve this answer





















  • 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
    – Alessar
    Nov 26 at 9:31






  • 2




    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.
    – Batominovski
    Nov 26 at 9:40










  • Thank you so much, it's the first time I study number theory
    – Alessar
    Nov 26 at 9:44











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Do you know Fermat‘s little theorem?



Consider the multiplicative group $Bbb Z^times_p$.






share|cite|improve this answer





















  • 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
    – Alessar
    Nov 26 at 9:31






  • 2




    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.
    – Batominovski
    Nov 26 at 9:40










  • Thank you so much, it's the first time I study number theory
    – Alessar
    Nov 26 at 9:44
















2














Do you know Fermat‘s little theorem?



Consider the multiplicative group $Bbb Z^times_p$.






share|cite|improve this answer





















  • 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
    – Alessar
    Nov 26 at 9:31






  • 2




    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.
    – Batominovski
    Nov 26 at 9:40










  • Thank you so much, it's the first time I study number theory
    – Alessar
    Nov 26 at 9:44














2












2








2






Do you know Fermat‘s little theorem?



Consider the multiplicative group $Bbb Z^times_p$.






share|cite|improve this answer












Do you know Fermat‘s little theorem?



Consider the multiplicative group $Bbb Z^times_p$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 at 9:12









Lukas Kofler

1,3552519




1,3552519












  • 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
    – Alessar
    Nov 26 at 9:31






  • 2




    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.
    – Batominovski
    Nov 26 at 9:40










  • Thank you so much, it's the first time I study number theory
    – Alessar
    Nov 26 at 9:44


















  • 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
    – Alessar
    Nov 26 at 9:31






  • 2




    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.
    – Batominovski
    Nov 26 at 9:40










  • Thank you so much, it's the first time I study number theory
    – Alessar
    Nov 26 at 9:44
















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
– Alessar
Nov 26 at 9:31




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
– Alessar
Nov 26 at 9:31




2




2




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.
– Batominovski
Nov 26 at 9:40




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.
– Batominovski
Nov 26 at 9:40












Thank you so much, it's the first time I study number theory
– Alessar
Nov 26 at 9:44




Thank you so much, it's the first time I study number theory
– Alessar
Nov 26 at 9:44


















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