Proving $m$ is prime under simple remainder-arithmetic conditions [closed]

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If $m = 4^n + 1$ for some natural $n$ and $$3^fracm - 12 equiv -1 bmod m,$$ why is $m$ necessarily prime?




elementary-number-theory prime-numbers modular-arithmetic




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closed as off-topic by amWhy, José Carlos Santos, Mostafa Ayaz, Arnaud Mortier, Xander Henderson Jul 24 at 0:44


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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Mostafa Ayaz, Arnaud Mortier, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Hint : $m$ is not divisible by $3$ (why?) The order of $3$ modulo $m$ mst be a power of two (again why?). Last step : Look at the congruence assuming the order is smaller than $m-1$
    – Peter
    Jul 23 at 15:48










  • A good idea is to think about the order of 3 in the multiplicative group mod m, and what that tells you about the size of the multiplicative group mod m, and what that tells you about the totient function at m.
    – m_t_
    Jul 23 at 15:50














up vote
3
down vote

favorite












If $m = 4^n + 1$ for some natural $n$ and $$3^fracm - 12 equiv -1 bmod m,$$ why is $m$ necessarily prime?




elementary-number-theory prime-numbers modular-arithmetic




share|cite|improve this question













closed as off-topic by amWhy, José Carlos Santos, Mostafa Ayaz, Arnaud Mortier, Xander Henderson Jul 24 at 0:44


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Mostafa Ayaz, Arnaud Mortier, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Hint : $m$ is not divisible by $3$ (why?) The order of $3$ modulo $m$ mst be a power of two (again why?). Last step : Look at the congruence assuming the order is smaller than $m-1$
    – Peter
    Jul 23 at 15:48










  • A good idea is to think about the order of 3 in the multiplicative group mod m, and what that tells you about the size of the multiplicative group mod m, and what that tells you about the totient function at m.
    – m_t_
    Jul 23 at 15:50












up vote
3
down vote

favorite









up vote
3
down vote

favorite











If $m = 4^n + 1$ for some natural $n$ and $$3^fracm - 12 equiv -1 bmod m,$$ why is $m$ necessarily prime?




elementary-number-theory prime-numbers modular-arithmetic




share|cite|improve this question













If $m = 4^n + 1$ for some natural $n$ and $$3^fracm - 12 equiv -1 bmod m,$$ why is $m$ necessarily prime?





elementary-number-theory prime-numbers modular-arithmetic





share|cite|improve this question












share|cite|improve this question




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edited Jul 23 at 20:40









Mr. Brooks

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asked Jul 23 at 15:28









Uri George Peterzil

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516




closed as off-topic by amWhy, José Carlos Santos, Mostafa Ayaz, Arnaud Mortier, Xander Henderson Jul 24 at 0:44


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Mostafa Ayaz, Arnaud Mortier, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, José Carlos Santos, Mostafa Ayaz, Arnaud Mortier, Xander Henderson Jul 24 at 0:44


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Mostafa Ayaz, Arnaud Mortier, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.











  • Hint : $m$ is not divisible by $3$ (why?) The order of $3$ modulo $m$ mst be a power of two (again why?). Last step : Look at the congruence assuming the order is smaller than $m-1$
    – Peter
    Jul 23 at 15:48










  • A good idea is to think about the order of 3 in the multiplicative group mod m, and what that tells you about the size of the multiplicative group mod m, and what that tells you about the totient function at m.
    – m_t_
    Jul 23 at 15:50
















  • Hint : $m$ is not divisible by $3$ (why?) The order of $3$ modulo $m$ mst be a power of two (again why?). Last step : Look at the congruence assuming the order is smaller than $m-1$
    – Peter
    Jul 23 at 15:48










  • A good idea is to think about the order of 3 in the multiplicative group mod m, and what that tells you about the size of the multiplicative group mod m, and what that tells you about the totient function at m.
    – m_t_
    Jul 23 at 15:50















Hint : $m$ is not divisible by $3$ (why?) The order of $3$ modulo $m$ mst be a power of two (again why?). Last step : Look at the congruence assuming the order is smaller than $m-1$
– Peter
Jul 23 at 15:48




Hint : $m$ is not divisible by $3$ (why?) The order of $3$ modulo $m$ mst be a power of two (again why?). Last step : Look at the congruence assuming the order is smaller than $m-1$
– Peter
Jul 23 at 15:48












A good idea is to think about the order of 3 in the multiplicative group mod m, and what that tells you about the size of the multiplicative group mod m, and what that tells you about the totient function at m.
– m_t_
Jul 23 at 15:50




A good idea is to think about the order of 3 in the multiplicative group mod m, and what that tells you about the size of the multiplicative group mod m, and what that tells you about the totient function at m.
– m_t_
Jul 23 at 15:50










1 Answer
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It suffices to show that $varphi(m)=m-1$; only prime numbers satisfy this property. Clearly it always holds that $varphi(m) leq m-1$ for $m>1$, so we reduce to showing that $m-1 leq varphi(m)$. Indeed, this will follow from Euler's theorem if we show that $m-1$ is the order of $3$ mod $m$. Clearly by the assumption, $3^m-1equiv 1 mod m$, so we know the order of $3$ divides $m-1$. But $m-1=4^n$ is a power of $2$, so any proper divisor of $m-1$ also divides $(m-1)/2$, and hence cannot be the order of $3$ since $3^(m-1)/2equiv-1 neq 1 mod m$. Thus $m-1$ is the order of $3$ and we are done.






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  • 1




    @lulu because, as I wrote, $3^m-1 equiv 1 mod m$. In general if $a^k equiv 1 mod n$ then the order of $a$ mod $n$ divides $k$.
    – Gal Porat
    Jul 23 at 16:18











  • @lulu well in your example, for $m=65$, it does not hold that $3^m-1 equiv 1 mod m$, so it is not a counterexample to my argument.
    – Gal Porat
    Jul 23 at 16:21











  • Perhaps it is your logic I am not following. I'll delete my comments and re-read.
    – lulu
    Jul 23 at 16:22










  • Yes, I misread. (+1)
    – lulu
    Jul 23 at 16:25

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
5
down vote













It suffices to show that $varphi(m)=m-1$; only prime numbers satisfy this property. Clearly it always holds that $varphi(m) leq m-1$ for $m>1$, so we reduce to showing that $m-1 leq varphi(m)$. Indeed, this will follow from Euler's theorem if we show that $m-1$ is the order of $3$ mod $m$. Clearly by the assumption, $3^m-1equiv 1 mod m$, so we know the order of $3$ divides $m-1$. But $m-1=4^n$ is a power of $2$, so any proper divisor of $m-1$ also divides $(m-1)/2$, and hence cannot be the order of $3$ since $3^(m-1)/2equiv-1 neq 1 mod m$. Thus $m-1$ is the order of $3$ and we are done.






share|cite|improve this answer

















  • 1




    @lulu because, as I wrote, $3^m-1 equiv 1 mod m$. In general if $a^k equiv 1 mod n$ then the order of $a$ mod $n$ divides $k$.
    – Gal Porat
    Jul 23 at 16:18











  • @lulu well in your example, for $m=65$, it does not hold that $3^m-1 equiv 1 mod m$, so it is not a counterexample to my argument.
    – Gal Porat
    Jul 23 at 16:21











  • Perhaps it is your logic I am not following. I'll delete my comments and re-read.
    – lulu
    Jul 23 at 16:22










  • Yes, I misread. (+1)
    – lulu
    Jul 23 at 16:25














up vote
5
down vote













It suffices to show that $varphi(m)=m-1$; only prime numbers satisfy this property. Clearly it always holds that $varphi(m) leq m-1$ for $m>1$, so we reduce to showing that $m-1 leq varphi(m)$. Indeed, this will follow from Euler's theorem if we show that $m-1$ is the order of $3$ mod $m$. Clearly by the assumption, $3^m-1equiv 1 mod m$, so we know the order of $3$ divides $m-1$. But $m-1=4^n$ is a power of $2$, so any proper divisor of $m-1$ also divides $(m-1)/2$, and hence cannot be the order of $3$ since $3^(m-1)/2equiv-1 neq 1 mod m$. Thus $m-1$ is the order of $3$ and we are done.






share|cite|improve this answer

















  • 1




    @lulu because, as I wrote, $3^m-1 equiv 1 mod m$. In general if $a^k equiv 1 mod n$ then the order of $a$ mod $n$ divides $k$.
    – Gal Porat
    Jul 23 at 16:18











  • @lulu well in your example, for $m=65$, it does not hold that $3^m-1 equiv 1 mod m$, so it is not a counterexample to my argument.
    – Gal Porat
    Jul 23 at 16:21











  • Perhaps it is your logic I am not following. I'll delete my comments and re-read.
    – lulu
    Jul 23 at 16:22










  • Yes, I misread. (+1)
    – lulu
    Jul 23 at 16:25












up vote
5
down vote










up vote
5
down vote









It suffices to show that $varphi(m)=m-1$; only prime numbers satisfy this property. Clearly it always holds that $varphi(m) leq m-1$ for $m>1$, so we reduce to showing that $m-1 leq varphi(m)$. Indeed, this will follow from Euler's theorem if we show that $m-1$ is the order of $3$ mod $m$. Clearly by the assumption, $3^m-1equiv 1 mod m$, so we know the order of $3$ divides $m-1$. But $m-1=4^n$ is a power of $2$, so any proper divisor of $m-1$ also divides $(m-1)/2$, and hence cannot be the order of $3$ since $3^(m-1)/2equiv-1 neq 1 mod m$. Thus $m-1$ is the order of $3$ and we are done.






share|cite|improve this answer













It suffices to show that $varphi(m)=m-1$; only prime numbers satisfy this property. Clearly it always holds that $varphi(m) leq m-1$ for $m>1$, so we reduce to showing that $m-1 leq varphi(m)$. Indeed, this will follow from Euler's theorem if we show that $m-1$ is the order of $3$ mod $m$. Clearly by the assumption, $3^m-1equiv 1 mod m$, so we know the order of $3$ divides $m-1$. But $m-1=4^n$ is a power of $2$, so any proper divisor of $m-1$ also divides $(m-1)/2$, and hence cannot be the order of $3$ since $3^(m-1)/2equiv-1 neq 1 mod m$. Thus $m-1$ is the order of $3$ and we are done.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 23 at 15:58









Gal Porat

718513




718513







  • 1




    @lulu because, as I wrote, $3^m-1 equiv 1 mod m$. In general if $a^k equiv 1 mod n$ then the order of $a$ mod $n$ divides $k$.
    – Gal Porat
    Jul 23 at 16:18











  • @lulu well in your example, for $m=65$, it does not hold that $3^m-1 equiv 1 mod m$, so it is not a counterexample to my argument.
    – Gal Porat
    Jul 23 at 16:21











  • Perhaps it is your logic I am not following. I'll delete my comments and re-read.
    – lulu
    Jul 23 at 16:22










  • Yes, I misread. (+1)
    – lulu
    Jul 23 at 16:25












  • 1




    @lulu because, as I wrote, $3^m-1 equiv 1 mod m$. In general if $a^k equiv 1 mod n$ then the order of $a$ mod $n$ divides $k$.
    – Gal Porat
    Jul 23 at 16:18











  • @lulu well in your example, for $m=65$, it does not hold that $3^m-1 equiv 1 mod m$, so it is not a counterexample to my argument.
    – Gal Porat
    Jul 23 at 16:21











  • Perhaps it is your logic I am not following. I'll delete my comments and re-read.
    – lulu
    Jul 23 at 16:22










  • Yes, I misread. (+1)
    – lulu
    Jul 23 at 16:25







1




1




@lulu because, as I wrote, $3^m-1 equiv 1 mod m$. In general if $a^k equiv 1 mod n$ then the order of $a$ mod $n$ divides $k$.
– Gal Porat
Jul 23 at 16:18





@lulu because, as I wrote, $3^m-1 equiv 1 mod m$. In general if $a^k equiv 1 mod n$ then the order of $a$ mod $n$ divides $k$.
– Gal Porat
Jul 23 at 16:18













@lulu well in your example, for $m=65$, it does not hold that $3^m-1 equiv 1 mod m$, so it is not a counterexample to my argument.
– Gal Porat
Jul 23 at 16:21





@lulu well in your example, for $m=65$, it does not hold that $3^m-1 equiv 1 mod m$, so it is not a counterexample to my argument.
– Gal Porat
Jul 23 at 16:21













Perhaps it is your logic I am not following. I'll delete my comments and re-read.
– lulu
Jul 23 at 16:22




Perhaps it is your logic I am not following. I'll delete my comments and re-read.
– lulu
Jul 23 at 16:22












Yes, I misread. (+1)
– lulu
Jul 23 at 16:25




Yes, I misread. (+1)
– lulu
Jul 23 at 16:25


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