Why does the Euler's totient function $phi(n)$ give the minimum of exponent s.t. $a^{phi(n)}equiv 1pmod n$,...












1












$begingroup$


I want to know whether it's possible that there would exist $1lt kltphi(n)$ s.t. $a^{phi(n)}equiv 1pmod n$, for a given $a$ and $n$? I need to prove/disprove it. I need some hints. (For title I meant minimum mod n.)



OK, seems it may be too easy for my question but may I also ask that how to find the minimum even if I know that $(a,n)=1$? i.e. I have to know the $k$ s.t. $a^kequiv 1pmod n,$ given $(a,n)=1$.






number-theory







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  • 1




    $begingroup$
    See this answer for order algorithms.
    $endgroup$
    – Bill Dubuque
    Jan 7 at 19:40










  • $begingroup$
    Please clarify whether you mean “minimum $k$ that works for all $a$” or for just a particular $a$.
    $endgroup$
    – Erick Wong
    Jan 7 at 20:14










  • $begingroup$
    @ErickWong: for the given $a$, sorry for confusing.
    $endgroup$
    – user7813604
    Jan 8 at 1:48
















1












$begingroup$


I want to know whether it's possible that there would exist $1lt kltphi(n)$ s.t. $a^{phi(n)}equiv 1pmod n$, for a given $a$ and $n$? I need to prove/disprove it. I need some hints. (For title I meant minimum mod n.)



OK, seems it may be too easy for my question but may I also ask that how to find the minimum even if I know that $(a,n)=1$? i.e. I have to know the $k$ s.t. $a^kequiv 1pmod n,$ given $(a,n)=1$.






number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    See this answer for order algorithms.
    $endgroup$
    – Bill Dubuque
    Jan 7 at 19:40










  • $begingroup$
    Please clarify whether you mean “minimum $k$ that works for all $a$” or for just a particular $a$.
    $endgroup$
    – Erick Wong
    Jan 7 at 20:14










  • $begingroup$
    @ErickWong: for the given $a$, sorry for confusing.
    $endgroup$
    – user7813604
    Jan 8 at 1:48














1












1








1





$begingroup$


I want to know whether it's possible that there would exist $1lt kltphi(n)$ s.t. $a^{phi(n)}equiv 1pmod n$, for a given $a$ and $n$? I need to prove/disprove it. I need some hints. (For title I meant minimum mod n.)



OK, seems it may be too easy for my question but may I also ask that how to find the minimum even if I know that $(a,n)=1$? i.e. I have to know the $k$ s.t. $a^kequiv 1pmod n,$ given $(a,n)=1$.






number-theory







share|cite|improve this question











$endgroup$




I want to know whether it's possible that there would exist $1lt kltphi(n)$ s.t. $a^{phi(n)}equiv 1pmod n$, for a given $a$ and $n$? I need to prove/disprove it. I need some hints. (For title I meant minimum mod n.)



OK, seems it may be too easy for my question but may I also ask that how to find the minimum even if I know that $(a,n)=1$? i.e. I have to know the $k$ s.t. $a^kequiv 1pmod n,$ given $(a,n)=1$.






number-theory




number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 2:54









kale

33




33










asked Jan 7 at 18:45









user7813604user7813604

15912




15912








  • 1




    $begingroup$
    See this answer for order algorithms.
    $endgroup$
    – Bill Dubuque
    Jan 7 at 19:40










  • $begingroup$
    Please clarify whether you mean “minimum $k$ that works for all $a$” or for just a particular $a$.
    $endgroup$
    – Erick Wong
    Jan 7 at 20:14










  • $begingroup$
    @ErickWong: for the given $a$, sorry for confusing.
    $endgroup$
    – user7813604
    Jan 8 at 1:48














  • 1




    $begingroup$
    See this answer for order algorithms.
    $endgroup$
    – Bill Dubuque
    Jan 7 at 19:40










  • $begingroup$
    Please clarify whether you mean “minimum $k$ that works for all $a$” or for just a particular $a$.
    $endgroup$
    – Erick Wong
    Jan 7 at 20:14










  • $begingroup$
    @ErickWong: for the given $a$, sorry for confusing.
    $endgroup$
    – user7813604
    Jan 8 at 1:48








1




1




$begingroup$
See this answer for order algorithms.
$endgroup$
– Bill Dubuque
Jan 7 at 19:40




$begingroup$
See this answer for order algorithms.
$endgroup$
– Bill Dubuque
Jan 7 at 19:40












$begingroup$
Please clarify whether you mean “minimum $k$ that works for all $a$” or for just a particular $a$.
$endgroup$
– Erick Wong
Jan 7 at 20:14




$begingroup$
Please clarify whether you mean “minimum $k$ that works for all $a$” or for just a particular $a$.
$endgroup$
– Erick Wong
Jan 7 at 20:14












$begingroup$
@ErickWong: for the given $a$, sorry for confusing.
$endgroup$
– user7813604
Jan 8 at 1:48




$begingroup$
@ErickWong: for the given $a$, sorry for confusing.
$endgroup$
– user7813604
Jan 8 at 1:48










3 Answers
3






active

oldest

votes


















2












$begingroup$

$phi(7)=6$ but $2^3equiv 1$(mod$7$).



Edit: As I know there is no general algorithm to find the least integer $k$ such that $a^kequiv 1$(mod$n$). Though a fact that can help is that this $k$ must divide $phi(n)$. Even more than that, this $k$ divides every integer $m$ such that $a^mequiv 1$(mod $m$). You can try to prove it, that's a pretty easy exercise. So for example, given that we know that $phi(7)=6$ we know that the minimum cannot be $4$ or $5$. It must be an integer which divides $6$. So that gives us less options which makes it a bit easier.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Well, in the example I gave you it is pretty fast. When it comes to big numbers it is obviously a very hard task. Even to find $phi(n)$ is very hard in general.
    $endgroup$
    – Mark
    Jan 7 at 19:06










  • $begingroup$
    that RSA is secure because $phi(n)$ is hard to find right? I just haven't connected these ideas...
    $endgroup$
    – user7813604
    Jan 7 at 19:07






  • 1




    $begingroup$
    Yes, exactly. And to find $phi(n)$ is hard because you need to find the prime factorization of $n$. And if the primes which are dividing $n$ are very big then the factorization might take years even for today's best computers. (maybe it might change in the future). So RSA is pretty safe.
    $endgroup$
    – Mark
    Jan 7 at 19:12










  • $begingroup$
    Very grateful for your kindness and detailed explanation.
    $endgroup$
    – user7813604
    Jan 7 at 19:15










  • $begingroup$
    You're welcome.
    $endgroup$
    – Mark
    Jan 7 at 19:17



















3












$begingroup$

Hint 1:



If $a^{phi(n)}equiv 1 pmod n$ and $phi(n)$ is composite and $k|phi(n)$ then Let $b = a^k$.



Then $b^{frac {phi(n)}k} = (a^k)^{frac {phi(n)}k}=a^{phi(n)} equiv 1 pmod n$.



Hint 2:



$(-1)^2 equiv 1 pmod n$. So if $phi(n) ne 2$....



Hint 3:



$a^3 equiv 1 pmod {a^3 -1}$. Can you find any $a^3 -1$ so that $phi(a^3 -1) > 3$?



=====



Euler's Theorem was never actually meant to be a way of finding the order of an element, least not directly. The fact that $a^{phi(n)} equiv 1 pmod n$ for $a$ relatively prime to $n$ in no way means that $phi(n)$ is the smallest order that such is true. In general, it rarely is!



Indeed, as $(a^k)^{frac {phi(n)}k}=a^k$ and $phi(n)$ is never prime for $phi(n) > 2$ for every element where $phi(n)$ is the smallest power there are $a^k; k|phi(n)$ where it isnt.



One useful tidbit. If $a^kequiv 1 pmod n$ then $k|phi(n)$. That is useful for finding an order.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Hint:



    $$frac{10^3-1}{37}=27$$






    share|cite|improve this answer









    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      $phi(7)=6$ but $2^3equiv 1$(mod$7$).



      Edit: As I know there is no general algorithm to find the least integer $k$ such that $a^kequiv 1$(mod$n$). Though a fact that can help is that this $k$ must divide $phi(n)$. Even more than that, this $k$ divides every integer $m$ such that $a^mequiv 1$(mod $m$). You can try to prove it, that's a pretty easy exercise. So for example, given that we know that $phi(7)=6$ we know that the minimum cannot be $4$ or $5$. It must be an integer which divides $6$. So that gives us less options which makes it a bit easier.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        Well, in the example I gave you it is pretty fast. When it comes to big numbers it is obviously a very hard task. Even to find $phi(n)$ is very hard in general.
        $endgroup$
        – Mark
        Jan 7 at 19:06










      • $begingroup$
        that RSA is secure because $phi(n)$ is hard to find right? I just haven't connected these ideas...
        $endgroup$
        – user7813604
        Jan 7 at 19:07






      • 1




        $begingroup$
        Yes, exactly. And to find $phi(n)$ is hard because you need to find the prime factorization of $n$. And if the primes which are dividing $n$ are very big then the factorization might take years even for today's best computers. (maybe it might change in the future). So RSA is pretty safe.
        $endgroup$
        – Mark
        Jan 7 at 19:12










      • $begingroup$
        Very grateful for your kindness and detailed explanation.
        $endgroup$
        – user7813604
        Jan 7 at 19:15










      • $begingroup$
        You're welcome.
        $endgroup$
        – Mark
        Jan 7 at 19:17
















      2












      $begingroup$

      $phi(7)=6$ but $2^3equiv 1$(mod$7$).



      Edit: As I know there is no general algorithm to find the least integer $k$ such that $a^kequiv 1$(mod$n$). Though a fact that can help is that this $k$ must divide $phi(n)$. Even more than that, this $k$ divides every integer $m$ such that $a^mequiv 1$(mod $m$). You can try to prove it, that's a pretty easy exercise. So for example, given that we know that $phi(7)=6$ we know that the minimum cannot be $4$ or $5$. It must be an integer which divides $6$. So that gives us less options which makes it a bit easier.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        Well, in the example I gave you it is pretty fast. When it comes to big numbers it is obviously a very hard task. Even to find $phi(n)$ is very hard in general.
        $endgroup$
        – Mark
        Jan 7 at 19:06










      • $begingroup$
        that RSA is secure because $phi(n)$ is hard to find right? I just haven't connected these ideas...
        $endgroup$
        – user7813604
        Jan 7 at 19:07






      • 1




        $begingroup$
        Yes, exactly. And to find $phi(n)$ is hard because you need to find the prime factorization of $n$. And if the primes which are dividing $n$ are very big then the factorization might take years even for today's best computers. (maybe it might change in the future). So RSA is pretty safe.
        $endgroup$
        – Mark
        Jan 7 at 19:12










      • $begingroup$
        Very grateful for your kindness and detailed explanation.
        $endgroup$
        – user7813604
        Jan 7 at 19:15










      • $begingroup$
        You're welcome.
        $endgroup$
        – Mark
        Jan 7 at 19:17














      2












      2








      2





      $begingroup$

      $phi(7)=6$ but $2^3equiv 1$(mod$7$).



      Edit: As I know there is no general algorithm to find the least integer $k$ such that $a^kequiv 1$(mod$n$). Though a fact that can help is that this $k$ must divide $phi(n)$. Even more than that, this $k$ divides every integer $m$ such that $a^mequiv 1$(mod $m$). You can try to prove it, that's a pretty easy exercise. So for example, given that we know that $phi(7)=6$ we know that the minimum cannot be $4$ or $5$. It must be an integer which divides $6$. So that gives us less options which makes it a bit easier.






      share|cite|improve this answer











      $endgroup$



      $phi(7)=6$ but $2^3equiv 1$(mod$7$).



      Edit: As I know there is no general algorithm to find the least integer $k$ such that $a^kequiv 1$(mod$n$). Though a fact that can help is that this $k$ must divide $phi(n)$. Even more than that, this $k$ divides every integer $m$ such that $a^mequiv 1$(mod $m$). You can try to prove it, that's a pretty easy exercise. So for example, given that we know that $phi(7)=6$ we know that the minimum cannot be $4$ or $5$. It must be an integer which divides $6$. So that gives us less options which makes it a bit easier.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 7 at 18:59

























      answered Jan 7 at 18:46









      MarkMark

      10.6k1622




      10.6k1622








      • 1




        $begingroup$
        Well, in the example I gave you it is pretty fast. When it comes to big numbers it is obviously a very hard task. Even to find $phi(n)$ is very hard in general.
        $endgroup$
        – Mark
        Jan 7 at 19:06










      • $begingroup$
        that RSA is secure because $phi(n)$ is hard to find right? I just haven't connected these ideas...
        $endgroup$
        – user7813604
        Jan 7 at 19:07






      • 1




        $begingroup$
        Yes, exactly. And to find $phi(n)$ is hard because you need to find the prime factorization of $n$. And if the primes which are dividing $n$ are very big then the factorization might take years even for today's best computers. (maybe it might change in the future). So RSA is pretty safe.
        $endgroup$
        – Mark
        Jan 7 at 19:12










      • $begingroup$
        Very grateful for your kindness and detailed explanation.
        $endgroup$
        – user7813604
        Jan 7 at 19:15










      • $begingroup$
        You're welcome.
        $endgroup$
        – Mark
        Jan 7 at 19:17














      • 1




        $begingroup$
        Well, in the example I gave you it is pretty fast. When it comes to big numbers it is obviously a very hard task. Even to find $phi(n)$ is very hard in general.
        $endgroup$
        – Mark
        Jan 7 at 19:06










      • $begingroup$
        that RSA is secure because $phi(n)$ is hard to find right? I just haven't connected these ideas...
        $endgroup$
        – user7813604
        Jan 7 at 19:07






      • 1




        $begingroup$
        Yes, exactly. And to find $phi(n)$ is hard because you need to find the prime factorization of $n$. And if the primes which are dividing $n$ are very big then the factorization might take years even for today's best computers. (maybe it might change in the future). So RSA is pretty safe.
        $endgroup$
        – Mark
        Jan 7 at 19:12










      • $begingroup$
        Very grateful for your kindness and detailed explanation.
        $endgroup$
        – user7813604
        Jan 7 at 19:15










      • $begingroup$
        You're welcome.
        $endgroup$
        – Mark
        Jan 7 at 19:17








      1




      1




      $begingroup$
      Well, in the example I gave you it is pretty fast. When it comes to big numbers it is obviously a very hard task. Even to find $phi(n)$ is very hard in general.
      $endgroup$
      – Mark
      Jan 7 at 19:06




      $begingroup$
      Well, in the example I gave you it is pretty fast. When it comes to big numbers it is obviously a very hard task. Even to find $phi(n)$ is very hard in general.
      $endgroup$
      – Mark
      Jan 7 at 19:06












      $begingroup$
      that RSA is secure because $phi(n)$ is hard to find right? I just haven't connected these ideas...
      $endgroup$
      – user7813604
      Jan 7 at 19:07




      $begingroup$
      that RSA is secure because $phi(n)$ is hard to find right? I just haven't connected these ideas...
      $endgroup$
      – user7813604
      Jan 7 at 19:07




      1




      1




      $begingroup$
      Yes, exactly. And to find $phi(n)$ is hard because you need to find the prime factorization of $n$. And if the primes which are dividing $n$ are very big then the factorization might take years even for today's best computers. (maybe it might change in the future). So RSA is pretty safe.
      $endgroup$
      – Mark
      Jan 7 at 19:12




      $begingroup$
      Yes, exactly. And to find $phi(n)$ is hard because you need to find the prime factorization of $n$. And if the primes which are dividing $n$ are very big then the factorization might take years even for today's best computers. (maybe it might change in the future). So RSA is pretty safe.
      $endgroup$
      – Mark
      Jan 7 at 19:12












      $begingroup$
      Very grateful for your kindness and detailed explanation.
      $endgroup$
      – user7813604
      Jan 7 at 19:15




      $begingroup$
      Very grateful for your kindness and detailed explanation.
      $endgroup$
      – user7813604
      Jan 7 at 19:15












      $begingroup$
      You're welcome.
      $endgroup$
      – Mark
      Jan 7 at 19:17




      $begingroup$
      You're welcome.
      $endgroup$
      – Mark
      Jan 7 at 19:17











      3












      $begingroup$

      Hint 1:



      If $a^{phi(n)}equiv 1 pmod n$ and $phi(n)$ is composite and $k|phi(n)$ then Let $b = a^k$.



      Then $b^{frac {phi(n)}k} = (a^k)^{frac {phi(n)}k}=a^{phi(n)} equiv 1 pmod n$.



      Hint 2:



      $(-1)^2 equiv 1 pmod n$. So if $phi(n) ne 2$....



      Hint 3:



      $a^3 equiv 1 pmod {a^3 -1}$. Can you find any $a^3 -1$ so that $phi(a^3 -1) > 3$?



      =====



      Euler's Theorem was never actually meant to be a way of finding the order of an element, least not directly. The fact that $a^{phi(n)} equiv 1 pmod n$ for $a$ relatively prime to $n$ in no way means that $phi(n)$ is the smallest order that such is true. In general, it rarely is!



      Indeed, as $(a^k)^{frac {phi(n)}k}=a^k$ and $phi(n)$ is never prime for $phi(n) > 2$ for every element where $phi(n)$ is the smallest power there are $a^k; k|phi(n)$ where it isnt.



      One useful tidbit. If $a^kequiv 1 pmod n$ then $k|phi(n)$. That is useful for finding an order.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        Hint 1:



        If $a^{phi(n)}equiv 1 pmod n$ and $phi(n)$ is composite and $k|phi(n)$ then Let $b = a^k$.



        Then $b^{frac {phi(n)}k} = (a^k)^{frac {phi(n)}k}=a^{phi(n)} equiv 1 pmod n$.



        Hint 2:



        $(-1)^2 equiv 1 pmod n$. So if $phi(n) ne 2$....



        Hint 3:



        $a^3 equiv 1 pmod {a^3 -1}$. Can you find any $a^3 -1$ so that $phi(a^3 -1) > 3$?



        =====



        Euler's Theorem was never actually meant to be a way of finding the order of an element, least not directly. The fact that $a^{phi(n)} equiv 1 pmod n$ for $a$ relatively prime to $n$ in no way means that $phi(n)$ is the smallest order that such is true. In general, it rarely is!



        Indeed, as $(a^k)^{frac {phi(n)}k}=a^k$ and $phi(n)$ is never prime for $phi(n) > 2$ for every element where $phi(n)$ is the smallest power there are $a^k; k|phi(n)$ where it isnt.



        One useful tidbit. If $a^kequiv 1 pmod n$ then $k|phi(n)$. That is useful for finding an order.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          Hint 1:



          If $a^{phi(n)}equiv 1 pmod n$ and $phi(n)$ is composite and $k|phi(n)$ then Let $b = a^k$.



          Then $b^{frac {phi(n)}k} = (a^k)^{frac {phi(n)}k}=a^{phi(n)} equiv 1 pmod n$.



          Hint 2:



          $(-1)^2 equiv 1 pmod n$. So if $phi(n) ne 2$....



          Hint 3:



          $a^3 equiv 1 pmod {a^3 -1}$. Can you find any $a^3 -1$ so that $phi(a^3 -1) > 3$?



          =====



          Euler's Theorem was never actually meant to be a way of finding the order of an element, least not directly. The fact that $a^{phi(n)} equiv 1 pmod n$ for $a$ relatively prime to $n$ in no way means that $phi(n)$ is the smallest order that such is true. In general, it rarely is!



          Indeed, as $(a^k)^{frac {phi(n)}k}=a^k$ and $phi(n)$ is never prime for $phi(n) > 2$ for every element where $phi(n)$ is the smallest power there are $a^k; k|phi(n)$ where it isnt.



          One useful tidbit. If $a^kequiv 1 pmod n$ then $k|phi(n)$. That is useful for finding an order.






          share|cite|improve this answer











          $endgroup$



          Hint 1:



          If $a^{phi(n)}equiv 1 pmod n$ and $phi(n)$ is composite and $k|phi(n)$ then Let $b = a^k$.



          Then $b^{frac {phi(n)}k} = (a^k)^{frac {phi(n)}k}=a^{phi(n)} equiv 1 pmod n$.



          Hint 2:



          $(-1)^2 equiv 1 pmod n$. So if $phi(n) ne 2$....



          Hint 3:



          $a^3 equiv 1 pmod {a^3 -1}$. Can you find any $a^3 -1$ so that $phi(a^3 -1) > 3$?



          =====



          Euler's Theorem was never actually meant to be a way of finding the order of an element, least not directly. The fact that $a^{phi(n)} equiv 1 pmod n$ for $a$ relatively prime to $n$ in no way means that $phi(n)$ is the smallest order that such is true. In general, it rarely is!



          Indeed, as $(a^k)^{frac {phi(n)}k}=a^k$ and $phi(n)$ is never prime for $phi(n) > 2$ for every element where $phi(n)$ is the smallest power there are $a^k; k|phi(n)$ where it isnt.



          One useful tidbit. If $a^kequiv 1 pmod n$ then $k|phi(n)$. That is useful for finding an order.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 3:18

























          answered Jan 8 at 2:38









          fleabloodfleablood

          1




          1























              2












              $begingroup$

              Hint:



              $$frac{10^3-1}{37}=27$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Hint:



                $$frac{10^3-1}{37}=27$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Hint:



                  $$frac{10^3-1}{37}=27$$






                  share|cite|improve this answer









                  $endgroup$



                  Hint:



                  $$frac{10^3-1}{37}=27$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 7 at 18:48









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