Trying to find a modulo for $2^{24}$












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$begingroup$


I was trying to figure out which modulo n would satisfy that $2^{24}$ is congruent to 1 mod n. One answer is 241, and I wonder if there is a good way to find it.



I tried to use Fermat's little theorem or Chinese Remainder theorem,but 24 is not a prime and 241 is not composite.






modular-arithmetic







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$endgroup$

















    1












    $begingroup$


    I was trying to figure out which modulo n would satisfy that $2^{24}$ is congruent to 1 mod n. One answer is 241, and I wonder if there is a good way to find it.



    I tried to use Fermat's little theorem or Chinese Remainder theorem,but 24 is not a prime and 241 is not composite.






    modular-arithmetic







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I was trying to figure out which modulo n would satisfy that $2^{24}$ is congruent to 1 mod n. One answer is 241, and I wonder if there is a good way to find it.



      I tried to use Fermat's little theorem or Chinese Remainder theorem,but 24 is not a prime and 241 is not composite.






      modular-arithmetic







      share|cite|improve this question











      $endgroup$




      I was trying to figure out which modulo n would satisfy that $2^{24}$ is congruent to 1 mod n. One answer is 241, and I wonder if there is a good way to find it.



      I tried to use Fermat's little theorem or Chinese Remainder theorem,but 24 is not a prime and 241 is not composite.






      modular-arithmetic




      modular-arithmetic






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 28 '18 at 17:18







      Jingting931015

















      asked Nov 28 '18 at 17:10









      Jingting931015Jingting931015

      828




      828






















          3 Answers
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          5












          $begingroup$

          Direct factoring gives:



          $$2^{24}-1=(2^{12}-1)(2^{12}+1)=(2^6-1)(2^6+1)(2^{12}+1)=(2^3-1)(2^3+1)(2^6+1)(2^{12}+1)$$



          So: $2^3-1=7$ will work (as will $3$ or any of the factors of $2^6+1$ or $2^{12}+1$).



          Indeed, a little effort shows that the full list of primes that work here is ${3,5,7, 13,17, 241}$






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            We want to find factors of



            begin{align}(2^{24}-1)&=(2^{12}-1)(2^{12}+1)\
            &= (2^6-1)(2^6+1)(2^{12}+1)
            end{align}



            Hence we can for example, pick our $n$ to be $2^6-1, 2^6+1$ or $2^{12}+1$. Notice that $241$ is a factor of $2^{12}+1$.



            Also note that $$2^{24}=(2^3)^8=(7+1)^8$$



            Hence we can pick $n=7$.



            Also $$2^{24}=(2^8)^3=((2^8-1)+1)^3$$



            Hence, we can pick $n=2^8-1$ or its factors and so on.






            share|cite|improve this answer











            $endgroup$





















              0












              $begingroup$

              the simplest is to find an odd prime $p$ such that $p-1$ is a divisor of $24$. You have at once:
              $$p= 3, 5, 7,13.$$
              You also have chances it works with a prime $p$ such that a divisor of $24$ ($>1$) is also a divisor of $p-1$. For instance $17$ works ($2$ has order $8$ modulo $17$).






              share|cite|improve this answer









              $endgroup$













                Your Answer





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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                5












                $begingroup$

                Direct factoring gives:



                $$2^{24}-1=(2^{12}-1)(2^{12}+1)=(2^6-1)(2^6+1)(2^{12}+1)=(2^3-1)(2^3+1)(2^6+1)(2^{12}+1)$$



                So: $2^3-1=7$ will work (as will $3$ or any of the factors of $2^6+1$ or $2^{12}+1$).



                Indeed, a little effort shows that the full list of primes that work here is ${3,5,7, 13,17, 241}$






                share|cite|improve this answer









                $endgroup$


















                  5












                  $begingroup$

                  Direct factoring gives:



                  $$2^{24}-1=(2^{12}-1)(2^{12}+1)=(2^6-1)(2^6+1)(2^{12}+1)=(2^3-1)(2^3+1)(2^6+1)(2^{12}+1)$$



                  So: $2^3-1=7$ will work (as will $3$ or any of the factors of $2^6+1$ or $2^{12}+1$).



                  Indeed, a little effort shows that the full list of primes that work here is ${3,5,7, 13,17, 241}$






                  share|cite|improve this answer









                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    Direct factoring gives:



                    $$2^{24}-1=(2^{12}-1)(2^{12}+1)=(2^6-1)(2^6+1)(2^{12}+1)=(2^3-1)(2^3+1)(2^6+1)(2^{12}+1)$$



                    So: $2^3-1=7$ will work (as will $3$ or any of the factors of $2^6+1$ or $2^{12}+1$).



                    Indeed, a little effort shows that the full list of primes that work here is ${3,5,7, 13,17, 241}$






                    share|cite|improve this answer









                    $endgroup$



                    Direct factoring gives:



                    $$2^{24}-1=(2^{12}-1)(2^{12}+1)=(2^6-1)(2^6+1)(2^{12}+1)=(2^3-1)(2^3+1)(2^6+1)(2^{12}+1)$$



                    So: $2^3-1=7$ will work (as will $3$ or any of the factors of $2^6+1$ or $2^{12}+1$).



                    Indeed, a little effort shows that the full list of primes that work here is ${3,5,7, 13,17, 241}$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 28 '18 at 17:14









                    lulululu

                    40.7k24879




                    40.7k24879























                        3












                        $begingroup$

                        We want to find factors of



                        begin{align}(2^{24}-1)&=(2^{12}-1)(2^{12}+1)\
                        &= (2^6-1)(2^6+1)(2^{12}+1)
                        end{align}



                        Hence we can for example, pick our $n$ to be $2^6-1, 2^6+1$ or $2^{12}+1$. Notice that $241$ is a factor of $2^{12}+1$.



                        Also note that $$2^{24}=(2^3)^8=(7+1)^8$$



                        Hence we can pick $n=7$.



                        Also $$2^{24}=(2^8)^3=((2^8-1)+1)^3$$



                        Hence, we can pick $n=2^8-1$ or its factors and so on.






                        share|cite|improve this answer











                        $endgroup$


















                          3












                          $begingroup$

                          We want to find factors of



                          begin{align}(2^{24}-1)&=(2^{12}-1)(2^{12}+1)\
                          &= (2^6-1)(2^6+1)(2^{12}+1)
                          end{align}



                          Hence we can for example, pick our $n$ to be $2^6-1, 2^6+1$ or $2^{12}+1$. Notice that $241$ is a factor of $2^{12}+1$.



                          Also note that $$2^{24}=(2^3)^8=(7+1)^8$$



                          Hence we can pick $n=7$.



                          Also $$2^{24}=(2^8)^3=((2^8-1)+1)^3$$



                          Hence, we can pick $n=2^8-1$ or its factors and so on.






                          share|cite|improve this answer











                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            We want to find factors of



                            begin{align}(2^{24}-1)&=(2^{12}-1)(2^{12}+1)\
                            &= (2^6-1)(2^6+1)(2^{12}+1)
                            end{align}



                            Hence we can for example, pick our $n$ to be $2^6-1, 2^6+1$ or $2^{12}+1$. Notice that $241$ is a factor of $2^{12}+1$.



                            Also note that $$2^{24}=(2^3)^8=(7+1)^8$$



                            Hence we can pick $n=7$.



                            Also $$2^{24}=(2^8)^3=((2^8-1)+1)^3$$



                            Hence, we can pick $n=2^8-1$ or its factors and so on.






                            share|cite|improve this answer











                            $endgroup$



                            We want to find factors of



                            begin{align}(2^{24}-1)&=(2^{12}-1)(2^{12}+1)\
                            &= (2^6-1)(2^6+1)(2^{12}+1)
                            end{align}



                            Hence we can for example, pick our $n$ to be $2^6-1, 2^6+1$ or $2^{12}+1$. Notice that $241$ is a factor of $2^{12}+1$.



                            Also note that $$2^{24}=(2^3)^8=(7+1)^8$$



                            Hence we can pick $n=7$.



                            Also $$2^{24}=(2^8)^3=((2^8-1)+1)^3$$



                            Hence, we can pick $n=2^8-1$ or its factors and so on.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 28 '18 at 17:22

























                            answered Nov 28 '18 at 17:14









                            Siong Thye GohSiong Thye Goh

                            101k1466117




                            101k1466117























                                0












                                $begingroup$

                                the simplest is to find an odd prime $p$ such that $p-1$ is a divisor of $24$. You have at once:
                                $$p= 3, 5, 7,13.$$
                                You also have chances it works with a prime $p$ such that a divisor of $24$ ($>1$) is also a divisor of $p-1$. For instance $17$ works ($2$ has order $8$ modulo $17$).






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  the simplest is to find an odd prime $p$ such that $p-1$ is a divisor of $24$. You have at once:
                                  $$p= 3, 5, 7,13.$$
                                  You also have chances it works with a prime $p$ such that a divisor of $24$ ($>1$) is also a divisor of $p-1$. For instance $17$ works ($2$ has order $8$ modulo $17$).






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    the simplest is to find an odd prime $p$ such that $p-1$ is a divisor of $24$. You have at once:
                                    $$p= 3, 5, 7,13.$$
                                    You also have chances it works with a prime $p$ such that a divisor of $24$ ($>1$) is also a divisor of $p-1$. For instance $17$ works ($2$ has order $8$ modulo $17$).






                                    share|cite|improve this answer









                                    $endgroup$



                                    the simplest is to find an odd prime $p$ such that $p-1$ is a divisor of $24$. You have at once:
                                    $$p= 3, 5, 7,13.$$
                                    You also have chances it works with a prime $p$ such that a divisor of $24$ ($>1$) is also a divisor of $p-1$. For instance $17$ works ($2$ has order $8$ modulo $17$).







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Nov 28 '18 at 17:39









                                    BernardBernard

                                    120k740113




                                    120k740113






























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