What is the power of a modular exponent expression? [closed]












0












$begingroup$


Is there any property to simplify expressions of the following type:
$(a^{x} mod N)^y$

The usual mathematical property $(a^{b})^c = a^{bc}$ doesn't seem to hold.






modular-arithmetic exponentiation







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closed as off-topic by mrtaurho, Ali Caglayan, Adrian Keister, clathratus, Andrew Jan 28 at 18:08


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Ali Caglayan, Adrian Keister, clathratus, Andrew

If this question can be reworded to fit the rules in the help center, please edit the question.





















    0












    $begingroup$


    Is there any property to simplify expressions of the following type:
    $(a^{x} mod N)^y$

    The usual mathematical property $(a^{b})^c = a^{bc}$ doesn't seem to hold.






    modular-arithmetic exponentiation







    share|cite|improve this question









    $endgroup$



    closed as off-topic by mrtaurho, Ali Caglayan, Adrian Keister, clathratus, Andrew Jan 28 at 18:08


    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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Ali Caglayan, Adrian Keister, clathratus, Andrew

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      0












      0








      0


      1



      $begingroup$


      Is there any property to simplify expressions of the following type:
      $(a^{x} mod N)^y$

      The usual mathematical property $(a^{b})^c = a^{bc}$ doesn't seem to hold.






      modular-arithmetic exponentiation







      share|cite|improve this question









      $endgroup$




      Is there any property to simplify expressions of the following type:
      $(a^{x} mod N)^y$

      The usual mathematical property $(a^{b})^c = a^{bc}$ doesn't seem to hold.






      modular-arithmetic exponentiation




      modular-arithmetic exponentiation






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 28 at 10:12









      Tabish MirTabish Mir

      12617




      12617




      closed as off-topic by mrtaurho, Ali Caglayan, Adrian Keister, clathratus, Andrew Jan 28 at 18:08


      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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Ali Caglayan, Adrian Keister, clathratus, Andrew

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by mrtaurho, Ali Caglayan, Adrian Keister, clathratus, Andrew Jan 28 at 18:08


      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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Ali Caglayan, Adrian Keister, clathratus, Andrew

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
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          $begingroup$

          I don't see any way to simplify this expression. It is the same as:



          $$
          left(a^x-left lfloorfrac{a^x}{N}right rfloor cdot N right)^y
          $$



          where $lfloor cdot rfloor$ is the floor function, and can be calculated using the generalized binomial theorem. But, clearly, this is not a simplification.






          share|cite|improve this answer









          $endgroup$




















            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            I don't see any way to simplify this expression. It is the same as:



            $$
            left(a^x-left lfloorfrac{a^x}{N}right rfloor cdot N right)^y
            $$



            where $lfloor cdot rfloor$ is the floor function, and can be calculated using the generalized binomial theorem. But, clearly, this is not a simplification.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              I don't see any way to simplify this expression. It is the same as:



              $$
              left(a^x-left lfloorfrac{a^x}{N}right rfloor cdot N right)^y
              $$



              where $lfloor cdot rfloor$ is the floor function, and can be calculated using the generalized binomial theorem. But, clearly, this is not a simplification.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                I don't see any way to simplify this expression. It is the same as:



                $$
                left(a^x-left lfloorfrac{a^x}{N}right rfloor cdot N right)^y
                $$



                where $lfloor cdot rfloor$ is the floor function, and can be calculated using the generalized binomial theorem. But, clearly, this is not a simplification.






                share|cite|improve this answer









                $endgroup$



                I don't see any way to simplify this expression. It is the same as:



                $$
                left(a^x-left lfloorfrac{a^x}{N}right rfloor cdot N right)^y
                $$



                where $lfloor cdot rfloor$ is the floor function, and can be calculated using the generalized binomial theorem. But, clearly, this is not a simplification.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 28 at 10:40









                Emilio NovatiEmilio Novati

                52.2k43474




                52.2k43474















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