about modular congruencecongruence proofCongruence Classes and the Chinese Remainder TheoremModulo and CongruenceQuestion about galois imaginary and modular arithmeticSimple congruence relation (modular arithmetic)Inverse modular arithmetic methodQuestion about repeated modular exponentiationsWhen can we take square root of one side in modular arithmetic?Smallest positive integer modular congruence problemConjecture about the system of powerful equations

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about modular congruence


congruence proofCongruence Classes and the Chinese Remainder TheoremModulo and CongruenceQuestion about galois imaginary and modular arithmeticSimple congruence relation (modular arithmetic)Inverse modular arithmetic methodQuestion about repeated modular exponentiationsWhen can we take square root of one side in modular arithmetic?Smallest positive integer modular congruence problemConjecture about the system of powerful equations













0












$begingroup$


Consider $p$ a prime and $(1+palpha)$ then clearly taking congruences mod $p^m$
$$
prod_i=0^p^m-1(1+palpha)=(1+palpha)^p^m equiv 1+(palpha)^p^m(equiv 1 ???)
$$

What we can say if we take more generally:



$$
prod_i=0^p^m-1(1+palpha_i)
$$
is $equiv 1$ mod $p^m$ ? ( $alpha_i$ in general for different $i$ are different integers)
thanks for clarifications!










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    Consider $p$ a prime and $(1+palpha)$ then clearly taking congruences mod $p^m$
    $$
    prod_i=0^p^m-1(1+palpha)=(1+palpha)^p^m equiv 1+(palpha)^p^m(equiv 1 ???)
    $$

    What we can say if we take more generally:



    $$
    prod_i=0^p^m-1(1+palpha_i)
    $$
    is $equiv 1$ mod $p^m$ ? ( $alpha_i$ in general for different $i$ are different integers)
    thanks for clarifications!










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      Consider $p$ a prime and $(1+palpha)$ then clearly taking congruences mod $p^m$
      $$
      prod_i=0^p^m-1(1+palpha)=(1+palpha)^p^m equiv 1+(palpha)^p^m(equiv 1 ???)
      $$

      What we can say if we take more generally:



      $$
      prod_i=0^p^m-1(1+palpha_i)
      $$
      is $equiv 1$ mod $p^m$ ? ( $alpha_i$ in general for different $i$ are different integers)
      thanks for clarifications!










      share|cite|improve this question











      $endgroup$




      Consider $p$ a prime and $(1+palpha)$ then clearly taking congruences mod $p^m$
      $$
      prod_i=0^p^m-1(1+palpha)=(1+palpha)^p^m equiv 1+(palpha)^p^m(equiv 1 ???)
      $$

      What we can say if we take more generally:



      $$
      prod_i=0^p^m-1(1+palpha_i)
      $$
      is $equiv 1$ mod $p^m$ ? ( $alpha_i$ in general for different $i$ are different integers)
      thanks for clarifications!







      modular-arithmetic arithmetic






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 13 at 5:28









      Jyrki Lahtonen

      110k13171385




      110k13171385










      asked Mar 12 at 11:29









      andresandres

      2439




      2439




















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