$a equiv b$ (mod $p$) implies $a^{p^n} equiv b^{p^n}$ (mod $p^n$)?












1












$begingroup$


Let $p$ be a prime number. If $a equiv b$ (mod $p$), does that imply $a^{p^n} equiv b^{p^n}$ (mod $p^n$)?



I think the answer will be yes, and I suspect that the way of proving it will involve writing $a^{p^n}-b^{p^n}$ as a multiple of $(a-b)^n$.



I also noticed that $n<p^n$, and I'm wondering if this will make the proof easier or not.






prime-numbers modular-arithmetic cyclotomic-polynomials







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $p$ be a prime number. If $a equiv b$ (mod $p$), does that imply $a^{p^n} equiv b^{p^n}$ (mod $p^n$)?



    I think the answer will be yes, and I suspect that the way of proving it will involve writing $a^{p^n}-b^{p^n}$ as a multiple of $(a-b)^n$.



    I also noticed that $n<p^n$, and I'm wondering if this will make the proof easier or not.






    prime-numbers modular-arithmetic cyclotomic-polynomials







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $p$ be a prime number. If $a equiv b$ (mod $p$), does that imply $a^{p^n} equiv b^{p^n}$ (mod $p^n$)?



      I think the answer will be yes, and I suspect that the way of proving it will involve writing $a^{p^n}-b^{p^n}$ as a multiple of $(a-b)^n$.



      I also noticed that $n<p^n$, and I'm wondering if this will make the proof easier or not.






      prime-numbers modular-arithmetic cyclotomic-polynomials







      share|cite|improve this question









      $endgroup$




      Let $p$ be a prime number. If $a equiv b$ (mod $p$), does that imply $a^{p^n} equiv b^{p^n}$ (mod $p^n$)?



      I think the answer will be yes, and I suspect that the way of proving it will involve writing $a^{p^n}-b^{p^n}$ as a multiple of $(a-b)^n$.



      I also noticed that $n<p^n$, and I'm wondering if this will make the proof easier or not.






      prime-numbers modular-arithmetic cyclotomic-polynomials




      prime-numbers modular-arithmetic cyclotomic-polynomials






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 8 '18 at 18:19









      Pascal's WagerPascal's Wager

      357315




      357315






















          2 Answers
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          4












          $begingroup$

          $a=b+kp$ where $k$ is an integer



          $(b+kp)^{p^n}=b^{p^n}+binom{p^n}1b^{p^n-1}kp+binom{p^n}2b^{p^n-2}(kp)^{2}+cdots+(kp)^{p^n}$



          $equiv b^{p^n}pmod{p^{n+1}}$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So simple yet so useful! Thank you!
            $endgroup$
            – Pascal's Wager
            Nov 8 '18 at 23:45



















          0












          $begingroup$

          There is following theorem (Lifting The Exponent Lemma).
          We will use notation $p^{alpha}||n$ for positive integer $n$, nonnegative integer $alpha$ and prime $p$, which equivalent to $p^{alpha}|n$ and $p^{alpha+1}nmid n$.



          Theorem.
          Let $a,b,n$ be a positive integers and $p^{alpha}||a-b$, $p^{beta}||n$. Then:




          • if $p>2$ and $alphageq 1$ then $p^{alpha+beta}||a^n-b^n$;

          • if $p=2$ and $alphageq 2$ then $2^{alpha+beta}||a^n-b^n$.


          Note. If $p=2$ and $alpha=1$ then $2^{beta+1}||a^n-b^n$.



          Your statement is a consequence of this theorem.



          Some links about LTE lemma:



          What can I do with the lifting the exponent lemma?



          https://brilliant.org/wiki/lifting-the-exponent/






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
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            4












            $begingroup$

            $a=b+kp$ where $k$ is an integer



            $(b+kp)^{p^n}=b^{p^n}+binom{p^n}1b^{p^n-1}kp+binom{p^n}2b^{p^n-2}(kp)^{2}+cdots+(kp)^{p^n}$



            $equiv b^{p^n}pmod{p^{n+1}}$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So simple yet so useful! Thank you!
              $endgroup$
              – Pascal's Wager
              Nov 8 '18 at 23:45
















            4












            $begingroup$

            $a=b+kp$ where $k$ is an integer



            $(b+kp)^{p^n}=b^{p^n}+binom{p^n}1b^{p^n-1}kp+binom{p^n}2b^{p^n-2}(kp)^{2}+cdots+(kp)^{p^n}$



            $equiv b^{p^n}pmod{p^{n+1}}$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So simple yet so useful! Thank you!
              $endgroup$
              – Pascal's Wager
              Nov 8 '18 at 23:45














            4












            4








            4





            $begingroup$

            $a=b+kp$ where $k$ is an integer



            $(b+kp)^{p^n}=b^{p^n}+binom{p^n}1b^{p^n-1}kp+binom{p^n}2b^{p^n-2}(kp)^{2}+cdots+(kp)^{p^n}$



            $equiv b^{p^n}pmod{p^{n+1}}$






            share|cite|improve this answer









            $endgroup$



            $a=b+kp$ where $k$ is an integer



            $(b+kp)^{p^n}=b^{p^n}+binom{p^n}1b^{p^n-1}kp+binom{p^n}2b^{p^n-2}(kp)^{2}+cdots+(kp)^{p^n}$



            $equiv b^{p^n}pmod{p^{n+1}}$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 8 '18 at 18:28









            lab bhattacharjeelab bhattacharjee

            225k15157275




            225k15157275












            • $begingroup$
              So simple yet so useful! Thank you!
              $endgroup$
              – Pascal's Wager
              Nov 8 '18 at 23:45


















            • $begingroup$
              So simple yet so useful! Thank you!
              $endgroup$
              – Pascal's Wager
              Nov 8 '18 at 23:45
















            $begingroup$
            So simple yet so useful! Thank you!
            $endgroup$
            – Pascal's Wager
            Nov 8 '18 at 23:45




            $begingroup$
            So simple yet so useful! Thank you!
            $endgroup$
            – Pascal's Wager
            Nov 8 '18 at 23:45











            0












            $begingroup$

            There is following theorem (Lifting The Exponent Lemma).
            We will use notation $p^{alpha}||n$ for positive integer $n$, nonnegative integer $alpha$ and prime $p$, which equivalent to $p^{alpha}|n$ and $p^{alpha+1}nmid n$.



            Theorem.
            Let $a,b,n$ be a positive integers and $p^{alpha}||a-b$, $p^{beta}||n$. Then:




            • if $p>2$ and $alphageq 1$ then $p^{alpha+beta}||a^n-b^n$;

            • if $p=2$ and $alphageq 2$ then $2^{alpha+beta}||a^n-b^n$.


            Note. If $p=2$ and $alpha=1$ then $2^{beta+1}||a^n-b^n$.



            Your statement is a consequence of this theorem.



            Some links about LTE lemma:



            What can I do with the lifting the exponent lemma?



            https://brilliant.org/wiki/lifting-the-exponent/






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              There is following theorem (Lifting The Exponent Lemma).
              We will use notation $p^{alpha}||n$ for positive integer $n$, nonnegative integer $alpha$ and prime $p$, which equivalent to $p^{alpha}|n$ and $p^{alpha+1}nmid n$.



              Theorem.
              Let $a,b,n$ be a positive integers and $p^{alpha}||a-b$, $p^{beta}||n$. Then:




              • if $p>2$ and $alphageq 1$ then $p^{alpha+beta}||a^n-b^n$;

              • if $p=2$ and $alphageq 2$ then $2^{alpha+beta}||a^n-b^n$.


              Note. If $p=2$ and $alpha=1$ then $2^{beta+1}||a^n-b^n$.



              Your statement is a consequence of this theorem.



              Some links about LTE lemma:



              What can I do with the lifting the exponent lemma?



              https://brilliant.org/wiki/lifting-the-exponent/






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                There is following theorem (Lifting The Exponent Lemma).
                We will use notation $p^{alpha}||n$ for positive integer $n$, nonnegative integer $alpha$ and prime $p$, which equivalent to $p^{alpha}|n$ and $p^{alpha+1}nmid n$.



                Theorem.
                Let $a,b,n$ be a positive integers and $p^{alpha}||a-b$, $p^{beta}||n$. Then:




                • if $p>2$ and $alphageq 1$ then $p^{alpha+beta}||a^n-b^n$;

                • if $p=2$ and $alphageq 2$ then $2^{alpha+beta}||a^n-b^n$.


                Note. If $p=2$ and $alpha=1$ then $2^{beta+1}||a^n-b^n$.



                Your statement is a consequence of this theorem.



                Some links about LTE lemma:



                What can I do with the lifting the exponent lemma?



                https://brilliant.org/wiki/lifting-the-exponent/






                share|cite|improve this answer









                $endgroup$



                There is following theorem (Lifting The Exponent Lemma).
                We will use notation $p^{alpha}||n$ for positive integer $n$, nonnegative integer $alpha$ and prime $p$, which equivalent to $p^{alpha}|n$ and $p^{alpha+1}nmid n$.



                Theorem.
                Let $a,b,n$ be a positive integers and $p^{alpha}||a-b$, $p^{beta}||n$. Then:




                • if $p>2$ and $alphageq 1$ then $p^{alpha+beta}||a^n-b^n$;

                • if $p=2$ and $alphageq 2$ then $2^{alpha+beta}||a^n-b^n$.


                Note. If $p=2$ and $alpha=1$ then $2^{beta+1}||a^n-b^n$.



                Your statement is a consequence of this theorem.



                Some links about LTE lemma:



                What can I do with the lifting the exponent lemma?



                https://brilliant.org/wiki/lifting-the-exponent/







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 12 at 15:00









                richrowrichrow

                17316




                17316






























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