If $a^3 equiv b^3pmod{11}$ then $a equiv b pmod{11}$. [closed]












2












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If $a^3 equiv b^3pmod{11}$ then $a equiv b pmod{11}$.




I verified the above statement for smaller values of $a$ and $b$ but I am unable to prove it in general. Please prove or disprove it. Can we replace $11$ by any other integer?






elementary-number-theory modular-arithmetic







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closed as off-topic by user21820, Did, amWhy, Lee David Chung Lin, KReiser Jan 8 at 1:36


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." – user21820, Did, amWhy, Lee David Chung Lin, KReiser

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


















    2












    $begingroup$



    If $a^3 equiv b^3pmod{11}$ then $a equiv b pmod{11}$.




    I verified the above statement for smaller values of $a$ and $b$ but I am unable to prove it in general. Please prove or disprove it. Can we replace $11$ by any other integer?






    elementary-number-theory modular-arithmetic







    share|cite|improve this question











    $endgroup$



    closed as off-topic by user21820, Did, amWhy, Lee David Chung Lin, KReiser Jan 8 at 1:36


    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." – user21820, Did, amWhy, Lee David Chung Lin, KReiser

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
















      2












      2








      2


      2



      $begingroup$



      If $a^3 equiv b^3pmod{11}$ then $a equiv b pmod{11}$.




      I verified the above statement for smaller values of $a$ and $b$ but I am unable to prove it in general. Please prove or disprove it. Can we replace $11$ by any other integer?






      elementary-number-theory modular-arithmetic







      share|cite|improve this question











      $endgroup$





      If $a^3 equiv b^3pmod{11}$ then $a equiv b pmod{11}$.




      I verified the above statement for smaller values of $a$ and $b$ but I am unable to prove it in general. Please prove or disprove it. Can we replace $11$ by any other integer?






      elementary-number-theory modular-arithmetic




      elementary-number-theory modular-arithmetic






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 7 at 14:12









      user21820

      38.8k543153




      38.8k543153










      asked Jan 7 at 10:07









      prashant sharmaprashant sharma

      756




      756




      closed as off-topic by user21820, Did, amWhy, Lee David Chung Lin, KReiser Jan 8 at 1:36


      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." – user21820, Did, amWhy, Lee David Chung Lin, KReiser

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




      closed as off-topic by user21820, Did, amWhy, Lee David Chung Lin, KReiser Jan 8 at 1:36


      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." – user21820, Did, amWhy, Lee David Chung Lin, KReiser

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






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          Here is a general statement:




          If $p$ is a prime and $m$ is coprime with $p-1$, then $a^m equiv b^m bmod p$ iff $a equiv bbmod p$.




          This follows from Fermat's theorem by writing $(p-1)(-u)+mv=1$ for $u,v in mathbb N$.



          In your example, $p=11$ and $m=3$. We can take $u=2$ and $v=7$ to get $10cdot (-2) + 3cdot(7)=1$. Write this as $1 + 10 cdot 2 = 3cdot 7 $. Then
          $$
          a equiv a cdot a^{10 cdot 2} = a^{1 + 10 cdot 2} = a^{3cdot7}
          equiv
          b^{3cdot7} = b^{1 + 10 cdot 2} = b cdot b^{10 cdot 2} equiv b
          bmod 11
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
            $endgroup$
            – prashant sharma
            Jan 7 at 10:33










          • $begingroup$
            @prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
            $endgroup$
            – lhf
            Jan 7 at 10:35












          • $begingroup$
            @Ihf Let's take $m = 3$, $p = 11$. Then one possible ordered pair of $(u, v)$ is $(1, -3)$. So we have $a^{10} equiv b^{10}$(mod $11$) and $a^{-9} equiv b^{-9}$(mod $11$) and we multiply both congruences then we get $a equiv b$(mod $11$). But second congruence is meaning less as exponents are negative.
            $endgroup$
            – prashant sharma
            Jan 7 at 16:13










          • $begingroup$
            @prashantsharma, negative exponents do make sense but anyway see my edited answer.
            $endgroup$
            – lhf
            Jan 7 at 22:58





















          1












          $begingroup$

          If $11mid ab$ we are done.



          Say $11not{mid} ab$. Then by Fermat we have $$a^{10}equiv b^{10}equiv 1pmod {11}$$



          Since $$a^{9}equiv b^{9}pmod {11}$$ we have $$a^{10}equiv ba^{9}pmod {11}$$



          so $$ 11mid a^9(a-b)implies 11mid a-b$$



          and we are done.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Note $$1^{3} equiv 2^{3}pmod{7}$$ doesnt imply $1equiv 2pmod{7}$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I don't understand this note. It is more like a comment, not a solution.
              $endgroup$
              – greedoid
              Jan 7 at 10:31










            • $begingroup$
              "Can we replace 11 by any other integer" is what i have answered
              $endgroup$
              – crskhr
              Jan 7 at 10:33




















            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            Here is a general statement:




            If $p$ is a prime and $m$ is coprime with $p-1$, then $a^m equiv b^m bmod p$ iff $a equiv bbmod p$.




            This follows from Fermat's theorem by writing $(p-1)(-u)+mv=1$ for $u,v in mathbb N$.



            In your example, $p=11$ and $m=3$. We can take $u=2$ and $v=7$ to get $10cdot (-2) + 3cdot(7)=1$. Write this as $1 + 10 cdot 2 = 3cdot 7 $. Then
            $$
            a equiv a cdot a^{10 cdot 2} = a^{1 + 10 cdot 2} = a^{3cdot7}
            equiv
            b^{3cdot7} = b^{1 + 10 cdot 2} = b cdot b^{10 cdot 2} equiv b
            bmod 11
            $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
              $endgroup$
              – prashant sharma
              Jan 7 at 10:33










            • $begingroup$
              @prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
              $endgroup$
              – lhf
              Jan 7 at 10:35












            • $begingroup$
              @Ihf Let's take $m = 3$, $p = 11$. Then one possible ordered pair of $(u, v)$ is $(1, -3)$. So we have $a^{10} equiv b^{10}$(mod $11$) and $a^{-9} equiv b^{-9}$(mod $11$) and we multiply both congruences then we get $a equiv b$(mod $11$). But second congruence is meaning less as exponents are negative.
              $endgroup$
              – prashant sharma
              Jan 7 at 16:13










            • $begingroup$
              @prashantsharma, negative exponents do make sense but anyway see my edited answer.
              $endgroup$
              – lhf
              Jan 7 at 22:58


















            4












            $begingroup$

            Here is a general statement:




            If $p$ is a prime and $m$ is coprime with $p-1$, then $a^m equiv b^m bmod p$ iff $a equiv bbmod p$.




            This follows from Fermat's theorem by writing $(p-1)(-u)+mv=1$ for $u,v in mathbb N$.



            In your example, $p=11$ and $m=3$. We can take $u=2$ and $v=7$ to get $10cdot (-2) + 3cdot(7)=1$. Write this as $1 + 10 cdot 2 = 3cdot 7 $. Then
            $$
            a equiv a cdot a^{10 cdot 2} = a^{1 + 10 cdot 2} = a^{3cdot7}
            equiv
            b^{3cdot7} = b^{1 + 10 cdot 2} = b cdot b^{10 cdot 2} equiv b
            bmod 11
            $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
              $endgroup$
              – prashant sharma
              Jan 7 at 10:33










            • $begingroup$
              @prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
              $endgroup$
              – lhf
              Jan 7 at 10:35












            • $begingroup$
              @Ihf Let's take $m = 3$, $p = 11$. Then one possible ordered pair of $(u, v)$ is $(1, -3)$. So we have $a^{10} equiv b^{10}$(mod $11$) and $a^{-9} equiv b^{-9}$(mod $11$) and we multiply both congruences then we get $a equiv b$(mod $11$). But second congruence is meaning less as exponents are negative.
              $endgroup$
              – prashant sharma
              Jan 7 at 16:13










            • $begingroup$
              @prashantsharma, negative exponents do make sense but anyway see my edited answer.
              $endgroup$
              – lhf
              Jan 7 at 22:58
















            4












            4








            4





            $begingroup$

            Here is a general statement:




            If $p$ is a prime and $m$ is coprime with $p-1$, then $a^m equiv b^m bmod p$ iff $a equiv bbmod p$.




            This follows from Fermat's theorem by writing $(p-1)(-u)+mv=1$ for $u,v in mathbb N$.



            In your example, $p=11$ and $m=3$. We can take $u=2$ and $v=7$ to get $10cdot (-2) + 3cdot(7)=1$. Write this as $1 + 10 cdot 2 = 3cdot 7 $. Then
            $$
            a equiv a cdot a^{10 cdot 2} = a^{1 + 10 cdot 2} = a^{3cdot7}
            equiv
            b^{3cdot7} = b^{1 + 10 cdot 2} = b cdot b^{10 cdot 2} equiv b
            bmod 11
            $$






            share|cite|improve this answer











            $endgroup$



            Here is a general statement:




            If $p$ is a prime and $m$ is coprime with $p-1$, then $a^m equiv b^m bmod p$ iff $a equiv bbmod p$.




            This follows from Fermat's theorem by writing $(p-1)(-u)+mv=1$ for $u,v in mathbb N$.



            In your example, $p=11$ and $m=3$. We can take $u=2$ and $v=7$ to get $10cdot (-2) + 3cdot(7)=1$. Write this as $1 + 10 cdot 2 = 3cdot 7 $. Then
            $$
            a equiv a cdot a^{10 cdot 2} = a^{1 + 10 cdot 2} = a^{3cdot7}
            equiv
            b^{3cdot7} = b^{1 + 10 cdot 2} = b cdot b^{10 cdot 2} equiv b
            bmod 11
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 7 at 22:58

























            answered Jan 7 at 10:27









            lhflhf

            163k10169393




            163k10169393












            • $begingroup$
              @Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
              $endgroup$
              – prashant sharma
              Jan 7 at 10:33










            • $begingroup$
              @prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
              $endgroup$
              – lhf
              Jan 7 at 10:35












            • $begingroup$
              @Ihf Let's take $m = 3$, $p = 11$. Then one possible ordered pair of $(u, v)$ is $(1, -3)$. So we have $a^{10} equiv b^{10}$(mod $11$) and $a^{-9} equiv b^{-9}$(mod $11$) and we multiply both congruences then we get $a equiv b$(mod $11$). But second congruence is meaning less as exponents are negative.
              $endgroup$
              – prashant sharma
              Jan 7 at 16:13










            • $begingroup$
              @prashantsharma, negative exponents do make sense but anyway see my edited answer.
              $endgroup$
              – lhf
              Jan 7 at 22:58




















            • $begingroup$
              @Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
              $endgroup$
              – prashant sharma
              Jan 7 at 10:33










            • $begingroup$
              @prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
              $endgroup$
              – lhf
              Jan 7 at 10:35












            • $begingroup$
              @Ihf Let's take $m = 3$, $p = 11$. Then one possible ordered pair of $(u, v)$ is $(1, -3)$. So we have $a^{10} equiv b^{10}$(mod $11$) and $a^{-9} equiv b^{-9}$(mod $11$) and we multiply both congruences then we get $a equiv b$(mod $11$). But second congruence is meaning less as exponents are negative.
              $endgroup$
              – prashant sharma
              Jan 7 at 16:13










            • $begingroup$
              @prashantsharma, negative exponents do make sense but anyway see my edited answer.
              $endgroup$
              – lhf
              Jan 7 at 22:58


















            $begingroup$
            @Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
            $endgroup$
            – prashant sharma
            Jan 7 at 10:33




            $begingroup$
            @Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
            $endgroup$
            – prashant sharma
            Jan 7 at 10:33












            $begingroup$
            @prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
            $endgroup$
            – lhf
            Jan 7 at 10:35






            $begingroup$
            @prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
            $endgroup$
            – lhf
            Jan 7 at 10:35














            $begingroup$
            @Ihf Let's take $m = 3$, $p = 11$. Then one possible ordered pair of $(u, v)$ is $(1, -3)$. So we have $a^{10} equiv b^{10}$(mod $11$) and $a^{-9} equiv b^{-9}$(mod $11$) and we multiply both congruences then we get $a equiv b$(mod $11$). But second congruence is meaning less as exponents are negative.
            $endgroup$
            – prashant sharma
            Jan 7 at 16:13




            $begingroup$
            @Ihf Let's take $m = 3$, $p = 11$. Then one possible ordered pair of $(u, v)$ is $(1, -3)$. So we have $a^{10} equiv b^{10}$(mod $11$) and $a^{-9} equiv b^{-9}$(mod $11$) and we multiply both congruences then we get $a equiv b$(mod $11$). But second congruence is meaning less as exponents are negative.
            $endgroup$
            – prashant sharma
            Jan 7 at 16:13












            $begingroup$
            @prashantsharma, negative exponents do make sense but anyway see my edited answer.
            $endgroup$
            – lhf
            Jan 7 at 22:58






            $begingroup$
            @prashantsharma, negative exponents do make sense but anyway see my edited answer.
            $endgroup$
            – lhf
            Jan 7 at 22:58













            1












            $begingroup$

            If $11mid ab$ we are done.



            Say $11not{mid} ab$. Then by Fermat we have $$a^{10}equiv b^{10}equiv 1pmod {11}$$



            Since $$a^{9}equiv b^{9}pmod {11}$$ we have $$a^{10}equiv ba^{9}pmod {11}$$



            so $$ 11mid a^9(a-b)implies 11mid a-b$$



            and we are done.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              If $11mid ab$ we are done.



              Say $11not{mid} ab$. Then by Fermat we have $$a^{10}equiv b^{10}equiv 1pmod {11}$$



              Since $$a^{9}equiv b^{9}pmod {11}$$ we have $$a^{10}equiv ba^{9}pmod {11}$$



              so $$ 11mid a^9(a-b)implies 11mid a-b$$



              and we are done.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                If $11mid ab$ we are done.



                Say $11not{mid} ab$. Then by Fermat we have $$a^{10}equiv b^{10}equiv 1pmod {11}$$



                Since $$a^{9}equiv b^{9}pmod {11}$$ we have $$a^{10}equiv ba^{9}pmod {11}$$



                so $$ 11mid a^9(a-b)implies 11mid a-b$$



                and we are done.






                share|cite|improve this answer









                $endgroup$



                If $11mid ab$ we are done.



                Say $11not{mid} ab$. Then by Fermat we have $$a^{10}equiv b^{10}equiv 1pmod {11}$$



                Since $$a^{9}equiv b^{9}pmod {11}$$ we have $$a^{10}equiv ba^{9}pmod {11}$$



                so $$ 11mid a^9(a-b)implies 11mid a-b$$



                and we are done.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 7 at 10:25









                greedoidgreedoid

                39.2k114797




                39.2k114797























                    0












                    $begingroup$

                    Note $$1^{3} equiv 2^{3}pmod{7}$$ doesnt imply $1equiv 2pmod{7}$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      I don't understand this note. It is more like a comment, not a solution.
                      $endgroup$
                      – greedoid
                      Jan 7 at 10:31










                    • $begingroup$
                      "Can we replace 11 by any other integer" is what i have answered
                      $endgroup$
                      – crskhr
                      Jan 7 at 10:33


















                    0












                    $begingroup$

                    Note $$1^{3} equiv 2^{3}pmod{7}$$ doesnt imply $1equiv 2pmod{7}$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      I don't understand this note. It is more like a comment, not a solution.
                      $endgroup$
                      – greedoid
                      Jan 7 at 10:31










                    • $begingroup$
                      "Can we replace 11 by any other integer" is what i have answered
                      $endgroup$
                      – crskhr
                      Jan 7 at 10:33
















                    0












                    0








                    0





                    $begingroup$

                    Note $$1^{3} equiv 2^{3}pmod{7}$$ doesnt imply $1equiv 2pmod{7}$






                    share|cite|improve this answer









                    $endgroup$



                    Note $$1^{3} equiv 2^{3}pmod{7}$$ doesnt imply $1equiv 2pmod{7}$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 7 at 10:13









                    crskhrcrskhr

                    3,873925




                    3,873925












                    • $begingroup$
                      I don't understand this note. It is more like a comment, not a solution.
                      $endgroup$
                      – greedoid
                      Jan 7 at 10:31










                    • $begingroup$
                      "Can we replace 11 by any other integer" is what i have answered
                      $endgroup$
                      – crskhr
                      Jan 7 at 10:33




















                    • $begingroup$
                      I don't understand this note. It is more like a comment, not a solution.
                      $endgroup$
                      – greedoid
                      Jan 7 at 10:31










                    • $begingroup$
                      "Can we replace 11 by any other integer" is what i have answered
                      $endgroup$
                      – crskhr
                      Jan 7 at 10:33


















                    $begingroup$
                    I don't understand this note. It is more like a comment, not a solution.
                    $endgroup$
                    – greedoid
                    Jan 7 at 10:31




                    $begingroup$
                    I don't understand this note. It is more like a comment, not a solution.
                    $endgroup$
                    – greedoid
                    Jan 7 at 10:31












                    $begingroup$
                    "Can we replace 11 by any other integer" is what i have answered
                    $endgroup$
                    – crskhr
                    Jan 7 at 10:33






                    $begingroup$
                    "Can we replace 11 by any other integer" is what i have answered
                    $endgroup$
                    – crskhr
                    Jan 7 at 10:33





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