Elementary Number Theory: Congruences












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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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    3















    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



























      3












      3








      3


      2






      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
















      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 2 hours ago









      greedoid

      38.2k114797




      38.2k114797










      asked 2 hours ago









      prashant sharma

      726




      726






















          3 Answers
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          active

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          2














          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 $u,v in mathbb Z$.






          share|cite|improve this answer





















          • @Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
            – prashant sharma
            2 hours ago










          • @prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
            – lhf
            2 hours ago





















          1














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






          share|cite|improve this answer





















          • I don't understand this note. It is more like a comment, not a solution.
            – greedoid
            2 hours ago










          • "Can we replace 11 by any other integer" is what i have answered
            – crskhr
            2 hours ago





















          1














          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





















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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            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 $u,v in mathbb Z$.






            share|cite|improve this answer





















            • @Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
              – prashant sharma
              2 hours ago










            • @prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
              – lhf
              2 hours ago


















            2














            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 $u,v in mathbb Z$.






            share|cite|improve this answer





















            • @Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
              – prashant sharma
              2 hours ago










            • @prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
              – lhf
              2 hours ago
















            2












            2








            2






            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 $u,v in mathbb Z$.






            share|cite|improve this answer












            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 $u,v in mathbb Z$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            lhf

            163k10167387




            163k10167387












            • @Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
              – prashant sharma
              2 hours ago










            • @prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
              – lhf
              2 hours ago




















            • @Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
              – prashant sharma
              2 hours ago










            • @prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
              – lhf
              2 hours ago


















            @Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
            – prashant sharma
            2 hours ago




            @Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
            – prashant sharma
            2 hours ago












            @prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
            – lhf
            2 hours ago






            @prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
            – lhf
            2 hours ago













            1














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






            share|cite|improve this answer





















            • I don't understand this note. It is more like a comment, not a solution.
              – greedoid
              2 hours ago










            • "Can we replace 11 by any other integer" is what i have answered
              – crskhr
              2 hours ago


















            1














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






            share|cite|improve this answer





















            • I don't understand this note. It is more like a comment, not a solution.
              – greedoid
              2 hours ago










            • "Can we replace 11 by any other integer" is what i have answered
              – crskhr
              2 hours ago
















            1












            1








            1






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






            share|cite|improve this answer












            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 2 hours ago









            crskhr

            3,875925




            3,875925












            • I don't understand this note. It is more like a comment, not a solution.
              – greedoid
              2 hours ago










            • "Can we replace 11 by any other integer" is what i have answered
              – crskhr
              2 hours ago




















            • I don't understand this note. It is more like a comment, not a solution.
              – greedoid
              2 hours ago










            • "Can we replace 11 by any other integer" is what i have answered
              – crskhr
              2 hours ago


















            I don't understand this note. It is more like a comment, not a solution.
            – greedoid
            2 hours ago




            I don't understand this note. It is more like a comment, not a solution.
            – greedoid
            2 hours ago












            "Can we replace 11 by any other integer" is what i have answered
            – crskhr
            2 hours ago






            "Can we replace 11 by any other integer" is what i have answered
            – crskhr
            2 hours ago













            1














            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


























              1














              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
























                1












                1








                1






                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












                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 2 hours ago









                greedoid

                38.2k114797




                38.2k114797






























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