Modular arithmetic power rule implications

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












Assume that $a equiv b text (mod m)$, the following:
$$a^nequiv b^n text (mod m)$$
is true.



However, does $a^nequiv b^n text (mod m)$ imply that $a equiv b text (mod m)$ is true when n is odd?



If n is odd, it seems like this is justifiable because negative numbers do not become positive numbers if raised to an odd power.



Also would it be true for when n is even if say $a^nequiv b^n text (mod m)$ imply $a equiv ±b text (mod m)$?




modular-arithmetic




share|cite|improve this question



















  • No, its is not true. For example, if $m=7$ and $n=3$, then $2^3equiv 1^3 pmod7$, but $2notequiv 1 pmod7$.
    – xarles
    Jul 29 at 6:52















up vote
0
down vote

favorite












Assume that $a equiv b text (mod m)$, the following:
$$a^nequiv b^n text (mod m)$$
is true.



However, does $a^nequiv b^n text (mod m)$ imply that $a equiv b text (mod m)$ is true when n is odd?



If n is odd, it seems like this is justifiable because negative numbers do not become positive numbers if raised to an odd power.



Also would it be true for when n is even if say $a^nequiv b^n text (mod m)$ imply $a equiv ±b text (mod m)$?




modular-arithmetic




share|cite|improve this question



















  • No, its is not true. For example, if $m=7$ and $n=3$, then $2^3equiv 1^3 pmod7$, but $2notequiv 1 pmod7$.
    – xarles
    Jul 29 at 6:52













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Assume that $a equiv b text (mod m)$, the following:
$$a^nequiv b^n text (mod m)$$
is true.



However, does $a^nequiv b^n text (mod m)$ imply that $a equiv b text (mod m)$ is true when n is odd?



If n is odd, it seems like this is justifiable because negative numbers do not become positive numbers if raised to an odd power.



Also would it be true for when n is even if say $a^nequiv b^n text (mod m)$ imply $a equiv ±b text (mod m)$?




modular-arithmetic




share|cite|improve this question











Assume that $a equiv b text (mod m)$, the following:
$$a^nequiv b^n text (mod m)$$
is true.



However, does $a^nequiv b^n text (mod m)$ imply that $a equiv b text (mod m)$ is true when n is odd?



If n is odd, it seems like this is justifiable because negative numbers do not become positive numbers if raised to an odd power.



Also would it be true for when n is even if say $a^nequiv b^n text (mod m)$ imply $a equiv ±b text (mod m)$?





modular-arithmetic





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 29 at 6:45









LHC2012

697




697











  • No, its is not true. For example, if $m=7$ and $n=3$, then $2^3equiv 1^3 pmod7$, but $2notequiv 1 pmod7$.
    – xarles
    Jul 29 at 6:52

















  • No, its is not true. For example, if $m=7$ and $n=3$, then $2^3equiv 1^3 pmod7$, but $2notequiv 1 pmod7$.
    – xarles
    Jul 29 at 6:52
















No, its is not true. For example, if $m=7$ and $n=3$, then $2^3equiv 1^3 pmod7$, but $2notequiv 1 pmod7$.
– xarles
Jul 29 at 6:52





No, its is not true. For example, if $m=7$ and $n=3$, then $2^3equiv 1^3 pmod7$, but $2notequiv 1 pmod7$.
– xarles
Jul 29 at 6:52











4 Answers
4






active

oldest

votes

















up vote
2
down vote



accepted










The above example by Lord Shark the Unknown indicates that no complete answer can be found. But sometimes, we do have that kind of higher cancellation:



We have, by telescopic sum,
$$
a^n - b^n = (a - b) sum_j=0^n-1 a^n-1-jb^j,
$$
so that $a^n equiv b^n mod m$ implies $a equiv b mod m$ if
$$
sum_j=0^n-1 a^n-1-jb^j
$$
is not a zero divisor.






share|cite|improve this answer




























    up vote
    0
    down vote













    $2^3equiv4^3pmod 7$, but $2notequiv4pmod7$.






    share|cite|improve this answer




























      up vote
      0
      down vote













      First, all you say about being positive or negative modulo $n$ makes really no sense, since $aequiv a-m pmodm$, so any positive number is equivalent to a negative number and viceversa.



      However, there is a natural condition to put on $n$ related to $m$, if $m$ is prime: if $n$ and $m-1$ are coprime (so they share no comon divisor) , then $a^nequiv b^n pmodm$ implies $a equiv b pmodm$.






      share|cite|improve this answer























      • Is there a reason they have to be coprime?
        – LHC2012
        Jul 29 at 7:08










      • See my posting for a counterexample to your final claim.
        – Lord Shark the Unknown
        Jul 29 at 7:11










      • Ups, I forgot to write the $varphi$ in front of $m$...
        – xarles
        Jul 29 at 7:16






      • 1




        OK, then $2^5equiv 6^5pmod 8$ and $2notequiv6pmod 8$. Here $n=5$ and $phi(m)=4$ are coprime.
        – Lord Shark the Unknown
        Jul 29 at 7:25






      • 1




        Sorry for the errors, now stated in the prime modulus case. If $m$ is not prime, the result is true for the invertible elements modulo $m$, changing $m-1$ by Euler's $varphi$ function.
        – xarles
        Jul 29 at 7:35

















      up vote
      0
      down vote













      For $gcd(n,phi(m))=1$, then you will have $a^nequiv b^n implies aequiv b bmod m.$ (Since $phi(m)$ is always even for $m>2$, $n$ will always be odd under this condition, but many odd numbers will not fulfill this).



      When $gcd(n,phi(m))>1$, you can have two (or more) different values producing the same $n$th power - for example, $4^3equiv 9^3equiv 29 bmod 35$.






      share|cite|improve this answer





















        Your Answer




        StackExchange.ifUsing("editor", function ()
        return StackExchange.using("mathjaxEditing", function ()
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        );
        );
        , "mathjax-editing");

        StackExchange.ready(function()
        var channelOptions =
        tags: "".split(" "),
        id: "69"
        ;
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function()
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled)
        StackExchange.using("snippets", function()
        createEditor();
        );

        else
        createEditor();

        );

        function createEditor()
        StackExchange.prepareEditor(
        heartbeatType: 'answer',
        convertImagesToLinks: true,
        noModals: false,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        );



        );








         

        draft saved


        draft discarded


















        StackExchange.ready(
        function ()
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865828%2fmodular-arithmetic-power-rule-implications%23new-answer', 'question_page');

        );

        Post as a guest

















        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        2
        down vote



        accepted










        The above example by Lord Shark the Unknown indicates that no complete answer can be found. But sometimes, we do have that kind of higher cancellation:



        We have, by telescopic sum,
        $$
        a^n - b^n = (a - b) sum_j=0^n-1 a^n-1-jb^j,
        $$
        so that $a^n equiv b^n mod m$ implies $a equiv b mod m$ if
        $$
        sum_j=0^n-1 a^n-1-jb^j
        $$
        is not a zero divisor.






        share|cite|improve this answer

























          up vote
          2
          down vote



          accepted










          The above example by Lord Shark the Unknown indicates that no complete answer can be found. But sometimes, we do have that kind of higher cancellation:



          We have, by telescopic sum,
          $$
          a^n - b^n = (a - b) sum_j=0^n-1 a^n-1-jb^j,
          $$
          so that $a^n equiv b^n mod m$ implies $a equiv b mod m$ if
          $$
          sum_j=0^n-1 a^n-1-jb^j
          $$
          is not a zero divisor.






          share|cite|improve this answer























            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            The above example by Lord Shark the Unknown indicates that no complete answer can be found. But sometimes, we do have that kind of higher cancellation:



            We have, by telescopic sum,
            $$
            a^n - b^n = (a - b) sum_j=0^n-1 a^n-1-jb^j,
            $$
            so that $a^n equiv b^n mod m$ implies $a equiv b mod m$ if
            $$
            sum_j=0^n-1 a^n-1-jb^j
            $$
            is not a zero divisor.






            share|cite|improve this answer













            The above example by Lord Shark the Unknown indicates that no complete answer can be found. But sometimes, we do have that kind of higher cancellation:



            We have, by telescopic sum,
            $$
            a^n - b^n = (a - b) sum_j=0^n-1 a^n-1-jb^j,
            $$
            so that $a^n equiv b^n mod m$ implies $a equiv b mod m$ if
            $$
            sum_j=0^n-1 a^n-1-jb^j
            $$
            is not a zero divisor.







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered Jul 29 at 7:12









            AlgebraicsAnonymous

            66611




            66611




















                up vote
                0
                down vote













                $2^3equiv4^3pmod 7$, but $2notequiv4pmod7$.






                share|cite|improve this answer

























                  up vote
                  0
                  down vote













                  $2^3equiv4^3pmod 7$, but $2notequiv4pmod7$.






                  share|cite|improve this answer























                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    $2^3equiv4^3pmod 7$, but $2notequiv4pmod7$.






                    share|cite|improve this answer













                    $2^3equiv4^3pmod 7$, but $2notequiv4pmod7$.







                    share|cite|improve this answer













                    share|cite|improve this answer



                    share|cite|improve this answer











                    answered Jul 29 at 6:49









                    Lord Shark the Unknown

                    84.5k950111




                    84.5k950111




















                        up vote
                        0
                        down vote













                        First, all you say about being positive or negative modulo $n$ makes really no sense, since $aequiv a-m pmodm$, so any positive number is equivalent to a negative number and viceversa.



                        However, there is a natural condition to put on $n$ related to $m$, if $m$ is prime: if $n$ and $m-1$ are coprime (so they share no comon divisor) , then $a^nequiv b^n pmodm$ implies $a equiv b pmodm$.






                        share|cite|improve this answer























                        • Is there a reason they have to be coprime?
                          – LHC2012
                          Jul 29 at 7:08










                        • See my posting for a counterexample to your final claim.
                          – Lord Shark the Unknown
                          Jul 29 at 7:11










                        • Ups, I forgot to write the $varphi$ in front of $m$...
                          – xarles
                          Jul 29 at 7:16






                        • 1




                          OK, then $2^5equiv 6^5pmod 8$ and $2notequiv6pmod 8$. Here $n=5$ and $phi(m)=4$ are coprime.
                          – Lord Shark the Unknown
                          Jul 29 at 7:25






                        • 1




                          Sorry for the errors, now stated in the prime modulus case. If $m$ is not prime, the result is true for the invertible elements modulo $m$, changing $m-1$ by Euler's $varphi$ function.
                          – xarles
                          Jul 29 at 7:35














                        up vote
                        0
                        down vote













                        First, all you say about being positive or negative modulo $n$ makes really no sense, since $aequiv a-m pmodm$, so any positive number is equivalent to a negative number and viceversa.



                        However, there is a natural condition to put on $n$ related to $m$, if $m$ is prime: if $n$ and $m-1$ are coprime (so they share no comon divisor) , then $a^nequiv b^n pmodm$ implies $a equiv b pmodm$.






                        share|cite|improve this answer























                        • Is there a reason they have to be coprime?
                          – LHC2012
                          Jul 29 at 7:08










                        • See my posting for a counterexample to your final claim.
                          – Lord Shark the Unknown
                          Jul 29 at 7:11










                        • Ups, I forgot to write the $varphi$ in front of $m$...
                          – xarles
                          Jul 29 at 7:16






                        • 1




                          OK, then $2^5equiv 6^5pmod 8$ and $2notequiv6pmod 8$. Here $n=5$ and $phi(m)=4$ are coprime.
                          – Lord Shark the Unknown
                          Jul 29 at 7:25






                        • 1




                          Sorry for the errors, now stated in the prime modulus case. If $m$ is not prime, the result is true for the invertible elements modulo $m$, changing $m-1$ by Euler's $varphi$ function.
                          – xarles
                          Jul 29 at 7:35












                        up vote
                        0
                        down vote










                        up vote
                        0
                        down vote









                        First, all you say about being positive or negative modulo $n$ makes really no sense, since $aequiv a-m pmodm$, so any positive number is equivalent to a negative number and viceversa.



                        However, there is a natural condition to put on $n$ related to $m$, if $m$ is prime: if $n$ and $m-1$ are coprime (so they share no comon divisor) , then $a^nequiv b^n pmodm$ implies $a equiv b pmodm$.






                        share|cite|improve this answer















                        First, all you say about being positive or negative modulo $n$ makes really no sense, since $aequiv a-m pmodm$, so any positive number is equivalent to a negative number and viceversa.



                        However, there is a natural condition to put on $n$ related to $m$, if $m$ is prime: if $n$ and $m-1$ are coprime (so they share no comon divisor) , then $a^nequiv b^n pmodm$ implies $a equiv b pmodm$.







                        share|cite|improve this answer















                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Jul 29 at 7:32


























                        answered Jul 29 at 7:01









                        xarles

                        83558




                        83558











                        • Is there a reason they have to be coprime?
                          – LHC2012
                          Jul 29 at 7:08










                        • See my posting for a counterexample to your final claim.
                          – Lord Shark the Unknown
                          Jul 29 at 7:11










                        • Ups, I forgot to write the $varphi$ in front of $m$...
                          – xarles
                          Jul 29 at 7:16






                        • 1




                          OK, then $2^5equiv 6^5pmod 8$ and $2notequiv6pmod 8$. Here $n=5$ and $phi(m)=4$ are coprime.
                          – Lord Shark the Unknown
                          Jul 29 at 7:25






                        • 1




                          Sorry for the errors, now stated in the prime modulus case. If $m$ is not prime, the result is true for the invertible elements modulo $m$, changing $m-1$ by Euler's $varphi$ function.
                          – xarles
                          Jul 29 at 7:35
















                        • Is there a reason they have to be coprime?
                          – LHC2012
                          Jul 29 at 7:08










                        • See my posting for a counterexample to your final claim.
                          – Lord Shark the Unknown
                          Jul 29 at 7:11










                        • Ups, I forgot to write the $varphi$ in front of $m$...
                          – xarles
                          Jul 29 at 7:16






                        • 1




                          OK, then $2^5equiv 6^5pmod 8$ and $2notequiv6pmod 8$. Here $n=5$ and $phi(m)=4$ are coprime.
                          – Lord Shark the Unknown
                          Jul 29 at 7:25






                        • 1




                          Sorry for the errors, now stated in the prime modulus case. If $m$ is not prime, the result is true for the invertible elements modulo $m$, changing $m-1$ by Euler's $varphi$ function.
                          – xarles
                          Jul 29 at 7:35















                        Is there a reason they have to be coprime?
                        – LHC2012
                        Jul 29 at 7:08




                        Is there a reason they have to be coprime?
                        – LHC2012
                        Jul 29 at 7:08












                        See my posting for a counterexample to your final claim.
                        – Lord Shark the Unknown
                        Jul 29 at 7:11




                        See my posting for a counterexample to your final claim.
                        – Lord Shark the Unknown
                        Jul 29 at 7:11












                        Ups, I forgot to write the $varphi$ in front of $m$...
                        – xarles
                        Jul 29 at 7:16




                        Ups, I forgot to write the $varphi$ in front of $m$...
                        – xarles
                        Jul 29 at 7:16




                        1




                        1




                        OK, then $2^5equiv 6^5pmod 8$ and $2notequiv6pmod 8$. Here $n=5$ and $phi(m)=4$ are coprime.
                        – Lord Shark the Unknown
                        Jul 29 at 7:25




                        OK, then $2^5equiv 6^5pmod 8$ and $2notequiv6pmod 8$. Here $n=5$ and $phi(m)=4$ are coprime.
                        – Lord Shark the Unknown
                        Jul 29 at 7:25




                        1




                        1




                        Sorry for the errors, now stated in the prime modulus case. If $m$ is not prime, the result is true for the invertible elements modulo $m$, changing $m-1$ by Euler's $varphi$ function.
                        – xarles
                        Jul 29 at 7:35




                        Sorry for the errors, now stated in the prime modulus case. If $m$ is not prime, the result is true for the invertible elements modulo $m$, changing $m-1$ by Euler's $varphi$ function.
                        – xarles
                        Jul 29 at 7:35










                        up vote
                        0
                        down vote













                        For $gcd(n,phi(m))=1$, then you will have $a^nequiv b^n implies aequiv b bmod m.$ (Since $phi(m)$ is always even for $m>2$, $n$ will always be odd under this condition, but many odd numbers will not fulfill this).



                        When $gcd(n,phi(m))>1$, you can have two (or more) different values producing the same $n$th power - for example, $4^3equiv 9^3equiv 29 bmod 35$.






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          For $gcd(n,phi(m))=1$, then you will have $a^nequiv b^n implies aequiv b bmod m.$ (Since $phi(m)$ is always even for $m>2$, $n$ will always be odd under this condition, but many odd numbers will not fulfill this).



                          When $gcd(n,phi(m))>1$, you can have two (or more) different values producing the same $n$th power - for example, $4^3equiv 9^3equiv 29 bmod 35$.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            For $gcd(n,phi(m))=1$, then you will have $a^nequiv b^n implies aequiv b bmod m.$ (Since $phi(m)$ is always even for $m>2$, $n$ will always be odd under this condition, but many odd numbers will not fulfill this).



                            When $gcd(n,phi(m))>1$, you can have two (or more) different values producing the same $n$th power - for example, $4^3equiv 9^3equiv 29 bmod 35$.






                            share|cite|improve this answer













                            For $gcd(n,phi(m))=1$, then you will have $a^nequiv b^n implies aequiv b bmod m.$ (Since $phi(m)$ is always even for $m>2$, $n$ will always be odd under this condition, but many odd numbers will not fulfill this).



                            When $gcd(n,phi(m))>1$, you can have two (or more) different values producing the same $n$th power - for example, $4^3equiv 9^3equiv 29 bmod 35$.







                            share|cite|improve this answer













                            share|cite|improve this answer



                            share|cite|improve this answer











                            answered Jul 29 at 17:09









                            Joffan

                            31.8k43169




                            31.8k43169






















                                 

                                draft saved


                                draft discarded


























                                 


                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function ()
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865828%2fmodular-arithmetic-power-rule-implications%23new-answer', 'question_page');

                                );

                                Post as a guest
































































                                Comments

                                Post a Comment

                                Popular posts from this blog

                                What is the equation of a 3D cone with generalised tilt?

                                Image
                                Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
                                Read more

                                Color the edges and diagonals of a regular polygon

                                Image
                                Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
                                Read more

                                Relationship between determinant of matrix and determinant of adjoint?

                                Image
                                Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
                                Read more