How to use modular arithmetic for $a^b-1 $












1












$begingroup$


The problem is: If $a,b ge 2$ are natural numbers s.t $a^b - 1$ is prime, then $a = 2$.



My ideas so far:



Since the we want to show that $a = 2$, we can take the the expression modulo $a$.



$a^b-1 equiv -1 pmod{a}$, so we know that there's a prime $p>2$ such that $p equiv -1 pmod{a}$ However I am not really sure where to go from here.






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think the intent might have been to look at this modulo $a-1$ - since $aequiv 1pmod{a-1}$, $a^b-1equiv 1^b-1=0pmod{a-1}$, so $a-1$ is a factor of $a^b-1$. This says nothing more than the answer below, but it's still worth pointing out.
    $endgroup$
    – Wojowu
    Dec 29 '18 at 21:40
















1












$begingroup$


The problem is: If $a,b ge 2$ are natural numbers s.t $a^b - 1$ is prime, then $a = 2$.



My ideas so far:



Since the we want to show that $a = 2$, we can take the the expression modulo $a$.



$a^b-1 equiv -1 pmod{a}$, so we know that there's a prime $p>2$ such that $p equiv -1 pmod{a}$ However I am not really sure where to go from here.






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think the intent might have been to look at this modulo $a-1$ - since $aequiv 1pmod{a-1}$, $a^b-1equiv 1^b-1=0pmod{a-1}$, so $a-1$ is a factor of $a^b-1$. This says nothing more than the answer below, but it's still worth pointing out.
    $endgroup$
    – Wojowu
    Dec 29 '18 at 21:40














1












1








1





$begingroup$


The problem is: If $a,b ge 2$ are natural numbers s.t $a^b - 1$ is prime, then $a = 2$.



My ideas so far:



Since the we want to show that $a = 2$, we can take the the expression modulo $a$.



$a^b-1 equiv -1 pmod{a}$, so we know that there's a prime $p>2$ such that $p equiv -1 pmod{a}$ However I am not really sure where to go from here.






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$




The problem is: If $a,b ge 2$ are natural numbers s.t $a^b - 1$ is prime, then $a = 2$.



My ideas so far:



Since the we want to show that $a = 2$, we can take the the expression modulo $a$.



$a^b-1 equiv -1 pmod{a}$, so we know that there's a prime $p>2$ such that $p equiv -1 pmod{a}$ However I am not really sure where to go from here.






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 Dec 29 '18 at 21:05









Namaste

1




1










asked Dec 29 '18 at 21:03









xAlyxAly

404




404








  • 1




    $begingroup$
    I think the intent might have been to look at this modulo $a-1$ - since $aequiv 1pmod{a-1}$, $a^b-1equiv 1^b-1=0pmod{a-1}$, so $a-1$ is a factor of $a^b-1$. This says nothing more than the answer below, but it's still worth pointing out.
    $endgroup$
    – Wojowu
    Dec 29 '18 at 21:40














  • 1




    $begingroup$
    I think the intent might have been to look at this modulo $a-1$ - since $aequiv 1pmod{a-1}$, $a^b-1equiv 1^b-1=0pmod{a-1}$, so $a-1$ is a factor of $a^b-1$. This says nothing more than the answer below, but it's still worth pointing out.
    $endgroup$
    – Wojowu
    Dec 29 '18 at 21:40








1




1




$begingroup$
I think the intent might have been to look at this modulo $a-1$ - since $aequiv 1pmod{a-1}$, $a^b-1equiv 1^b-1=0pmod{a-1}$, so $a-1$ is a factor of $a^b-1$. This says nothing more than the answer below, but it's still worth pointing out.
$endgroup$
– Wojowu
Dec 29 '18 at 21:40




$begingroup$
I think the intent might have been to look at this modulo $a-1$ - since $aequiv 1pmod{a-1}$, $a^b-1equiv 1^b-1=0pmod{a-1}$, so $a-1$ is a factor of $a^b-1$. This says nothing more than the answer below, but it's still worth pointing out.
$endgroup$
– Wojowu
Dec 29 '18 at 21:40










2 Answers
2






active

oldest

votes


















3












$begingroup$

Suppose, there is an $a>2$, and a $b$ such that $a^b-1$ is prime. Observe that $a-1mid a^b-1$, $2leq a-1<a^b-1$ establishes, $a^b-1$ is not a prime.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    What's the point in taking $q$? $a-1mid a^b-1$ already proves $a^b-1$ is not prime.
    $endgroup$
    – Wojowu
    Dec 29 '18 at 21:10










  • $begingroup$
    No need to. I was thinking as I was writing. Condensed it.
    $endgroup$
    – Aaron
    Dec 29 '18 at 21:17



















1












$begingroup$

Hmm, If $b ge 2$ then $a^b -1 = (a-1)(a^{b-1}+ .... + 1)$ and $a ge 2$ then $a^{b-1}+ .... + 1> 1$ so the only way $a^b - 1$ is prime is if $a-1 = 1$ or $a = 2$.



The real question is under what conditions of $b$ can we assume $2^b - 1$ is prime. If $b= nk$ is not prime then you have $2^b - 1 = 2^n)^k - 1 = (2^n-1)((2^n)^{k-1} + ..... +1)$ so $b$ must be prime if $2^b -1$ is to be prime.



These are the Mersenne primes. Which are well researched.






share|cite|improve this answer









$endgroup$













    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',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    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%2f3056263%2fhow-to-use-modular-arithmetic-for-ab-1%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Suppose, there is an $a>2$, and a $b$ such that $a^b-1$ is prime. Observe that $a-1mid a^b-1$, $2leq a-1<a^b-1$ establishes, $a^b-1$ is not a prime.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      What's the point in taking $q$? $a-1mid a^b-1$ already proves $a^b-1$ is not prime.
      $endgroup$
      – Wojowu
      Dec 29 '18 at 21:10










    • $begingroup$
      No need to. I was thinking as I was writing. Condensed it.
      $endgroup$
      – Aaron
      Dec 29 '18 at 21:17
















    3












    $begingroup$

    Suppose, there is an $a>2$, and a $b$ such that $a^b-1$ is prime. Observe that $a-1mid a^b-1$, $2leq a-1<a^b-1$ establishes, $a^b-1$ is not a prime.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      What's the point in taking $q$? $a-1mid a^b-1$ already proves $a^b-1$ is not prime.
      $endgroup$
      – Wojowu
      Dec 29 '18 at 21:10










    • $begingroup$
      No need to. I was thinking as I was writing. Condensed it.
      $endgroup$
      – Aaron
      Dec 29 '18 at 21:17














    3












    3








    3





    $begingroup$

    Suppose, there is an $a>2$, and a $b$ such that $a^b-1$ is prime. Observe that $a-1mid a^b-1$, $2leq a-1<a^b-1$ establishes, $a^b-1$ is not a prime.






    share|cite|improve this answer











    $endgroup$



    Suppose, there is an $a>2$, and a $b$ such that $a^b-1$ is prime. Observe that $a-1mid a^b-1$, $2leq a-1<a^b-1$ establishes, $a^b-1$ is not a prime.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 29 '18 at 21:18

























    answered Dec 29 '18 at 21:07









    AaronAaron

    1,922415




    1,922415








    • 1




      $begingroup$
      What's the point in taking $q$? $a-1mid a^b-1$ already proves $a^b-1$ is not prime.
      $endgroup$
      – Wojowu
      Dec 29 '18 at 21:10










    • $begingroup$
      No need to. I was thinking as I was writing. Condensed it.
      $endgroup$
      – Aaron
      Dec 29 '18 at 21:17














    • 1




      $begingroup$
      What's the point in taking $q$? $a-1mid a^b-1$ already proves $a^b-1$ is not prime.
      $endgroup$
      – Wojowu
      Dec 29 '18 at 21:10










    • $begingroup$
      No need to. I was thinking as I was writing. Condensed it.
      $endgroup$
      – Aaron
      Dec 29 '18 at 21:17








    1




    1




    $begingroup$
    What's the point in taking $q$? $a-1mid a^b-1$ already proves $a^b-1$ is not prime.
    $endgroup$
    – Wojowu
    Dec 29 '18 at 21:10




    $begingroup$
    What's the point in taking $q$? $a-1mid a^b-1$ already proves $a^b-1$ is not prime.
    $endgroup$
    – Wojowu
    Dec 29 '18 at 21:10












    $begingroup$
    No need to. I was thinking as I was writing. Condensed it.
    $endgroup$
    – Aaron
    Dec 29 '18 at 21:17




    $begingroup$
    No need to. I was thinking as I was writing. Condensed it.
    $endgroup$
    – Aaron
    Dec 29 '18 at 21:17











    1












    $begingroup$

    Hmm, If $b ge 2$ then $a^b -1 = (a-1)(a^{b-1}+ .... + 1)$ and $a ge 2$ then $a^{b-1}+ .... + 1> 1$ so the only way $a^b - 1$ is prime is if $a-1 = 1$ or $a = 2$.



    The real question is under what conditions of $b$ can we assume $2^b - 1$ is prime. If $b= nk$ is not prime then you have $2^b - 1 = 2^n)^k - 1 = (2^n-1)((2^n)^{k-1} + ..... +1)$ so $b$ must be prime if $2^b -1$ is to be prime.



    These are the Mersenne primes. Which are well researched.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hmm, If $b ge 2$ then $a^b -1 = (a-1)(a^{b-1}+ .... + 1)$ and $a ge 2$ then $a^{b-1}+ .... + 1> 1$ so the only way $a^b - 1$ is prime is if $a-1 = 1$ or $a = 2$.



      The real question is under what conditions of $b$ can we assume $2^b - 1$ is prime. If $b= nk$ is not prime then you have $2^b - 1 = 2^n)^k - 1 = (2^n-1)((2^n)^{k-1} + ..... +1)$ so $b$ must be prime if $2^b -1$ is to be prime.



      These are the Mersenne primes. Which are well researched.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hmm, If $b ge 2$ then $a^b -1 = (a-1)(a^{b-1}+ .... + 1)$ and $a ge 2$ then $a^{b-1}+ .... + 1> 1$ so the only way $a^b - 1$ is prime is if $a-1 = 1$ or $a = 2$.



        The real question is under what conditions of $b$ can we assume $2^b - 1$ is prime. If $b= nk$ is not prime then you have $2^b - 1 = 2^n)^k - 1 = (2^n-1)((2^n)^{k-1} + ..... +1)$ so $b$ must be prime if $2^b -1$ is to be prime.



        These are the Mersenne primes. Which are well researched.






        share|cite|improve this answer









        $endgroup$



        Hmm, If $b ge 2$ then $a^b -1 = (a-1)(a^{b-1}+ .... + 1)$ and $a ge 2$ then $a^{b-1}+ .... + 1> 1$ so the only way $a^b - 1$ is prime is if $a-1 = 1$ or $a = 2$.



        The real question is under what conditions of $b$ can we assume $2^b - 1$ is prime. If $b= nk$ is not prime then you have $2^b - 1 = 2^n)^k - 1 = (2^n-1)((2^n)^{k-1} + ..... +1)$ so $b$ must be prime if $2^b -1$ is to be prime.



        These are the Mersenne primes. Which are well researched.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 21:43









        fleabloodfleablood

        72.9k22789




        72.9k22789






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056263%2fhow-to-use-modular-arithmetic-for-ab-1%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            To store a contact into the json file from server.js file using a class in NodeJS

            Image
            0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
            Read more

            Redirect URL with Chrome Remote Debugging Android Devices

            Image
            1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more