Induction on powers of powers.












0












$begingroup$


So I would like to prove that $ a_n ; | ; a_{n+1} - 2 $, where $a_n = 6^{2^n} + 1 $ Now I know I need to do this by induction and so I begin by showing this is true for the base case.




(Proposition): P(n) := $ 6^{2^{n+1}} -1 = (6^{2^n} + 1)q ;$, where $q in mathbb{Z}$




(Base Case): P(1) is true as $ 6^{2^{2}} -1 = (6^{2} + 1)35 ;$




(Hypothesis): We assume that P(k) is True, where P(k):= $ 6^{2^{k+1}} -1 = (6^{2^k} + 1)q ;$




(Induction Step): Need to show that $P(k) implies P(k+1)$ Now I know I need to write $ 6^{2^{k+2}} -1$ in terms of my hypothesis and end up with something along the lines of $(6^{2^{k+1}} + 1)r$, where $r in mathbb{Z}$. My issue: I have and the reason I'm here is I don't feel comfortable working with powers of powers. I'm not sure how to use my hypothesis. In short looking for some guidance to get me started.



Best Regards
and thanks for your time.






induction exponentiation







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    So I would like to prove that $ a_n ; | ; a_{n+1} - 2 $, where $a_n = 6^{2^n} + 1 $ Now I know I need to do this by induction and so I begin by showing this is true for the base case.




    (Proposition): P(n) := $ 6^{2^{n+1}} -1 = (6^{2^n} + 1)q ;$, where $q in mathbb{Z}$




    (Base Case): P(1) is true as $ 6^{2^{2}} -1 = (6^{2} + 1)35 ;$




    (Hypothesis): We assume that P(k) is True, where P(k):= $ 6^{2^{k+1}} -1 = (6^{2^k} + 1)q ;$




    (Induction Step): Need to show that $P(k) implies P(k+1)$ Now I know I need to write $ 6^{2^{k+2}} -1$ in terms of my hypothesis and end up with something along the lines of $(6^{2^{k+1}} + 1)r$, where $r in mathbb{Z}$. My issue: I have and the reason I'm here is I don't feel comfortable working with powers of powers. I'm not sure how to use my hypothesis. In short looking for some guidance to get me started.



    Best Regards
    and thanks for your time.






    induction exponentiation







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      So I would like to prove that $ a_n ; | ; a_{n+1} - 2 $, where $a_n = 6^{2^n} + 1 $ Now I know I need to do this by induction and so I begin by showing this is true for the base case.




      (Proposition): P(n) := $ 6^{2^{n+1}} -1 = (6^{2^n} + 1)q ;$, where $q in mathbb{Z}$




      (Base Case): P(1) is true as $ 6^{2^{2}} -1 = (6^{2} + 1)35 ;$




      (Hypothesis): We assume that P(k) is True, where P(k):= $ 6^{2^{k+1}} -1 = (6^{2^k} + 1)q ;$




      (Induction Step): Need to show that $P(k) implies P(k+1)$ Now I know I need to write $ 6^{2^{k+2}} -1$ in terms of my hypothesis and end up with something along the lines of $(6^{2^{k+1}} + 1)r$, where $r in mathbb{Z}$. My issue: I have and the reason I'm here is I don't feel comfortable working with powers of powers. I'm not sure how to use my hypothesis. In short looking for some guidance to get me started.



      Best Regards
      and thanks for your time.






      induction exponentiation







      share|cite|improve this question











      $endgroup$




      So I would like to prove that $ a_n ; | ; a_{n+1} - 2 $, where $a_n = 6^{2^n} + 1 $ Now I know I need to do this by induction and so I begin by showing this is true for the base case.




      (Proposition): P(n) := $ 6^{2^{n+1}} -1 = (6^{2^n} + 1)q ;$, where $q in mathbb{Z}$




      (Base Case): P(1) is true as $ 6^{2^{2}} -1 = (6^{2} + 1)35 ;$




      (Hypothesis): We assume that P(k) is True, where P(k):= $ 6^{2^{k+1}} -1 = (6^{2^k} + 1)q ;$




      (Induction Step): Need to show that $P(k) implies P(k+1)$ Now I know I need to write $ 6^{2^{k+2}} -1$ in terms of my hypothesis and end up with something along the lines of $(6^{2^{k+1}} + 1)r$, where $r in mathbb{Z}$. My issue: I have and the reason I'm here is I don't feel comfortable working with powers of powers. I'm not sure how to use my hypothesis. In short looking for some guidance to get me started.



      Best Regards
      and thanks for your time.






      induction exponentiation




      induction exponentiation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 7 '18 at 19:25









      Bernard

      119k740113




      119k740113










      asked Dec 7 '18 at 19:11









      Dead_Ling0Dead_Ling0

      476




      476






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          You're on the right track. The key thing to use here is the difference of squares formula. In particular,



          $$6^{2^{k + 1}} - 1 = (6^{2^k})^2 - 1 = (6^{2^k} - 1)(6^{2^k} + 1)$$



          from which your result should follow easily.



          However, induction isn't really necessary here... this gives a direct proof of divisibility.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah can't believe i didn't spot this thank you for the clarity kicking myself here !!!!
            $endgroup$
            – Dead_Ling0
            Dec 7 '18 at 20:10











          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%2f3030263%2finduction-on-powers-of-powers%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          You're on the right track. The key thing to use here is the difference of squares formula. In particular,



          $$6^{2^{k + 1}} - 1 = (6^{2^k})^2 - 1 = (6^{2^k} - 1)(6^{2^k} + 1)$$



          from which your result should follow easily.



          However, induction isn't really necessary here... this gives a direct proof of divisibility.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah can't believe i didn't spot this thank you for the clarity kicking myself here !!!!
            $endgroup$
            – Dead_Ling0
            Dec 7 '18 at 20:10
















          2












          $begingroup$

          You're on the right track. The key thing to use here is the difference of squares formula. In particular,



          $$6^{2^{k + 1}} - 1 = (6^{2^k})^2 - 1 = (6^{2^k} - 1)(6^{2^k} + 1)$$



          from which your result should follow easily.



          However, induction isn't really necessary here... this gives a direct proof of divisibility.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah can't believe i didn't spot this thank you for the clarity kicking myself here !!!!
            $endgroup$
            – Dead_Ling0
            Dec 7 '18 at 20:10














          2












          2








          2





          $begingroup$

          You're on the right track. The key thing to use here is the difference of squares formula. In particular,



          $$6^{2^{k + 1}} - 1 = (6^{2^k})^2 - 1 = (6^{2^k} - 1)(6^{2^k} + 1)$$



          from which your result should follow easily.



          However, induction isn't really necessary here... this gives a direct proof of divisibility.






          share|cite|improve this answer









          $endgroup$



          You're on the right track. The key thing to use here is the difference of squares formula. In particular,



          $$6^{2^{k + 1}} - 1 = (6^{2^k})^2 - 1 = (6^{2^k} - 1)(6^{2^k} + 1)$$



          from which your result should follow easily.



          However, induction isn't really necessary here... this gives a direct proof of divisibility.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 19:21









          T. BongersT. Bongers

          23.1k54662




          23.1k54662












          • $begingroup$
            Ah can't believe i didn't spot this thank you for the clarity kicking myself here !!!!
            $endgroup$
            – Dead_Ling0
            Dec 7 '18 at 20:10


















          • $begingroup$
            Ah can't believe i didn't spot this thank you for the clarity kicking myself here !!!!
            $endgroup$
            – Dead_Ling0
            Dec 7 '18 at 20:10
















          $begingroup$
          Ah can't believe i didn't spot this thank you for the clarity kicking myself here !!!!
          $endgroup$
          – Dead_Ling0
          Dec 7 '18 at 20:10




          $begingroup$
          Ah can't believe i didn't spot this thank you for the clarity kicking myself here !!!!
          $endgroup$
          – Dead_Ling0
          Dec 7 '18 at 20:10


















          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%2f3030263%2finduction-on-powers-of-powers%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

          Ellipse (mathématiques)

          Image
          Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
          Read more

          Quarter-circle Tiles

          Image
          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
          Read more

          Mont Emei

          Image
          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
          Read more