Proving that $p ^ {p + 2} + (p + 2)^{p} equiv 0pmod{2p + 2}$ for a prime $p>2$












0












$begingroup$


So, the task is to prove the following:



$$p ^ {p + 2} + (p + 2)^{p} equiv 0 pmod{2p + 2}$$



For all prime numbers $p > 2$.



I haven't got any clue on how to do this, but, for some reason, I feel like it is somehow related to the Euler's theorem or its 'friend' Fermat's little theorem.



One observation I have made is that this expression can be refactored as
$a ^ b + b ^ a equiv 0 pmod{a + b}$ which is maybe another way to think about the problem.



Any help would be appreciated!






number-theory prime-numbers







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    So, the task is to prove the following:



    $$p ^ {p + 2} + (p + 2)^{p} equiv 0 pmod{2p + 2}$$



    For all prime numbers $p > 2$.



    I haven't got any clue on how to do this, but, for some reason, I feel like it is somehow related to the Euler's theorem or its 'friend' Fermat's little theorem.



    One observation I have made is that this expression can be refactored as
    $a ^ b + b ^ a equiv 0 pmod{a + b}$ which is maybe another way to think about the problem.



    Any help would be appreciated!






    number-theory prime-numbers







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      So, the task is to prove the following:



      $$p ^ {p + 2} + (p + 2)^{p} equiv 0 pmod{2p + 2}$$



      For all prime numbers $p > 2$.



      I haven't got any clue on how to do this, but, for some reason, I feel like it is somehow related to the Euler's theorem or its 'friend' Fermat's little theorem.



      One observation I have made is that this expression can be refactored as
      $a ^ b + b ^ a equiv 0 pmod{a + b}$ which is maybe another way to think about the problem.



      Any help would be appreciated!






      number-theory prime-numbers







      share|cite|improve this question











      $endgroup$




      So, the task is to prove the following:



      $$p ^ {p + 2} + (p + 2)^{p} equiv 0 pmod{2p + 2}$$



      For all prime numbers $p > 2$.



      I haven't got any clue on how to do this, but, for some reason, I feel like it is somehow related to the Euler's theorem or its 'friend' Fermat's little theorem.



      One observation I have made is that this expression can be refactored as
      $a ^ b + b ^ a equiv 0 pmod{a + b}$ which is maybe another way to think about the problem.



      Any help would be appreciated!






      number-theory prime-numbers




      number-theory prime-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 11 '18 at 17:48









      Asaf Karagila♦

      305k33435765




      305k33435765










      asked Dec 11 '18 at 17:45









      AlexAlex

      1134




      1134






















          4 Answers
          4






          active

          oldest

          votes


















          2












          $begingroup$

          We can assume $p$ odd.



          Since $p^{p+2}+(p+2)^p$ is a sum of two odd numbers, it is even. Thus, it is enough to prove that the sum is divisible by $p+1$. Modulo $p+1$, the expression is $(-1)^{p+2}+1^p=0$ since $p$ is odd. Hence the original expression can be divided by $2p+2$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            As $p,p+2$ are odd, so it's sufficient to prove $p+1$ divides
            $p^{2m+1}+(p+2)^n$



            Now $pequiv-1pmod{p+1}implies p^{2m+1}equiv?$



            and $p+2equiv1pmod{p+1}implies(p+2)^nequiv?$






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Let $p$ be an odd prime. Note that by Fermat's Little Theorem, $2p+2$ divides $p^{p} + (p+2)^{p}$, thus it suffices to prove that $2p+2$ divides $p^{p+2}-p^p$ since $$p^{p+2}+(p+2)^p=(p^{p+2}-p^p)+p^p+(p+2)^p.$$ Notice that the task is now easy. Hint: $p^{p+2}-p^p=p^{p}(p+1)(p-1)$.






              share|cite|improve this answer











              $endgroup$





















                0












                $begingroup$

                $p^{p+2} +(p+2)^p equiv p^{p+2} + (-p)^p equiv p^{p+2}-p^pequiv p^p(p^2 -1)$



                $p^p(p+1)(p-1)equiv p^p(2p + 2)frac {p-1}2 equiv 0pmod {2p+2}$



                Unless I did something horribly wrong this will be true for any odd number $p$ whether prime or not.






                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%2f3035555%2fproving-that-p-p-2-p-2p-equiv-0-pmod2p-2-for-a-prime-p2%23new-answer', 'question_page');
                  }
                  );

                  Post as a guest















                  Required, but never shown

























                  4 Answers
                  4






                  active

                  oldest

                  votes








                  4 Answers
                  4






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  2












                  $begingroup$

                  We can assume $p$ odd.



                  Since $p^{p+2}+(p+2)^p$ is a sum of two odd numbers, it is even. Thus, it is enough to prove that the sum is divisible by $p+1$. Modulo $p+1$, the expression is $(-1)^{p+2}+1^p=0$ since $p$ is odd. Hence the original expression can be divided by $2p+2$.






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    We can assume $p$ odd.



                    Since $p^{p+2}+(p+2)^p$ is a sum of two odd numbers, it is even. Thus, it is enough to prove that the sum is divisible by $p+1$. Modulo $p+1$, the expression is $(-1)^{p+2}+1^p=0$ since $p$ is odd. Hence the original expression can be divided by $2p+2$.






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      We can assume $p$ odd.



                      Since $p^{p+2}+(p+2)^p$ is a sum of two odd numbers, it is even. Thus, it is enough to prove that the sum is divisible by $p+1$. Modulo $p+1$, the expression is $(-1)^{p+2}+1^p=0$ since $p$ is odd. Hence the original expression can be divided by $2p+2$.






                      share|cite|improve this answer









                      $endgroup$



                      We can assume $p$ odd.



                      Since $p^{p+2}+(p+2)^p$ is a sum of two odd numbers, it is even. Thus, it is enough to prove that the sum is divisible by $p+1$. Modulo $p+1$, the expression is $(-1)^{p+2}+1^p=0$ since $p$ is odd. Hence the original expression can be divided by $2p+2$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 11 '18 at 17:52









                      Leo163Leo163

                      1,670512




                      1,670512























                          0












                          $begingroup$

                          As $p,p+2$ are odd, so it's sufficient to prove $p+1$ divides
                          $p^{2m+1}+(p+2)^n$



                          Now $pequiv-1pmod{p+1}implies p^{2m+1}equiv?$



                          and $p+2equiv1pmod{p+1}implies(p+2)^nequiv?$






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            As $p,p+2$ are odd, so it's sufficient to prove $p+1$ divides
                            $p^{2m+1}+(p+2)^n$



                            Now $pequiv-1pmod{p+1}implies p^{2m+1}equiv?$



                            and $p+2equiv1pmod{p+1}implies(p+2)^nequiv?$






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              As $p,p+2$ are odd, so it's sufficient to prove $p+1$ divides
                              $p^{2m+1}+(p+2)^n$



                              Now $pequiv-1pmod{p+1}implies p^{2m+1}equiv?$



                              and $p+2equiv1pmod{p+1}implies(p+2)^nequiv?$






                              share|cite|improve this answer









                              $endgroup$



                              As $p,p+2$ are odd, so it's sufficient to prove $p+1$ divides
                              $p^{2m+1}+(p+2)^n$



                              Now $pequiv-1pmod{p+1}implies p^{2m+1}equiv?$



                              and $p+2equiv1pmod{p+1}implies(p+2)^nequiv?$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 11 '18 at 17:53









                              lab bhattacharjeelab bhattacharjee

                              226k15157275




                              226k15157275























                                  0












                                  $begingroup$

                                  Let $p$ be an odd prime. Note that by Fermat's Little Theorem, $2p+2$ divides $p^{p} + (p+2)^{p}$, thus it suffices to prove that $2p+2$ divides $p^{p+2}-p^p$ since $$p^{p+2}+(p+2)^p=(p^{p+2}-p^p)+p^p+(p+2)^p.$$ Notice that the task is now easy. Hint: $p^{p+2}-p^p=p^{p}(p+1)(p-1)$.






                                  share|cite|improve this answer











                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Let $p$ be an odd prime. Note that by Fermat's Little Theorem, $2p+2$ divides $p^{p} + (p+2)^{p}$, thus it suffices to prove that $2p+2$ divides $p^{p+2}-p^p$ since $$p^{p+2}+(p+2)^p=(p^{p+2}-p^p)+p^p+(p+2)^p.$$ Notice that the task is now easy. Hint: $p^{p+2}-p^p=p^{p}(p+1)(p-1)$.






                                    share|cite|improve this answer











                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Let $p$ be an odd prime. Note that by Fermat's Little Theorem, $2p+2$ divides $p^{p} + (p+2)^{p}$, thus it suffices to prove that $2p+2$ divides $p^{p+2}-p^p$ since $$p^{p+2}+(p+2)^p=(p^{p+2}-p^p)+p^p+(p+2)^p.$$ Notice that the task is now easy. Hint: $p^{p+2}-p^p=p^{p}(p+1)(p-1)$.






                                      share|cite|improve this answer











                                      $endgroup$



                                      Let $p$ be an odd prime. Note that by Fermat's Little Theorem, $2p+2$ divides $p^{p} + (p+2)^{p}$, thus it suffices to prove that $2p+2$ divides $p^{p+2}-p^p$ since $$p^{p+2}+(p+2)^p=(p^{p+2}-p^p)+p^p+(p+2)^p.$$ Notice that the task is now easy. Hint: $p^{p+2}-p^p=p^{p}(p+1)(p-1)$.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Dec 11 '18 at 18:11

























                                      answered Dec 11 '18 at 18:02







                                      user507152






























                                          0












                                          $begingroup$

                                          $p^{p+2} +(p+2)^p equiv p^{p+2} + (-p)^p equiv p^{p+2}-p^pequiv p^p(p^2 -1)$



                                          $p^p(p+1)(p-1)equiv p^p(2p + 2)frac {p-1}2 equiv 0pmod {2p+2}$



                                          Unless I did something horribly wrong this will be true for any odd number $p$ whether prime or not.






                                          share|cite|improve this answer









                                          $endgroup$


















                                            0












                                            $begingroup$

                                            $p^{p+2} +(p+2)^p equiv p^{p+2} + (-p)^p equiv p^{p+2}-p^pequiv p^p(p^2 -1)$



                                            $p^p(p+1)(p-1)equiv p^p(2p + 2)frac {p-1}2 equiv 0pmod {2p+2}$



                                            Unless I did something horribly wrong this will be true for any odd number $p$ whether prime or not.






                                            share|cite|improve this answer









                                            $endgroup$
















                                              0












                                              0








                                              0





                                              $begingroup$

                                              $p^{p+2} +(p+2)^p equiv p^{p+2} + (-p)^p equiv p^{p+2}-p^pequiv p^p(p^2 -1)$



                                              $p^p(p+1)(p-1)equiv p^p(2p + 2)frac {p-1}2 equiv 0pmod {2p+2}$



                                              Unless I did something horribly wrong this will be true for any odd number $p$ whether prime or not.






                                              share|cite|improve this answer









                                              $endgroup$



                                              $p^{p+2} +(p+2)^p equiv p^{p+2} + (-p)^p equiv p^{p+2}-p^pequiv p^p(p^2 -1)$



                                              $p^p(p+1)(p-1)equiv p^p(2p + 2)frac {p-1}2 equiv 0pmod {2p+2}$



                                              Unless I did something horribly wrong this will be true for any odd number $p$ whether prime or not.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Dec 11 '18 at 18:28









                                              fleabloodfleablood

                                              71.2k22686




                                              71.2k22686






























                                                  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%2f3035555%2fproving-that-p-p-2-p-2p-equiv-0-pmod2p-2-for-a-prime-p2%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

                                                  Le Mesnil-Réaume

                                                  Image
                                                  Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
                                                  Read more

                                                  Ida-Boy-Ed-Garten

                                                  Image
                                                  Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                                                  Read more

                                                  web3.py web3.isConnected() returns false always

                                                  Image
                                                  1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
                                                  Read more