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







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$endgroup$

















    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






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      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 11 '18 at 17:48









      Asaf Karagila♦

      305k33435765




      305k33435765










      asked Dec 11 '18 at 17:45









      AlexAlex

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      1134






















          4 Answers
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          $begingroup$

          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









                $endgroup$













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                  4 Answers
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                  4 Answers
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                  $begingroup$

                  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






























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