Show that $sumlimits_i=1^p i^p$ is divisible by $p$ for all primes $p > 2$.

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5
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Show that
$$
sumlimits_i=1^p i^p
$$
is divisible by $p$ for all primes $p > 2$.




I think this has something to do with Fermat's Theorem and I have tried using congruences modulus to do it.




elementary-number-theory summation modular-arithmetic




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  • elcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Aug 26 at 9:46










  • See math.stackexchange.com/questions/529790/…
    – lab bhattacharjee
    Aug 26 at 10:26














up vote
5
down vote

favorite













Show that
$$
sumlimits_i=1^p i^p
$$
is divisible by $p$ for all primes $p > 2$.




I think this has something to do with Fermat's Theorem and I have tried using congruences modulus to do it.




elementary-number-theory summation modular-arithmetic




share|cite|improve this question






















  • elcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Aug 26 at 9:46










  • See math.stackexchange.com/questions/529790/…
    – lab bhattacharjee
    Aug 26 at 10:26












up vote
5
down vote

favorite









up vote
5
down vote

favorite












Show that
$$
sumlimits_i=1^p i^p
$$
is divisible by $p$ for all primes $p > 2$.




I think this has something to do with Fermat's Theorem and I have tried using congruences modulus to do it.




elementary-number-theory summation modular-arithmetic




share|cite|improve this question















Show that
$$
sumlimits_i=1^p i^p
$$
is divisible by $p$ for all primes $p > 2$.




I think this has something to do with Fermat's Theorem and I have tried using congruences modulus to do it.





elementary-number-theory summation modular-arithmetic





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 26 at 10:15









rtybase

9,04221433




9,04221433










asked Aug 26 at 9:40









T.J

261




261











  • elcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Aug 26 at 9:46










  • See math.stackexchange.com/questions/529790/…
    – lab bhattacharjee
    Aug 26 at 10:26
















  • elcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Aug 26 at 9:46










  • See math.stackexchange.com/questions/529790/…
    – lab bhattacharjee
    Aug 26 at 10:26















elcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 26 at 9:46




elcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 26 at 9:46












See math.stackexchange.com/questions/529790/…
– lab bhattacharjee
Aug 26 at 10:26




See math.stackexchange.com/questions/529790/…
– lab bhattacharjee
Aug 26 at 10:26










2 Answers
2






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5
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Guide:



Notice that $p$ is odd.



$$1^p + (p-1)^p equiv 1^p+(-1)^p equiv 1-1 equiv0 pmodp$$



$$2^p + (p-2)^p equiv 2^p+(-2)^p equiv 2^p-2^p equiv0 pmodp$$



Something to think about:



  • How important do you think $p$ being prime is here, or is $p$ being odd positive number sufficient.





share|cite|improve this answer





























    up vote
    2
    down vote













    And yes, it can be solved using LFT as well, since
    $$i^pequiv i pmodp$$
    thus
    $$sumlimits_i=1^pi^p equiv sumlimits_i=1^pi pmodp tag1$$
    but
    $$sumlimits_i=1^pi = fracp(p+1)2$$
    and $forall p>2$ primes are odd, thus $2 mid p+1$
    which means
    $$sumlimits_i=1^pi = fracp(p+1)2 equiv 0 pmodp$$
    and from $(1)$
    $$sumlimits_i=1^pi^p equiv 0 pmodp$$






    share|cite|improve this answer




















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      5
      down vote













      Guide:



      Notice that $p$ is odd.



      $$1^p + (p-1)^p equiv 1^p+(-1)^p equiv 1-1 equiv0 pmodp$$



      $$2^p + (p-2)^p equiv 2^p+(-2)^p equiv 2^p-2^p equiv0 pmodp$$



      Something to think about:



      • How important do you think $p$ being prime is here, or is $p$ being odd positive number sufficient.





      share|cite|improve this answer


























        up vote
        5
        down vote













        Guide:



        Notice that $p$ is odd.



        $$1^p + (p-1)^p equiv 1^p+(-1)^p equiv 1-1 equiv0 pmodp$$



        $$2^p + (p-2)^p equiv 2^p+(-2)^p equiv 2^p-2^p equiv0 pmodp$$



        Something to think about:



        • How important do you think $p$ being prime is here, or is $p$ being odd positive number sufficient.





        share|cite|improve this answer
























          up vote
          5
          down vote










          up vote
          5
          down vote









          Guide:



          Notice that $p$ is odd.



          $$1^p + (p-1)^p equiv 1^p+(-1)^p equiv 1-1 equiv0 pmodp$$



          $$2^p + (p-2)^p equiv 2^p+(-2)^p equiv 2^p-2^p equiv0 pmodp$$



          Something to think about:



          • How important do you think $p$ being prime is here, or is $p$ being odd positive number sufficient.





          share|cite|improve this answer














          Guide:



          Notice that $p$ is odd.



          $$1^p + (p-1)^p equiv 1^p+(-1)^p equiv 1-1 equiv0 pmodp$$



          $$2^p + (p-2)^p equiv 2^p+(-2)^p equiv 2^p-2^p equiv0 pmodp$$



          Something to think about:



          • How important do you think $p$ being prime is here, or is $p$ being odd positive number sufficient.






          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 26 at 10:07

























          answered Aug 26 at 9:45









          Siong Thye Goh

          80.7k1453102




          80.7k1453102




















              up vote
              2
              down vote













              And yes, it can be solved using LFT as well, since
              $$i^pequiv i pmodp$$
              thus
              $$sumlimits_i=1^pi^p equiv sumlimits_i=1^pi pmodp tag1$$
              but
              $$sumlimits_i=1^pi = fracp(p+1)2$$
              and $forall p>2$ primes are odd, thus $2 mid p+1$
              which means
              $$sumlimits_i=1^pi = fracp(p+1)2 equiv 0 pmodp$$
              and from $(1)$
              $$sumlimits_i=1^pi^p equiv 0 pmodp$$






              share|cite|improve this answer
























                up vote
                2
                down vote













                And yes, it can be solved using LFT as well, since
                $$i^pequiv i pmodp$$
                thus
                $$sumlimits_i=1^pi^p equiv sumlimits_i=1^pi pmodp tag1$$
                but
                $$sumlimits_i=1^pi = fracp(p+1)2$$
                and $forall p>2$ primes are odd, thus $2 mid p+1$
                which means
                $$sumlimits_i=1^pi = fracp(p+1)2 equiv 0 pmodp$$
                and from $(1)$
                $$sumlimits_i=1^pi^p equiv 0 pmodp$$






                share|cite|improve this answer






















                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  And yes, it can be solved using LFT as well, since
                  $$i^pequiv i pmodp$$
                  thus
                  $$sumlimits_i=1^pi^p equiv sumlimits_i=1^pi pmodp tag1$$
                  but
                  $$sumlimits_i=1^pi = fracp(p+1)2$$
                  and $forall p>2$ primes are odd, thus $2 mid p+1$
                  which means
                  $$sumlimits_i=1^pi = fracp(p+1)2 equiv 0 pmodp$$
                  and from $(1)$
                  $$sumlimits_i=1^pi^p equiv 0 pmodp$$






                  share|cite|improve this answer












                  And yes, it can be solved using LFT as well, since
                  $$i^pequiv i pmodp$$
                  thus
                  $$sumlimits_i=1^pi^p equiv sumlimits_i=1^pi pmodp tag1$$
                  but
                  $$sumlimits_i=1^pi = fracp(p+1)2$$
                  and $forall p>2$ primes are odd, thus $2 mid p+1$
                  which means
                  $$sumlimits_i=1^pi = fracp(p+1)2 equiv 0 pmodp$$
                  and from $(1)$
                  $$sumlimits_i=1^pi^p equiv 0 pmodp$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 26 at 10:14









                  rtybase

                  9,04221433




                  9,04221433



























                       

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