For which $p$ primes is $p^{p-1}$ a divisor of $(p-1)^p + 1$?












4












$begingroup$


$p = 2$ and $p = 3$ definitely are solutions. I think these are all the solutions, but how can I prove it?






number-theory prime-numbers divisibility







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  • $begingroup$
    Use binomial theorem to compute the reminder of what is left after diving by $p^2$
    $endgroup$
    – rtybase
    Jan 7 at 11:43


















4












$begingroup$


$p = 2$ and $p = 3$ definitely are solutions. I think these are all the solutions, but how can I prove it?






number-theory prime-numbers divisibility







share|cite|improve this question









$endgroup$












  • $begingroup$
    Use binomial theorem to compute the reminder of what is left after diving by $p^2$
    $endgroup$
    – rtybase
    Jan 7 at 11:43
















4












4








4





$begingroup$


$p = 2$ and $p = 3$ definitely are solutions. I think these are all the solutions, but how can I prove it?






number-theory prime-numbers divisibility







share|cite|improve this question









$endgroup$




$p = 2$ and $p = 3$ definitely are solutions. I think these are all the solutions, but how can I prove it?






number-theory prime-numbers divisibility




number-theory prime-numbers divisibility






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Jan 7 at 11:34









Krisztián KissKrisztián Kiss

484




484












  • $begingroup$
    Use binomial theorem to compute the reminder of what is left after diving by $p^2$
    $endgroup$
    – rtybase
    Jan 7 at 11:43




















  • $begingroup$
    Use binomial theorem to compute the reminder of what is left after diving by $p^2$
    $endgroup$
    – rtybase
    Jan 7 at 11:43


















$begingroup$
Use binomial theorem to compute the reminder of what is left after diving by $p^2$
$endgroup$
– rtybase
Jan 7 at 11:43






$begingroup$
Use binomial theorem to compute the reminder of what is left after diving by $p^2$
$endgroup$
– rtybase
Jan 7 at 11:43












2 Answers
2






active

oldest

votes


















3












$begingroup$

We consider the given number $N(p) =(p-1)^p+1$ modulo $p^3$. In the following, $p$ is a prime $>3$. Using the binomial formula we have:
$$
begin{aligned}
N(p)&equiv
1+
(-1)^p
+ pbinom p1(-1)^{p-1}
+ p^2binom p2(-1)^{p-2}
\
&=
1+
(-1)
+ p^2
- p^2frac 12p(p-1)
\
&=
p^2(text{Number no divisible by $p$}) .
end{aligned}
$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    @rtybase... yes, did not see the comment, started it as an answer...
    $endgroup$
    – dan_fulea
    Jan 7 at 11:55



















2












$begingroup$

According to binomial theorem and assuming $p$-odd, thus $(-1)^p=-1$ and $(-1)^{p-1}=1$
$$(p-1)^p+1=sumlimits_{k=0}^p binom{p}{k}p^k(-1)^{p-k}+1=\
-1+sumlimits_{color{red}{k=1}}^p binom{p}{k}p^k(-1)^{p-k} +1=p^2+sumlimits_{color{red}{k=2}}^p binom{p}{k}p^k(-1)^{p-k}=\
p^2left(1+sumlimits_{k=2}^p binom{p}{k}p^{color{red}{k-2}}(-1)^{p-k}right)=...$$
also $binom{p}{2}=frac{p(p-1)}{2}$ so we can extract a $p$ from it, thus
$$...=p^{color{red}{2}}left(1+pcdot Qright)$$
and the answer follows.






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









    3












    $begingroup$

    We consider the given number $N(p) =(p-1)^p+1$ modulo $p^3$. In the following, $p$ is a prime $>3$. Using the binomial formula we have:
    $$
    begin{aligned}
    N(p)&equiv
    1+
    (-1)^p
    + pbinom p1(-1)^{p-1}
    + p^2binom p2(-1)^{p-2}
    \
    &=
    1+
    (-1)
    + p^2
    - p^2frac 12p(p-1)
    \
    &=
    p^2(text{Number no divisible by $p$}) .
    end{aligned}
    $$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      @rtybase... yes, did not see the comment, started it as an answer...
      $endgroup$
      – dan_fulea
      Jan 7 at 11:55
















    3












    $begingroup$

    We consider the given number $N(p) =(p-1)^p+1$ modulo $p^3$. In the following, $p$ is a prime $>3$. Using the binomial formula we have:
    $$
    begin{aligned}
    N(p)&equiv
    1+
    (-1)^p
    + pbinom p1(-1)^{p-1}
    + p^2binom p2(-1)^{p-2}
    \
    &=
    1+
    (-1)
    + p^2
    - p^2frac 12p(p-1)
    \
    &=
    p^2(text{Number no divisible by $p$}) .
    end{aligned}
    $$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      @rtybase... yes, did not see the comment, started it as an answer...
      $endgroup$
      – dan_fulea
      Jan 7 at 11:55














    3












    3








    3





    $begingroup$

    We consider the given number $N(p) =(p-1)^p+1$ modulo $p^3$. In the following, $p$ is a prime $>3$. Using the binomial formula we have:
    $$
    begin{aligned}
    N(p)&equiv
    1+
    (-1)^p
    + pbinom p1(-1)^{p-1}
    + p^2binom p2(-1)^{p-2}
    \
    &=
    1+
    (-1)
    + p^2
    - p^2frac 12p(p-1)
    \
    &=
    p^2(text{Number no divisible by $p$}) .
    end{aligned}
    $$






    share|cite|improve this answer









    $endgroup$



    We consider the given number $N(p) =(p-1)^p+1$ modulo $p^3$. In the following, $p$ is a prime $>3$. Using the binomial formula we have:
    $$
    begin{aligned}
    N(p)&equiv
    1+
    (-1)^p
    + pbinom p1(-1)^{p-1}
    + p^2binom p2(-1)^{p-2}
    \
    &=
    1+
    (-1)
    + p^2
    - p^2frac 12p(p-1)
    \
    &=
    p^2(text{Number no divisible by $p$}) .
    end{aligned}
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 7 at 11:53









    dan_fuleadan_fulea

    7,1781513




    7,1781513








    • 1




      $begingroup$
      @rtybase... yes, did not see the comment, started it as an answer...
      $endgroup$
      – dan_fulea
      Jan 7 at 11:55














    • 1




      $begingroup$
      @rtybase... yes, did not see the comment, started it as an answer...
      $endgroup$
      – dan_fulea
      Jan 7 at 11:55








    1




    1




    $begingroup$
    @rtybase... yes, did not see the comment, started it as an answer...
    $endgroup$
    – dan_fulea
    Jan 7 at 11:55




    $begingroup$
    @rtybase... yes, did not see the comment, started it as an answer...
    $endgroup$
    – dan_fulea
    Jan 7 at 11:55











    2












    $begingroup$

    According to binomial theorem and assuming $p$-odd, thus $(-1)^p=-1$ and $(-1)^{p-1}=1$
    $$(p-1)^p+1=sumlimits_{k=0}^p binom{p}{k}p^k(-1)^{p-k}+1=\
    -1+sumlimits_{color{red}{k=1}}^p binom{p}{k}p^k(-1)^{p-k} +1=p^2+sumlimits_{color{red}{k=2}}^p binom{p}{k}p^k(-1)^{p-k}=\
    p^2left(1+sumlimits_{k=2}^p binom{p}{k}p^{color{red}{k-2}}(-1)^{p-k}right)=...$$
    also $binom{p}{2}=frac{p(p-1)}{2}$ so we can extract a $p$ from it, thus
    $$...=p^{color{red}{2}}left(1+pcdot Qright)$$
    and the answer follows.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      According to binomial theorem and assuming $p$-odd, thus $(-1)^p=-1$ and $(-1)^{p-1}=1$
      $$(p-1)^p+1=sumlimits_{k=0}^p binom{p}{k}p^k(-1)^{p-k}+1=\
      -1+sumlimits_{color{red}{k=1}}^p binom{p}{k}p^k(-1)^{p-k} +1=p^2+sumlimits_{color{red}{k=2}}^p binom{p}{k}p^k(-1)^{p-k}=\
      p^2left(1+sumlimits_{k=2}^p binom{p}{k}p^{color{red}{k-2}}(-1)^{p-k}right)=...$$
      also $binom{p}{2}=frac{p(p-1)}{2}$ so we can extract a $p$ from it, thus
      $$...=p^{color{red}{2}}left(1+pcdot Qright)$$
      and the answer follows.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        According to binomial theorem and assuming $p$-odd, thus $(-1)^p=-1$ and $(-1)^{p-1}=1$
        $$(p-1)^p+1=sumlimits_{k=0}^p binom{p}{k}p^k(-1)^{p-k}+1=\
        -1+sumlimits_{color{red}{k=1}}^p binom{p}{k}p^k(-1)^{p-k} +1=p^2+sumlimits_{color{red}{k=2}}^p binom{p}{k}p^k(-1)^{p-k}=\
        p^2left(1+sumlimits_{k=2}^p binom{p}{k}p^{color{red}{k-2}}(-1)^{p-k}right)=...$$
        also $binom{p}{2}=frac{p(p-1)}{2}$ so we can extract a $p$ from it, thus
        $$...=p^{color{red}{2}}left(1+pcdot Qright)$$
        and the answer follows.






        share|cite|improve this answer











        $endgroup$



        According to binomial theorem and assuming $p$-odd, thus $(-1)^p=-1$ and $(-1)^{p-1}=1$
        $$(p-1)^p+1=sumlimits_{k=0}^p binom{p}{k}p^k(-1)^{p-k}+1=\
        -1+sumlimits_{color{red}{k=1}}^p binom{p}{k}p^k(-1)^{p-k} +1=p^2+sumlimits_{color{red}{k=2}}^p binom{p}{k}p^k(-1)^{p-k}=\
        p^2left(1+sumlimits_{k=2}^p binom{p}{k}p^{color{red}{k-2}}(-1)^{p-k}right)=...$$
        also $binom{p}{2}=frac{p(p-1)}{2}$ so we can extract a $p$ from it, thus
        $$...=p^{color{red}{2}}left(1+pcdot Qright)$$
        and the answer follows.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 7 at 12:12

























        answered Jan 7 at 12:06









        rtybasertybase

        11.6k31534




        11.6k31534






























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