Prove $sum_{n = 1}^{p - 1} n^{p - 1} equiv (p - 1)! + p pmod {p^2}$ for $p$ being an odd prime












12












$begingroup$


I need to prove the following:




$$sum_{n = 1}^{p - 1} n^{p - 1} equiv (p - 1)! + p pmod {p^2}$$




...with $p$ being an odd prime number. The statement is obviously true for$pmod p$ because left-hand side is congruent to $-1 pmod p$ by Fermat's little theorem, and the right-hand side also turns out to be congruent to $-1 pmod p$ by Wilson's theorem. Now, I am not sure how to make a jump from$pmod p$ to$pmod {p^2}$, if that is even possible. Maybe the sum on the left could somehow be modified using the existence of the primitive root$pmod {p^2}$.



EDIT: Elementary solution can be found here: https://mathoverflow.net/a/319824/134054






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    @MohammadZuhairKhan Yes, thanks. I have corrected the formatting
    $endgroup$
    – Oldboy
    Dec 29 '18 at 11:56










  • $begingroup$
    This is probably true for odd $p$
    $endgroup$
    – greedoid
    Dec 29 '18 at 12:10










  • $begingroup$
    My bad, I have corrected the statement. I apologize for that.
    $endgroup$
    – Oldboy
    Dec 29 '18 at 12:13






  • 1




    $begingroup$
    @Oldboy Isn't Fermat's little theorem, only $a^p equiv a pmod p$ and not for the series?
    $endgroup$
    – toric_actions
    Dec 29 '18 at 12:46






  • 3




    $begingroup$
    Elementary proof can be found here: mathoverflow.net/a/319824/134054
    $endgroup$
    – Oldboy
    Dec 31 '18 at 17:29
















12












$begingroup$


I need to prove the following:




$$sum_{n = 1}^{p - 1} n^{p - 1} equiv (p - 1)! + p pmod {p^2}$$




...with $p$ being an odd prime number. The statement is obviously true for$pmod p$ because left-hand side is congruent to $-1 pmod p$ by Fermat's little theorem, and the right-hand side also turns out to be congruent to $-1 pmod p$ by Wilson's theorem. Now, I am not sure how to make a jump from$pmod p$ to$pmod {p^2}$, if that is even possible. Maybe the sum on the left could somehow be modified using the existence of the primitive root$pmod {p^2}$.



EDIT: Elementary solution can be found here: https://mathoverflow.net/a/319824/134054






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    @MohammadZuhairKhan Yes, thanks. I have corrected the formatting
    $endgroup$
    – Oldboy
    Dec 29 '18 at 11:56










  • $begingroup$
    This is probably true for odd $p$
    $endgroup$
    – greedoid
    Dec 29 '18 at 12:10










  • $begingroup$
    My bad, I have corrected the statement. I apologize for that.
    $endgroup$
    – Oldboy
    Dec 29 '18 at 12:13






  • 1




    $begingroup$
    @Oldboy Isn't Fermat's little theorem, only $a^p equiv a pmod p$ and not for the series?
    $endgroup$
    – toric_actions
    Dec 29 '18 at 12:46






  • 3




    $begingroup$
    Elementary proof can be found here: mathoverflow.net/a/319824/134054
    $endgroup$
    – Oldboy
    Dec 31 '18 at 17:29














12












12








12


8



$begingroup$


I need to prove the following:




$$sum_{n = 1}^{p - 1} n^{p - 1} equiv (p - 1)! + p pmod {p^2}$$




...with $p$ being an odd prime number. The statement is obviously true for$pmod p$ because left-hand side is congruent to $-1 pmod p$ by Fermat's little theorem, and the right-hand side also turns out to be congruent to $-1 pmod p$ by Wilson's theorem. Now, I am not sure how to make a jump from$pmod p$ to$pmod {p^2}$, if that is even possible. Maybe the sum on the left could somehow be modified using the existence of the primitive root$pmod {p^2}$.



EDIT: Elementary solution can be found here: https://mathoverflow.net/a/319824/134054






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$




I need to prove the following:




$$sum_{n = 1}^{p - 1} n^{p - 1} equiv (p - 1)! + p pmod {p^2}$$




...with $p$ being an odd prime number. The statement is obviously true for$pmod p$ because left-hand side is congruent to $-1 pmod p$ by Fermat's little theorem, and the right-hand side also turns out to be congruent to $-1 pmod p$ by Wilson's theorem. Now, I am not sure how to make a jump from$pmod p$ to$pmod {p^2}$, if that is even possible. Maybe the sum on the left could somehow be modified using the existence of the primitive root$pmod {p^2}$.



EDIT: Elementary solution can be found here: https://mathoverflow.net/a/319824/134054






elementary-number-theory modular-arithmetic




elementary-number-theory modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 6:05







Oldboy

















asked Dec 29 '18 at 11:52









OldboyOldboy

8,4521936




8,4521936












  • $begingroup$
    @MohammadZuhairKhan Yes, thanks. I have corrected the formatting
    $endgroup$
    – Oldboy
    Dec 29 '18 at 11:56










  • $begingroup$
    This is probably true for odd $p$
    $endgroup$
    – greedoid
    Dec 29 '18 at 12:10










  • $begingroup$
    My bad, I have corrected the statement. I apologize for that.
    $endgroup$
    – Oldboy
    Dec 29 '18 at 12:13






  • 1




    $begingroup$
    @Oldboy Isn't Fermat's little theorem, only $a^p equiv a pmod p$ and not for the series?
    $endgroup$
    – toric_actions
    Dec 29 '18 at 12:46






  • 3




    $begingroup$
    Elementary proof can be found here: mathoverflow.net/a/319824/134054
    $endgroup$
    – Oldboy
    Dec 31 '18 at 17:29


















  • $begingroup$
    @MohammadZuhairKhan Yes, thanks. I have corrected the formatting
    $endgroup$
    – Oldboy
    Dec 29 '18 at 11:56










  • $begingroup$
    This is probably true for odd $p$
    $endgroup$
    – greedoid
    Dec 29 '18 at 12:10










  • $begingroup$
    My bad, I have corrected the statement. I apologize for that.
    $endgroup$
    – Oldboy
    Dec 29 '18 at 12:13






  • 1




    $begingroup$
    @Oldboy Isn't Fermat's little theorem, only $a^p equiv a pmod p$ and not for the series?
    $endgroup$
    – toric_actions
    Dec 29 '18 at 12:46






  • 3




    $begingroup$
    Elementary proof can be found here: mathoverflow.net/a/319824/134054
    $endgroup$
    – Oldboy
    Dec 31 '18 at 17:29
















$begingroup$
@MohammadZuhairKhan Yes, thanks. I have corrected the formatting
$endgroup$
– Oldboy
Dec 29 '18 at 11:56




$begingroup$
@MohammadZuhairKhan Yes, thanks. I have corrected the formatting
$endgroup$
– Oldboy
Dec 29 '18 at 11:56












$begingroup$
This is probably true for odd $p$
$endgroup$
– greedoid
Dec 29 '18 at 12:10




$begingroup$
This is probably true for odd $p$
$endgroup$
– greedoid
Dec 29 '18 at 12:10












$begingroup$
My bad, I have corrected the statement. I apologize for that.
$endgroup$
– Oldboy
Dec 29 '18 at 12:13




$begingroup$
My bad, I have corrected the statement. I apologize for that.
$endgroup$
– Oldboy
Dec 29 '18 at 12:13




1




1




$begingroup$
@Oldboy Isn't Fermat's little theorem, only $a^p equiv a pmod p$ and not for the series?
$endgroup$
– toric_actions
Dec 29 '18 at 12:46




$begingroup$
@Oldboy Isn't Fermat's little theorem, only $a^p equiv a pmod p$ and not for the series?
$endgroup$
– toric_actions
Dec 29 '18 at 12:46




3




3




$begingroup$
Elementary proof can be found here: mathoverflow.net/a/319824/134054
$endgroup$
– Oldboy
Dec 31 '18 at 17:29




$begingroup$
Elementary proof can be found here: mathoverflow.net/a/319824/134054
$endgroup$
– Oldboy
Dec 31 '18 at 17:29










1 Answer
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$begingroup$

A Hint: notice that




$$sum^{p-1}_{n=1}n^{p-1}=frac{1}{p}sum^{p}_{n=1}C^{p-n}_{p}B_{p-n}p^n=sum^{p}_{n=1}C^{p-n}_{p}B_{p-n}p^{n-1}$$
where $B_k$ is the $k$-th Bernoulli number.




(For further information about Bernoulli number: https://www.bernoulli.org/)






share|cite|improve this answer











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    $begingroup$

    A Hint: notice that




    $$sum^{p-1}_{n=1}n^{p-1}=frac{1}{p}sum^{p}_{n=1}C^{p-n}_{p}B_{p-n}p^n=sum^{p}_{n=1}C^{p-n}_{p}B_{p-n}p^{n-1}$$
    where $B_k$ is the $k$-th Bernoulli number.




    (For further information about Bernoulli number: https://www.bernoulli.org/)






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      A Hint: notice that




      $$sum^{p-1}_{n=1}n^{p-1}=frac{1}{p}sum^{p}_{n=1}C^{p-n}_{p}B_{p-n}p^n=sum^{p}_{n=1}C^{p-n}_{p}B_{p-n}p^{n-1}$$
      where $B_k$ is the $k$-th Bernoulli number.




      (For further information about Bernoulli number: https://www.bernoulli.org/)






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        A Hint: notice that




        $$sum^{p-1}_{n=1}n^{p-1}=frac{1}{p}sum^{p}_{n=1}C^{p-n}_{p}B_{p-n}p^n=sum^{p}_{n=1}C^{p-n}_{p}B_{p-n}p^{n-1}$$
        where $B_k$ is the $k$-th Bernoulli number.




        (For further information about Bernoulli number: https://www.bernoulli.org/)






        share|cite|improve this answer











        $endgroup$



        A Hint: notice that




        $$sum^{p-1}_{n=1}n^{p-1}=frac{1}{p}sum^{p}_{n=1}C^{p-n}_{p}B_{p-n}p^n=sum^{p}_{n=1}C^{p-n}_{p}B_{p-n}p^{n-1}$$
        where $B_k$ is the $k$-th Bernoulli number.




        (For further information about Bernoulli number: https://www.bernoulli.org/)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 31 '18 at 10:37









        Martin Sleziak

        44.7k10119272




        44.7k10119272










        answered Dec 29 '18 at 13:42









        LauLau

        517315




        517315






























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