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I need to prove that:

If $p$ is prime greater than or equal to five, then $pB_{p-1}$ belongs to the p-integers and more over:

$$pB_{p-1} \equiv -1 \pmod p$$ Hint:Put $N=p$ in the Faulhaber´s formula: $S_{j}(N-1)= \displaystyle\sum_{k=o}^j \frac{1}{(j-k+1)}{j \choose k}N^{j-k+1}B_{k}$ and reduce everything mod p and use the Fermat´s little theorem.

But I do not know how to reduce this: $S_{j}(p-1)= \displaystyle\sum_{k=o}^j \frac{1}{(j-k+1)}{j \choose k}p^{j-k+1}B_{k}$ to mod p, and I have problems trying to prove that $pB_{p-1}$ belongs to the p-integers, Can you help to prove the theorem please, Thank you.

user162343
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    This is a special case of Von Staudt Clausen theorem. Wikipedia has a proof. – Sungjin Kim Sep 16 '14 at 20:08
  • Do you mean this one :) One has from Fermat's little theorem,

    $m^{p-1}\equiv 1\pmod p !$ for $m=1,2,...,p-1 !$. If p-1 divides 2n then one has,

    $ m^{2n}\equiv 1\pmod p !$ for $m=1,2,...,p-1 !$. Thereafter one has,

    $\sum_{m=1}^{p-1}{(-1)^m{p-1\choose m} m^{2n}}\equiv \sum_{m=1}^{p-1}{(-1)^m{p-1\choose m}}\pmod p!$ from which (1) follows immediately.

     – user162343 Sep 16 '14 at 20:10
  • Actually, for your case, just put $j=p-1$. Then reduce mod $p$. The LHS is by Fermat's little theorem, $p-1$, and the RHS is $pB_{p-1}$. – Sungjin Kim Sep 16 '14 at 20:19
  • I do not understand you, can you be more explicit please :) thank you I apreciate your help, and can you help me with the other stuff please :) – user162343 Sep 16 '14 at 20:21

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