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The statement of this problem is elementary... How about the proof ? Let $p$ be an odd prime number. For every integer $a$ , define the number

$ S_{a}=\frac{a}{1}+\frac{a^{2}}{2}+....+\frac{a^{p-1}}{p-1} $.

Let $m$ and $n$ be integers such that $ S_{3}+S_{4}-3S_{2}=\frac{m}{n} $. Prove that $p$ divides $m$.

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    $\begingroup$ I would like to see the proof. Voted to reopen. $\endgroup$
    – GH from MO
    Apr 29, 2012 at 15:34
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    $\begingroup$ $S_a(p) \equiv ((a-1)^p + 1 - a^p)/p \bmod p$ [proof: expand $(a-1)^p$ in a binomial series, subtract first and last term, divide by $p$, and evaluate termwise]. So $S_3 + S_4 - 3S_2 \equiv (4 \cdot 2^p - 4 - 4^p)/p$. But that's $-(2^p-2)^2/p$, which is a multiple of $p$, QED. $\endgroup$
    – Noam D. Elkies
    Apr 29, 2012 at 16:07
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    $\begingroup$ Well, the punchline is related with the fact that a nontrivial homomorphism from $({\bf Z}/p^2{\bf Z}, \times)$ to $({\bf Z}/p{\bf Z}, +)$ is given by $a \mapsto (a^{p-1}-1)/p$. (The formula for $S_a$ looks like a logarithm...) $\endgroup$
    – Noam D. Elkies
    Apr 29, 2012 at 22:57
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    $\begingroup$ @Gerry: If I find a question interesting and nontrivial, then I like to see the answer for it. All this is highly subjective, I know. $\endgroup$
    – GH from MO
    Apr 29, 2012 at 23:20
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    $\begingroup$ @Noam: Can you elaborate on your last comment? It sounds rather interesting. Thanks. $\endgroup$
    – GH from MO
    Apr 29, 2012 at 23:21

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