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Let $V_p$ be the $p$-adic valuation. We know that $(p - 1)! + 1\equiv0\mod p$ for the prime $p$ by Wilson's theorem. I wonder if there is an upper bound for $V_p((p - 1)! + 1)$.

Also I do not know how to prove the following statement: Let $p\equiv 7\mod8$ be a prime. Then $\sum\limits_{r = 1}^{\frac{p - 1}{2}}r(\frac{r}{p}) = 0$, where $(\frac{\cdot}{\cdot})$ is the Legendre symbol.

Could anybody help me to answer these questions? Thanks.

  • 0
    Might not be very interesting, but there's always an easy upper bound for the valuation of any integer $n \in \mathbb{N}^*$, namely : $$V_p(n) \le \log_p(n)$$ So here you could easily prove $$V_p((p-1)!+1) \le \log_p((p-1)!+1) \le \log_p(p^p) \le p$$2012-02-27
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    Thanks for your reply. This is an easy bound for the fixed prime $p$. Is there any uniform bound, i.e., which is independent of $p$?2012-02-27
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    For the first 10,000 primes, we have only three primes $p = 5, 13, 563$ with $\mathrm{ord}_p ((p-1)!+1) = 2$ and $\mathrm{ord}_p ((p-1)!+1) = 1$ otherwise. I suspect that $\mathrm{ord}_p ((p-1)!+1) = 1$ except for very rare occasion, though neither I can prove it nor disprove it.2012-02-27
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    Thanks. That is what I mean. Although I do not know if it is true or false.2012-02-27
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    I thought of this question when I worked on the exercise that the equation $(p - 1)! + 1 = p^k$ has positive integer solutions only if $p\leq 5$.2012-02-27
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    The next such prime after 563, if there is one, exceeds $4\times10^{11}$ - see http://oeis.org/A007540.2012-02-27
  • 0
    Please ask one question at a time, or directly related questions. (This would also have avoided the problem that [no-one responded to the second question](http://math.stackexchange.com/questions/114293).)2016-06-28
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    @GerryMyerson: The lower bound in the OEIS entry has meanwhile been updated to $2\cdot10^{13}$.2016-06-28

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