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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.

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  • $\begingroup$ 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 $$V_p((p-1)!+1) \le \log_p((p-1)!+1) \le \log_p(p^p) \le p$$ $\endgroup$
    – Joel Cohen
    Feb 27, 2012 at 10:50
  • $\begingroup$ 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 $p$? $\endgroup$
    – ksj03
    Feb 27, 2012 at 10:58
  • $\begingroup$ 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. $\endgroup$
    – Sangchul Lee
    Feb 27, 2012 at 11:03
  • $\begingroup$ Thanks. That is what I mean. Although I do not know if it is true or false. $\endgroup$
    – ksj03
    Feb 27, 2012 at 11:07
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    $\begingroup$ @GerryMyerson: The lower bound in the OEIS entry has meanwhile been updated to $2\cdot10^{13}$. $\endgroup$
    – joriki
    Jun 28, 2016 at 8:43

1 Answer 1

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$$\lfloor \frac{n-1}{p-1} \rfloor - k \leq v_p(n !) \leq \lfloor \frac{n-1}{p-1} \rfloor$$

with $k$ defined by $ p^k \leq n < p^{k+1} $

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  • $\begingroup$ You can make $\lfloor$ and $\rfloor$ (or any pair of delimiters) adapt to the size of their content by preceding them with \left and \right, respectively. $\endgroup$
    – joriki
    Jun 28, 2016 at 8:40

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