arXiv 1108.41782011-08-21math.NT

Wolstenholme素数的同余

Congruences for Wolstenholme primes

Romeo Mestrovic

备注：: pages 16
发表信息：: Czechoslovak Mathematical Journal 65, No 1 (2015), 237-253

摘要：

一个素数$p$称为Wolstenholme素数，如果它满足同余式${2p-1\choose p-1} \equiv 1 \,\,(\bmod{\,\,p^4})$。对于这样一个素数$p$，我们建立了${2p-1\choose p-1}\,\,(\bmod{\,\,p^8})$的表达式，该表达式用和$R_i:=\sum_{k=1}^{p-1}1/k^i$（$i=1,2,3,4,5,6)$）给出。此外，该同余式中的表达式根据和$R_i$（$i=1,3,4,5$）来简化。利用这个同余，我们证明了对任何Wolstenholme素数，$$ {2p-1\choose p-1}\equiv 1 -2p \sum_{k=1}^{p-1}\frac{1}{k} -2p^2\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^7}. $$.此外，利用作者\cite{Me}最近的一个结果，我们证明了上述同余蕴含一个素数$p$必然是Wolstenholme素数.应用Helou和Terjanian \cite{HT}的技巧，给出了上述同余式的包含Bernoulli数的表达式.

英文摘要：

A prime number $p$ is said to be a Wolstenholme prime if it satisfies the congruence ${2p-1\choose p-1} \equiv 1 \,\,(\bmod{\,\,p^4})$. For such a prime $p$, we establish the expression for ${2p-1\choose p-1}\,\,(\bmod{\,\,p^8})$ given in terms of the sums $R_i:=\sum_{k=1}^{p-1}1/k^i$ ($i=1,2,3,4,5,6)$. Further, the expression in this congruence is reduced in terms of the sums $R_i$ ($i=1,3,4,5$). Using this congruence, we prove that for any Wolstenholme prime, $$ {2p-1\choose p-1}\equiv 1 -2p \sum_{k=1}^{p-1}\frac{1}{k} -2p^2\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^7}. $$ Moreover, using a recent result of the author \cite{Me}, we prove that the above congruence implies that a prime $p$ necessarily must be a Wolstenholme prime. Applying a technique of Helou and Terjanian \cite{HT}, the above congruence is given as the expression involving the Bernoulli numbers.

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