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Given a positive integer $n$ and a prime number $p\geq 5$. Prove that \begin{align*} &\dfrac{1}{0p+1}+\dfrac{1}{0p+2}+\dots+\dfrac{1}{0p+(p-1)}+\\ &\dfrac{1}{p+1}+\dfrac{1}{p+2}+\dots+\dfrac{1}{p+(p-1)}+\\ &\dfrac{1}{2p+1}+\dfrac{1}{2p+2}+\dots+\dfrac{1}{2p+(p-1)}+\\ &\dots +\\ &\dfrac{1}{(n-1)p+1}+\dfrac{1}{(n-1)p+2}+\dots+\dfrac{1}{(n-1)p+(p-1)}\equiv 0 \pmod{p^{2+v_p(n)}}. \end{align*} (consider rational congruence). This problem poped up while I was trying to prove $\displaystyle\binom{p^{k-1}}{n}\equiv \displaystyle\binom{p^{k}}{pn}\pmod{p^{2k+1}}$ after a sequence of algebraic and modular manipulation with lagrange polynomial ($f(x)=(x-1)(x-2)\cdots(x-(p-1))$ or something similar to that). We can easily checked that the case when $v_p(n)=0$ is true due to Wolstenhome theorem. For the general problem I'm trying to imitate Wolstenhome prove by consider the sum $\dfrac{1}{ip+j}+\dfrac{1}{ip+p-j}$ but don't have any progress, and there are multiple sums to consider. I'm wondering if anyone can help solve it or help to make some progress.

Edit: $v_p(n)$ denotes the highest power of p that divides n. We say $v_p(n)=k$ when $p^k\mid n$ but $p^{k+1}\nmid n.$ About rational congruence, we say $\dfrac{a}{b}\equiv \dfrac{c}{d}\pmod{m}$ with $a,b,c,d$ are integers satisfying $\gcd(b,m)=\gcd(d,m)=1$ and $m\mid ad-bc.$

Bill Dubuque
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Nguyễn Hữu Phi Vũ CLQĐ_ĐN
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1 Answers1

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We have $\sum_{j=1}^{p-1}\frac{1}{ip+j}=\sum_{j=1}^{p-1}\frac{1}{j}\frac{1}{\frac{ip}{j}+1}=\sum_{j=1}^{p-1}\frac{1}{j}\sum_{k\ge0}(-ip)^k(\frac{1}{j})^k=\sum_{k\ge0}(-ip)^kH_{p-1}^{(k+1)}$ where $H_{p-1}^{(k+1)}:=\sum_{j=1}^{p-1}\frac{1}{j^{k+1}}$.

Let $S:=\sum_{i=0}^{n-1}\sum_{j=1}^{p-1}\frac{1}{ip+j}$, then $$S:=\sum_{k\ge0}(-p)^k\Big(\sum_{i=0}^{n-1}i^k\Big)H_{p-1}^{(k+1)}= nH_{p-1}-p\frac{n(n-1)}{2}H_{p-1}^{(2)}+\sum_{k\ge2}(-p)^k\Big(\sum_{i=1}^{n-1}i^k\Big)H_{p-1}^{(k+1)}.$$

and since $p\ge 5 $,

$v_p(nH_{p-1}) \ge 2+ v_p(n)$ by Wolstenholmes theorem.

$v_p(p\frac{n(n-1)}{2}H_{p-1}^{(2)}) \ge 2+ v_p(n)$ by Wolstenholmes theorem.

For $k\ge2$, we write $v_p(\sum_{i=1}^{n-1}i^k)=v_p(\sum_{i=1}^{n}i^k-n^k) $.

But $v_p(\sum_{i=1}^{n}i^k) \ge v_p(n)$ when $k$ is odd and $v_p(\sum_{i=1}^{n}i^k) \ge v_p(n)-1$ when $k$ is even (see this answer),

then $v_p(\sum_{i=1}^{n-1}i^k) \ge v_p(n)$ when $k$ is odd and $v_p(\sum_{i=1}^{n-1}i^k) \ge v_p(n)-1$ when $k$ is even.

But when $k$ is even $v_p(H_{p-1}^{(k+1)}) \ge 1$, and then $v_p(\sum_{k\ge2}(-p)^k(\sum_{i=1}^{n-1}i^k)H_{p-1}^{(k+1)}) \ge 2+v_p(n)$

and finally $v_p(S) \ge 2+v_p(n). \square$

René Gy
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  • Oh wow, that's the first time I've ever seen the general binomial theorem (the 1/(ip/j+1) part) elsewhere than in generating function to prove identities and in real analysis. Thanks for your solution, I wonder if there is any document where I can find some proof of how to use the infinite expansion directly in number theory to prove congruence relation. – Nguyễn Hữu Phi Vũ CLQĐ_ĐN Jun 22 '25 at 13:54
  • @NguyễnHữuPhiVũCLQĐĐN If a $p$-adic integer $x$ is a multiple of $p$, then $x$ is "as good as infinitesimal", meaning that $x$ can be substituted into any formal power series with coefficients in $\mathbb{Z}_p$ (and even in some with coefficients in $\mathbb{Q}_p$, but this is trickier and more slippery). Here, $x=-ip$ is substituted into the identity $\dfrac{1}{X+1}=\sum\limits{k\geq 0}\left(-X\right)^k$. That this all is legit should be an easy consequence of the theory of topological rings, particularly of its easy case where the topology is defined by an ultrametric. – darij grinberg Jun 22 '25 at 15:16