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A well known congruence of Wolstenholme states that $$\frac1{1^2}+\frac1{2^2}+\cdots+\frac1{(p-1)^2}\equiv0\pmod{p}$$ for any prime $p>3$. For each $n=3,4,\ldots$ we clearly have $$\frac1{1\times2}+\frac1{2\times3}+\cdots+\frac1{(n-1)n}+\frac1{n\times1} = 1.$$

Motivated by the above, here I ask a new question.

Question. Is it true that for each prime $p>3$ there is a permutation $\pi\in S_{p-1}$ with $\pi(p-1)=p-1$ and $\pi(p-2)=p-2$ such that the congruence $$\frac1{\pi(1)\pi(2)}+\frac1{\pi(2)\pi(3)}+\cdots+\frac1{\pi(p-2)\pi(p-1)}+\frac1{\pi(p-1)\pi(1)}\equiv0\pmod{p^2}$$ holds?

For $p=5$, there is a unique permutation $\pi\in S_4$ meeting the requirement, namely, $$\frac 1{2\times1}+\frac1{1\times3}+\frac1{3\times4}+\frac1{4\times2}=\frac{25}{24}\equiv0\pmod{5^2}.$$ For $p=7$, there is also a unique permutation $\pi\in S_6$ meeting the requirement, namely, $$\frac1{2\times3}+\frac1{3\times4}+\frac1{4\times1}+\frac1{1\times5}+\frac1{5\times6}+\frac1{6\times2}=\frac{49}{60}\equiv0\pmod{7^2}.$$ For $p=11$ there are totally $323$ permutations $\pi\in S_{10}$ meeting the requirement. For $p=13$, the permutation $$(\pi(1),\ldots,\pi(12))=(1,2,3,7,4,9,5,8,10,6,11,12)$$ meets our purpose. Based on these data, I conjecture that the question has a positive answer.

Your comments are welcome!

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    $\begingroup$ Did you check the repartition of the numbers $\sum_{i=1}^{p-1}1/(\pi(i)\pi(i+1))$ (with cyclic conditions) into different classes modulo $p^2$? Since $p^2$ is much smaller than $p!$ for $p$ large, there should be only a finite number of counterexamples except if the repartition of the above numbers into classes modulo $p^2$ is very unequal (which would pe surprising and interesting in itself). $\endgroup$ Jul 15 at 7:40
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Let $$ (\pi(1),\pi(2),\ldots,\pi(p-1))=(2,3,\ldots,p-3,1,p-2,p-1). $$ Then \begin{align*} \frac1{\pi(1)\pi(2)}+\frac1{\pi(2)\pi(3)}+\cdots+\frac1{\pi(p-2)\pi(p-1)}+\frac1{\pi(p-1)\pi(1)}=&\frac{p^2}{2(p-1)(p-2)}\\ \equiv&0\pmod{p^2}. \end{align*}

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  • $\begingroup$ So, I guess this means that this the OPs conjecture is not only true for each prime $>3$ but each odd number $>3$.. $\endgroup$ Jul 16 at 15:39
  • $\begingroup$ Of course!If we replace $p$ by any odd integer $n>3$ in the construction, then the sum equals n^2/(2(n-1)(n-2)). $\endgroup$
    – C. WANG
    Jul 16 at 15:44
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Not a proof, but some numerical evidence and an alternative conjecture:

It is straightforward to search permutations that do the trick for each $p$. When doing that for all $5\leq p\leq101$ I have found permissible permutations for all $p$ except $\\{6, 8, 30, 60, 70, 78, 88, 90\\}$, all of which are non-prime. I dumped the solutions here. I hope the format is clear; the first few lines are:

5 25/24 = 5^2*1/24 [2, 1, 3, 4]
6 None
7 49/60 = 7^2*1/60 [2, 3, 4, 1, 5, 6]
8 None
9 81/112 = 9^2*1/112 [2, 3, 4, 5, 6, 1, 7, 8]
10 200/189 = 10^2*2/189 [6, 3, 2, 1, 5, 4, 7, 8, 9]
11 121/180 = 11^2*1/180 [2, 3, 4, 5, 6, 7, 8, 1, 9, 10]
12 288/385 = 12^2*2/385 [6, 1, 5, 2, 7, 4, 3, 8, 9, 10, 11]
13 169/264 = 13^2*1/264 [2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 11, 12]
14 20776/19305 = 14^2*106/19305 [11, 10, 8, 6, 1, 2, 3, 4, 5, 7, 9, 12, 13]

For $p=6$ and $p=8$, I searched all permutations exhaustively, so I'm sure that none exists. For $p\in\\{30, 60, 70, 78, 88, 90\\}$, it might just be that I didn't search long enough.

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  • $\begingroup$ Please note that in the original question, the author required that $\pi(p-2)=p-2$ and $\pi(p-1)=p-1$. However, some of your examples do not meet the requirement. $\endgroup$
    – C. WANG
    Jul 16 at 13:55
  • $\begingroup$ thanks for the heads up, I will make sure to add that requirement... $\endgroup$ Jul 16 at 14:18
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    $\begingroup$ Based on your numerical evidence, I find a constructive solution of original question. $\endgroup$
    – C. WANG
    Jul 16 at 15:29
  • $\begingroup$ @C.WANG nice work! $\endgroup$ Jul 16 at 15:30

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