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Let $q = \frac{3p-5}{2}$ where $p$ is an odd prime, and let $S_q = \frac{1}{2\cdot 3 \cdot 4} + \frac{1}{5\cdot 6 \cdot 7} + \cdots + \frac{1}{q(q+1)(q+2)} $

Prove that if $\frac{1}{p}-2S_q = \frac{m}{n}$ for coprime integers $m$ and $n$, then $m - n$ is divisible by $p$.

My Progress till now: $$2S_q = 2\sum_{x=1}^{\frac{q+1}{3}} \frac{1}{(3x-1)(3x)(3x+1)} = \sum_{x=1}^{\frac{p-1}{2}} \left[\frac{1}{3x(3x-1)}-\frac{1}{3x(3x+1)}\right]\\ =\sum_{x=1}^{\frac{p-1}{2}} \left[ \frac{1}{3x-1} - \frac{2}{3x} +\frac{1}{3x+1}\right]\\ =\sum_{x=1}^{\frac{p-1}{2}}\left[ \frac{1}{3x-1} + \frac{1}{3x} +\frac{1}{3x+1}\right] - \sum_{x=1}^{\frac{p-1}{2}} \frac{1}{x} $$

With the help of @user10354138 , I have got $\frac{1}{p} - 2S_q = \frac{1}{p} + \frac{1}{1} - \sum_{k=\frac{p+1}{2}}^{\frac{3p-1}{2}}\frac{1}{k} = \frac{m}{n}$

But then I am stuck.

Please give me some hints rather than a solution. Thanks in advance.

PS: I didn't post it in AOPS, because there we don't get any guidance.

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    $\begingroup$ It would be nice if you join some chat rooms which are mainly developed for problem solving,this room can be helpful chat.stackexchange.com/rooms/77161/basic-mathematics,, this one down is a little higher level for a high schooler but anyway explore it,chat.stackexchange.com/rooms/36/mathematics $\endgroup$
    – Arjun
    Jul 28 '20 at 6:44
  • $\begingroup$ I am sorry , I am new to MSE, what is a chat room ? $\endgroup$ Jul 28 '20 at 6:47
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    $\begingroup$ People interact there, about any topic they find interesting, i am sorry i cannot introduce you to others there, i am currently suspended from chatting $\endgroup$
    – Arjun
    Jul 28 '20 at 6:49
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(Original) Hint: You are almost there with the simplification. Note that you are summing over $\frac1n$ from $n=2$ to $\frac{3p-1}2$ in the first. So $$ 2S_q+1=\sum_{n=(p+1)/2}^{(3p-1)/2}\frac1n $$ If you tweak the RHS slightly, you would be summing over $\frac1n$ as $n$ runs through representative of each of the nonzero residue classes mod $p$. So ...


Addendum (2020-07-29): As discussed in the comments, \begin{align*} \frac1p-2S_q-1&=-\left(\sum_{n=(p+1)/2}^{p-1}\frac1n+\sum_{n=p+1}^{p+(p-1)/2}\frac1n\right)\\ &=-\sum_{i=1}^{(p-1)/2}\left(\frac1{p-i}+\frac1{p+i}\right) \end{align*} and now $$ \frac1{p-i}+\frac1{p+i}=\frac{p}{(p-i)(p+i)} $$ so the numerators are divisible by $p$ and the denominators are not. So putting everything over a common denominator, we see $$ \frac{m-n}{n}=-\sum_{i=1}^{(p-1)/2}\frac{p}{(p-i)(p+i)}=\frac{p\times \text{some integer}}{\text{some integer not divisible by }p}. $$ That is, every representation of $\frac{m-n}{n}$ must have more factors of $p$ in the numerator than in the denominator, hence $m-n$ is divisible by $p$.

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  • $\begingroup$ aren't we summing 1/n from n=2 to 3(p-1)/2 +1 ? $\endgroup$ Jul 27 '20 at 4:05
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    $\begingroup$ Oops, yes it should be $(3p-1)/2$, corrected. $\endgroup$ Jul 27 '20 at 4:07
  • $\begingroup$ So we have ,$\frac{1}{p} - 2S_q = \frac{1}{p} + \frac{1}{1} - \sum_{k=\frac{p+1}{2}}^{\frac{3p-1}{2}}\frac{1}{k} = \frac{m}{n}$ $\endgroup$ Jul 27 '20 at 4:11
  • $\begingroup$ also can you explain this line , " you would be summing over 1/𝑛 as 𝑛 runs through representative of each of the nonzero residue classes mod 𝑝" $\endgroup$ Jul 27 '20 at 4:13
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    $\begingroup$ As $k$ runs through $(p+1)/2$ to $(3p-1)/2$ it is $p$ consecutive numbers you are taking reciprocal of. If you cancel out the $1/p$ then those $k$s leave remainders $1,2,\dots,p-1$ when divided by $p$, in some order. $\endgroup$ Jul 27 '20 at 4:25
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With the help of @user10354138 's hints, I think I got the solution.I will be grateful if someone proof reads it.

Note that $$2S_q = 2\sum_{x=1}^{\frac{q+1}{3}} \frac{1}{(3x-1)(3x)(3x+1)} = \sum_{x=1}^{\frac{p-1}{2}} \left[\frac{1}{3x(3x-1)}-\frac{1}{3x(3x+1)}\right]\\ =\sum_{x=1}^{\frac{p-1}{2}} \left[ \frac{1}{3x-1} - \frac{2}{3x} +\frac{1}{3x+1}\right]\\ =\sum_{x=1}^{\frac{p-1}{2}}\left[ \frac{1}{3x-1} + \frac{1}{3x} +\frac{1}{3x+1}\right] - \sum_{x=1}^{\frac{p-1}{2}} \frac{1}{x}$$ .

Proceeding further we get that,$$\frac{1}{p} - 2S_q = \frac{1}{p} + \frac{1}{1} - \sum_{k=\frac{p+1}{2}}^{\frac{3p-1}{2}}\frac{1}{k} = \frac{m}{n}$$

or we get that $$- 2S_q = \frac{1}{1} - \sum_{k=\frac{p+1}{2}}^{\frac{3p-1}{2}}\frac{1}{k} = \frac{m}{n}-\frac{1}{p}$$

Now, note that $$\sum_{k=\frac{p+1}{2}}^{\frac{3p-1}{2}}\frac{1}{k}\equiv \sum_{k=1}^{p-1}\frac1k \equiv \sum_{k=1}^{p-1}k \equiv 0$$ mod $p$

So we get that $$\frac{m}{n}\equiv 1$$ mod $p$ .

Hence we have $$1-\frac{m}{n}\equiv 0$$ mod p $$\implies m-n \equiv 0 $$ mod $p$.

And we are done!

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  • $\begingroup$ How did you get $\sum_{k=\frac{p+1}{2}}^{\frac{3p-1}{2}}\frac{1}{k}\equiv \sum_{k=1}^{p-1}\frac1k $ mod p ? I understood the rest part though.. $\endgroup$
    – Raheel
    Jul 29 '20 at 6:24
  • $\begingroup$ @Raheel , Note that $\sum_{k=\frac{p+1}{2}}^{\frac{3p-1}{2}}\frac{1}{k}$ is nothing but p consecutive integers in their multiplicative inverse modulo p . $\endgroup$ Jul 29 '20 at 6:54
  • $\begingroup$ oh..I see, thanks $\endgroup$
    – Raheel
    Jul 29 '20 at 9:14
  • $\begingroup$ You need to be careful about what you mean by taking a fraction in $\mathbb{Q}$ modulo $p$. $\endgroup$ Jul 29 '20 at 12:01

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