2010 USAMO #5:Prove that if $frac{1}{p}-2S_q = frac{m}{n}$ for integers $m$ and $n$, then $m - n$ is divisible by $p$.

Question

Let $q = frac{3p-5}{2}$ where $p$ is an odd prime, and let $S_q = frac{1}{2cdot 3 cdot 4} + frac{1}{5cdot 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: $$2S_q = 2sum_{x=1}^{frac{q+1}{3}} frac{1}{(3x-1)(3x)(3x+1)} = sum_{x=1}^{frac{p-1}{2}} left[frac{1}{3x(3x-1)}-frac{1}{3x(3x+1)}right]\
=sum_{x=1}^{frac{p-1}{2}} left[ frac{1}{3x-1} - frac{2}{3x} +frac{1}{3x+1}right]\
=sum_{x=1}^{frac{p-1}{2}}left[ frac{1}{3x-1} + frac{1}{3x} +frac{1}{3x+1}right] - sum_{x=1}^{frac{p-1}{2}} frac{1}{x} $$
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.

user10354138 · Accepted Answer

(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}frac1nright)\
&=-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
$$
frac{m-n}{n}=-sum_{i=1}^{(p-1)/2}frac{p}{(p-i)(p+i)}=frac{ptimes text{some integer}}{text{some integer not divisible by }p}.
$$
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 $p$.

Shubhangi · Answer

With the help of @user10354138 's hints, I think I got the solution.I will be grateful if someone proof reads it.
Note that $$2S_q = 2sum_{x=1}^{frac{q+1}{3}} frac{1}{(3x-1)(3x)(3x+1)} = sum_{x=1}^{frac{p-1}{2}} left[frac{1}{3x(3x-1)}-frac{1}{3x(3x+1)}right]\
=sum_{x=1}^{frac{p-1}{2}} left[ frac{1}{3x-1} - frac{2}{3x} +frac{1}{3x+1}right]\
=sum_{x=1}^{frac{p-1}{2}}left[ frac{1}{3x-1} + frac{1}{3x} +frac{1}{3x+1}right] - sum_{x=1}^{frac{p-1}{2}} frac{1}{x}$$ .
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!

2010 USAMO #5:Prove that if $frac{1}{p}-2S_q = frac{m}{n}$ for integers $m$ and $n$, then $m - n$ is divisible by $p$.

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