# 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 $p$.

Mathematics Asked by Shubhangi on September 7, 2020

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.

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

(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 $$p$$.

Correct answer by user10354138 on September 7, 2020

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!

Answered by Shubhangi on September 7, 2020

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