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I encountered some other problem and I found a beautiful proof here Write $1/1 + 1/2 + ...1/ (p-1)=a/b$ with $(a,b)=1$. Show that $p^2 \mid a$ if $p\geq 5$. (see Thomas Andrew's post)

But I thought he may miss the proof that $(a_1,(p-1)!)=1,$ which is not trivial for me. (As for the definitions of $a_1$ and $p$ please see the link above. The definitions are simple and clear there.)

My problem is just that how to prove $(a_1,(p-1)!)=1$.

I tried to use the property that $(n,m)=(n.n+km)$ for any integers $n,m,k.$ But it turns out it makes the expression messy and dirty. And I can't go further.

Any help will be thanked.

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    $\begingroup$ Would you mind explaining why should we prove $\gcd(a_1,(p-1)!)=1$ in this one you posted? $\endgroup$ – Oolong milk tea May 11 at 6:29
  • $\begingroup$ @Oolongmilktea I found that they are not necessarily coprime. And with the explanations of Tarit and awllower, I am clear now. $\endgroup$ – maths learner May 11 at 8:06
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It is not neceassary to have $\text{gcd}(a_1,(p-1)!)= 1$. From Vieta's formula, in $f(x)$, coeff. of $x$ is sum of all products of $(p-2)$ roots of $f(x)$ choosen at a time. $$a_1=2\cdot3\cdots(p-1)+1\cdot 3\cdot \cdots(p-1)+\cdots +1\cdot 2\cdots (p-2)\\ =(p-1)!\big(1+\frac{1}{2}+\frac13+\cdots + \frac1{p-1}\big) $$

hence, the required sum is equal to $$\frac{a_1}{(p-1)!}$$ If some common factors between $a_1$ and $(p-1)!$ cancels out, then also $p^2|a_1$, as all terms of $a_1$ contains products of numbers below $p$.

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  • $\begingroup$ Oh yea, the crucial fact is that $(p-1)!$ doesn't have a factor as $p$. Thanks! $\endgroup$ – maths learner May 11 at 8:03
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By definition $$a_1=\sum_{n=1}^{p-1}(\prod_{m\ne n}m).$$ So in fact $a_1$ and $(p-1)!$ need not be co-prime (for $p=7$, we have $a_1=1764$ and $(p-1)!$ is even). The proof in the linked question only requires that $p$ does not divide $(p-1)!$, which is clear: $(p-1)!$ is divisible only by primes $<p$.

P.S. In case you are wondering why the proof only requires that $p$ does not divide $(p-1)!$, notice that we can reduce the fraction $\frac{a_1}{(p-1)!}$ to the lowest terms $\frac ab$, where $a=\frac{a_1}{\gcd(a_1,(p-1)!)}$. Since $p$ does not divide $(p-1)!$, $\frac{a_1}{\gcd(a_1,(p-1)!)}$is divisible by $p^2$, hence concluding the proof.


Hope this helps.

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    $\begingroup$ Thank you so much!! It does help. I think @tarit's post is simpler so I took his post as an answer. But your post is very systematic and helpful. :) $\endgroup$ – maths learner May 11 at 8:03

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