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Actually, I posted this long ago in MO but did not get a reply as it was unfit.

Now this is an exercise in some textbook (I think Apostol), and I would be happy to receive some answers.

Let $P(n)$ be the product of positive integers which are $\leq n$ and relatively prime to $n$. Prove that $ \displaystyle P(n) = n^{\phi(n)} \prod\limits_{d \mid n} \left(\frac{d!}{d^d} \right)^{\mu(n/d)}.$

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    @Mariano: I think i have got it now.!2010-10-30

2 Answers 2

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Success finally!

Let,

$f( n) = \sum_{(k,n)=1;1\leq k\leq n} \log\Bigl(\frac{k}{n}\Bigr)$ therefore we have

$\sum_{d|n}f(d) =\log\Bigl(\frac{1}{n}\Bigr)+...+\log\Bigl(\frac{n}{n}\Bigr)=\log\left(\frac{n!}{n^n}\right)$

Thus by Moebius Inversion Formula:

$f(n) = \sum_{d|n}\log\left(\frac{d!}{d^d}\right)\cdot \mu\left(\frac{n}{d}\right) = \log\left(\prod_{d|n}\left(\frac{d!}{d^d}\right)^{\mu\left(\frac{n}{d}\right) }\right)$

$f(n) = \sum_{(k,n)=1;1\leq k\leq n} {\log(k)} -\phi(n)\cdot \log( n) = \log(P(n))-\log(n^{\phi(n)})$

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Start by classifying $[n]$ according to GCD:

$n! = \prod_{d|n} \prod_{(q,n)=d} q$

where $q$ ranges from $1$ to $n.$ This is

$n! = \prod_{d|n} \prod_{(r,n/d)=1} (dr) = \prod_{d|n} d^{\varphi(n/d)} \prod_{(r,n/d)=1} r \\ = \prod_{d|n} d^{\varphi(n/d)} P(n/d). $

This becomes

$n! = \prod_{d|n} (n/d)^{\varphi(d)} P(d) = \prod_{d|n} n^{\varphi(d)} \prod_{d|n} d^{-\varphi(d)} P(d) \\ = n^n \prod_{d|n} d^{-\varphi(d)} P(d).$

so that we find

$\prod_{d|n} d^{-\varphi(d)} P(d) = \frac{n!}{n^n}.$

By Mobius inversion we thus have

$n^{\Large -\varphi(n)} P(n) = \prod_{d|n} \left(\frac{d!}{d^d}\right)^{\Large \mu(n/d)}.$

This finally yields

$\bbox[5px,border:2px solid #00A000]{ P(n) = n^{\Large \varphi(n)} \prod_{d|n} \left(\frac{d!}{d^d}\right)^{\Large \mu(n/d)}.}$

as claimed.