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For a discrete math course I'm taking, I was solving the following question:

Given that $\mathbb{Z}_{n}^{*}=\left\{a \in \mathbb{Z}_{n} \mid g \operatorname{cd}(a, n)=1\right\} . \text { Let } \varphi(n)=\left|\mathbb{Z}_{n}^{*}\right|$, show that for every $n$, we have $\varphi(n) = n \prod_{primes\ p|n} \left(1 - \frac{1}{p}\right)$

My approach was as follows:

Let $\mathbb{P}_i$ be the multiset whose elements represent the prime factorization of i. Then, the set $\mathbb{Z}_{i}^*$ is composed of elements, $x$ which satisfy $x \not\equiv 0 \text{ (mod p) } \forall p \in \mathbb{P}_i, x \in \mathbb{Z}_i$. By the Chinese Remainder Theorem, because the elements p are prime, and thus by definition also pairwise coprime, the total number of elements in $\mathbb{Z}_{i}^*$ is the product of the number of solutions for each congruence (mod p). The number of solutions to the congruence for a prime number $p$ for $x \in \mathbb{Z}_{n}$ is given by $n \left(1 - \frac{1}{p}\right)$ (shown in a different part of the problem set). This directly gives the desired expression.

My question is:

Is my application of the Chinese Remainder Theorem valid? Does the Chinese Remainder Theorem apply for "incongruence" expressions as well as congruence expressions? If it is invalid, how can I correct the proof to account for this?

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    $\begingroup$ I'm not quite able to follow your argument, so I'm unsure if it is valid. I've written out how I use the CRT to prove the underlying theorem. See the addendum re math.stackexchange.com/questions/3858851/… $\endgroup$ Oct 13 '20 at 4:07
  • $\begingroup$ Thank you! This proof is a really helpful reference. $\endgroup$ Oct 13 '20 at 4:12
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Well, it's not clear what you mean by the "incongruence" version of the Chinese remainder theorem.

But one thing we can definitely say is the following. Say we know $x \not\equiv 0 \pmod{p_1}$, $x \not\equiv 0 \pmod{p_2}$, and so on. Then there are $p_1 -1$ options for what $x$ could be modulo $p_1$; $p_2 - 1$ options modulo $p_2$, and so on.

For every one of the $(p_1 - 1)(p_2-1)\cdots$ combinations $(b_1, b_2, \dots)$ of these options, you could write down the equations $x \equiv b_1 \pmod{p_1}$, $x \equiv b_2 \pmod{p_2}$, and so on. Here, the ordinary Chinese remainder theorem applies, telling us that there is a unique answer modulo $p_1p_2\cdots$.


We need a slightly different (but very similar) argument when higher powers of a prime divide $n$, so watch out for that.

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Clearly for prime $p$ we have $\varphi(p^n)=p^n(1-1/p)$, by counting.

But, by CRT, $\varphi$ is multiplicative. That is, $(m,n)=1\implies \varphi(mn)=\varphi(m)\cdot\varphi(n)$.

The result follows.

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