---
tag: 83WR72D96X
kind: proof
name: Proof of the multiplicativity of the Möbius function
msc: [11A25]
proves: 0VB9G5B034
uses: [0C8QENHYZW, 0AP80TT660, 004JW7A2N5, 0YVVXT3A0B, 051VYKA1EX, 04VCPS30ZE, 0850JJMJYZ, 0GFKP8F8AG]
sources:
  - url: https://proofwiki.org/wiki/M%C3%B6bius_Function_is_Multiplicative
---

## Proof

1. No prime square divides $1$, so $1$ is [[square-free|004JW7A2N5]]. No prime divides $1$, so $\omega(1) = 0$, where $\omega$ is the [[number of distinct prime factors|0YVVXT3A0B]]. By the definition of the [[Möbius function|0C8QENHYZW]], $\mu(1) = (-1)^0 = 1$.
2. Let $m$ and $n$ be [[coprime|0GFKP8F8AG]] positive integers.
3. If $m$ is not square-free, then $q^2 \mid m$ for some prime $q$, so $q^2 \mid mn$. Then $\mu(m) = 0$ and $\mu(mn) = 0$, so $\mu(mn) = \mu(m)\mu(n)$. The case where $n$ is not square-free is the same.
4. Now let $m$ and $n$ be square-free. Suppose $q^2 \mid mn$ for a prime $q$. By [[Euclid's lemma|051VYKA1EX]], $q \mid m$ or $q \mid n$, and not both, because $\gcd(m, n) = 1$. Say $q \mid m$ and $q \nmid n$.
5. With the [[p-adic valuation|04VCPS30ZE]] and the [[valuation of a product|0850JJMJYZ]], $\nu_q(m) = \nu_q(m) + \nu_q(n) = \nu_q(mn) \ge 2$. So $q^2 \mid m$, a contradiction. Hence $mn$ is square-free.
6. A prime divides $mn$ if and only if it divides $m$ or $n$, by Euclid's lemma. No prime divides both, since $\gcd(m, n) = 1$. So $\omega(mn) = \omega(m) + \omega(n)$.
7. Therefore $\mu(mn) = (-1)^{\omega(m) + \omega(n)} = (-1)^{\omega(m)} (-1)^{\omega(n)} = \mu(m) \mu(n)$.
8. With step 1, $\mu$ is a [[multiplicative function|0AP80TT660]].

## Idea

A square factor in $m$ or $n$ stays in $mn$, so both sides vanish. If $m$ and $n$ are square-free and coprime, the primes of $mn$ are those of $m$ together with those of $n$, each once. So the signs multiply.