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  "body": "\n## Proof\n\n1. 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]], $\\mu(1) = (-1)^0 = 1$.\n2. Let $m$ and $n$ be [[coprime|0GFKP8F8AG]] positive integers.\n3. 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.\n4. 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 $q \\nmid n$.\n5. 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.\n6. 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 $\\omega(mn) = \\omega(m) + \\omega(n)$.\n7. Therefore $\\mu(mn) = (-1)^{\\omega(m) + \\omega(n)} = (-1)^{\\omega(m)} (-1)^{\\omega(n)} = \\mu(m) \\mu(n)$.\n8. With step 1, $\\mu$ is a [[multiplicative function|0AP80TT660]].\n\n## Idea\n\nA 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.\n",
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