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  "body": "\n## Proof\n\n1. Let $n$ be a positive integer. By the definition of the [[Dirichlet convolution|0KH1N66QWW]], $(\\mu * \\mathbf{1})(n) = \\sum_{d \\mid n} \\mu(d) \\mathbf{1}(n/d) = \\sum_{d \\mid n} \\mu(d)$.\n2. Case $n = 1$. The only divisor is $d = 1$. No prime square divides $1$, so $1$ is [[squarefree|004JW7A2N5]], and $\\omega(1) = 0$. So the sum is $\\mu(1) = (-1)^0 = 1$.\n3. Case $n \u003e 1$. By the [[fundamental theorem of arithmetic|084HH1NDCB]], $n = p_1^{e_1} \\cdots p_k^{e_k}$ with distinct primes $p_i$, exponents $e_i \\ge 1$, and $k \\ge 1$.\n4. By the definition of the [[Möbius function|0C8QENHYZW]], $\\mu(d) = 0$ unless $d$ is squarefree. So only squarefree divisors $d$ of $n$ contribute to the sum.\n5. A divisor of $n$ has the form $p_1^{c_1} \\cdots p_k^{c_k}$ with $0 \\le c_i \\le e_i$. It is squarefree exactly when each $c_i$ is $0$ or $1$.\n6. So the squarefree divisors of $n$ are the products $d_S = \\prod_{i \\in S} p_i$, one for each subset $S$ of $\\set{1, \\dots, k}$. Distinct subsets give distinct divisors, by uniqueness of factorization.\n7. The divisor $d_S$ has exactly $\\abs{S}$ distinct prime factors, so $\\omega(d_S) = \\abs{S}$ (see [[the number of distinct prime factors|0YVVXT3A0B]]). Hence $\\mu(d_S) = (-1)^{\\abs{S}}$.\n8. There are $\\binom{k}{j}$ subsets $S$ with $\\abs{S} = j$. So\n$$\\sum_{d \\mid n} \\mu(d) = \\sum_{j=0}^{k} \\binom{k}{j} (-1)^j.$$\n9. By the [[binomial theorem|0EHWPWGHB2]] with $x = -1$ and $y = 1$, this sum equals $(1 - 1)^k = 0$, since $k \\ge 1$.\n10. So $(\\mu * \\mathbf{1})(n)$ is $1$ for $n = 1$ and $0$ for $n \u003e 1$. That is, $\\mu * \\mathbf{1} = \\varepsilon$.\n\n## Idea\n\nOnly squarefree divisors count, and these correspond to subsets of the distinct primes of $n$. The sign $(-1)^{\\abs{S}}$ then turns the sum into $(1-1)^k$ by the binomial theorem. This vanishes as soon as $n$ has a prime factor.\n",
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