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  "body": "\n## Proof\n\n1. Let $f$ and $g$ be [[multiplicative functions|0AP80TT660]], and let $h = f * g$ be their [[Dirichlet convolution|0KH1N66QWW]].\n2. First, $1$ is the only [[positive divisor|0FCWA16FZE]] of $1$, so $h(1) = f(1) g(1) = 1 \\cdot 1 = 1$.\n3. Now let $m$ and $n$ be [[coprime|0GFKP8F8AG]] positive integers. We show $h(mn) = h(m) h(n)$.\n4. Since $\\gcd(m, n) = 1$, no prime divides both $m$ and $n$. For a prime $p$, let $\\nu_p$ be the [[p-adic valuation|04VCPS30ZE]].\n5. Let $P$ be the set of pairs $(d_1, d_2)$ with $d_1 \\mid m$ and $d_2 \\mid n$. Define $\\Phi(d_1, d_2) = d_1 d_2$. Then $d_1 d_2$ divides $mn$, so $\\Phi$ maps $P$ to the divisors of $mn$.\n6. $\\Phi$ is onto. Let $d \\mid mn$. Put $d_1 = \\prod_{p \\mid m} p^{\\nu_p(d)}$ and $d_2 = \\prod_{p \\mid n} p^{\\nu_p(d)}$, products over primes.\n7. Every prime factor $p$ of $d$ divides $mn$, hence $m$ or $n$, but not both. So $d = d_1 d_2$ by the [[fundamental theorem of arithmetic|084HH1NDCB]].\n8. For $p \\mid m$ we have $\\nu_p(d) \\le \\nu_p(mn) = \\nu_p(m) + \\nu_p(n) = \\nu_p(m)$, by the [[valuation of a product|0850JJMJYZ]]. Hence $d_1 \\mid m$, and in the same way $d_2 \\mid n$.\n9. $\\Phi$ is one-to-one. Let $d = d_1 d_2$ with $(d_1, d_2) \\in P$, and let $p \\mid m$. Then $p \\nmid d_2$, since $d_2 \\mid n$. So $\\nu_p(d_1) = \\nu_p(d)$.\n10. All prime factors of $d_1$ divide $m$. So step 9 gives $d_1 = \\prod_{p \\mid m} p^{\\nu_p(d)}$, which depends only on $d$. Similarly $d_2$ depends only on $d$.\n11. For $(d_1, d_2) \\in P$, a common divisor of $d_1$ and $d_2$ divides $m$ and $n$, hence divides $1$. So $d_1$ and $d_2$ are coprime.\n12. In the same way $m / d_1$ divides $m$ and $n / d_2$ divides $n$, so they are coprime. Also $mn / (d_1 d_2) = (m/d_1)(n/d_2)$.\n13. By steps 5 to 10, we can sum over $P$ instead of over the divisors of $mn$:\n$$h(mn) = \\sum_{(d_1, d_2) \\in P} f(d_1 d_2) \\, g\\!\\left(\\frac{m}{d_1} \\cdot \\frac{n}{d_2}\\right).$$\n14. By steps 11 and 12 and the multiplicativity of $f$ and $g$, each term equals $f(d_1) g(m/d_1) \\cdot f(d_2) g(n/d_2)$.\n15. The sum over $P$ of these products splits as a product of two sums:\n$$h(mn) = \\Bigl(\\sum_{d_1 \\mid m} f(d_1) g(m/d_1)\\Bigr) \\Bigl(\\sum_{d_2 \\mid n} f(d_2) g(n/d_2)\\Bigr) = h(m) h(n).$$\n16. With step 2, this shows that $f * g$ is multiplicative.\n\n## Idea\n\nWhen $m$ and $n$ are coprime, each divisor of $mn$ splits in exactly one way into a divisor of $m$ times a divisor of $n$. The two pieces are coprime, so $f$ and $g$ split over them. The double sum then factors into the product of the two convolutions.\n",
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