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  "body": "\n## Statement\n\nLet $m$ and $n$ be positive integers that are [[coprime|0GFKP8F8AG]].\nThen $\\varphi(m n) = \\varphi(m) \\varphi(n)$, where $\\varphi$ is [[Euler's totient function|0VV04M41AN]].\n\n## Remarks\n\nThe proof uses the [[Chinese remainder theorem|0FNN9B4W64]]. It gives a bijection between the residues modulo $m n$ and the pairs of residues modulo $m$ and modulo $n$. A residue is coprime to $m n$ if and only if both parts are coprime to their moduli.\nThe hypothesis is necessary: $\\varphi(4) = 2$, but $\\varphi(2) \\varphi(2) = 1$.\nSo $\\varphi$ is a [[multiplicative function|0AP80TT660]].\nFor example, $\\varphi(36) = \\varphi(4) \\varphi(9) = 2 \\cdot 6 = 12$.\n\n## Explanations\n\n### L1 human\n\nWhen $m$ and $n$ have no common factor, a number is coprime to $m n$ exactly when it is coprime to $m$ and to $n$.\nThe two conditions are independent, so the counts multiply.\nFor example, $\\varphi(15) = \\varphi(3) \\varphi(5) = 2 \\cdot 4 = 8$.\n",
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  "msc": [
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  "name": "Multiplicativity of the totient function",
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