{
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  "body": "\n## Statement\n\nLet $n$ be a positive integer.\nThen $\\sum_{d \\mid n} \\varphi(d) = n$, where the sum runs over the positive divisors $d$ of $n$, and $\\varphi$ is [[Euler's totient function|0VV04M41AN]].\n\n## Remarks\n\nThe proof: sort the fractions $\\frac{1}{n}, \\frac{2}{n}, \\dots, \\frac{n}{n}$ by their denominators in lowest terms. For each divisor $d$ of $n$, exactly $\\varphi(d)$ of them have the denominator $d$.\nFor example, for $n = 12$: $\\varphi(1) + \\varphi(2) + \\varphi(3) + \\varphi(4) + \\varphi(6) + \\varphi(12) = 1 + 1 + 2 + 2 + 2 + 4 = 12$.\nWith the [[Möbius inversion formula|07Q4TBTFVB]], the result gives $\\varphi(n) = \\sum_{d \\mid n} \\mu(d) \\, n / d$.\n\n## History\n\nGauss proved this identity in his Disquisitiones Arithmeticae of 1801.\n\n## Explanations\n\n### L1 human\n\nWrite the fractions $\\frac{1}{6}, \\frac{2}{6}, \\dots, \\frac{6}{6}$ in lowest terms: $\\frac16, \\frac13, \\frac12, \\frac23, \\frac56, \\frac11$.\nThe denominators $1, 2, 3, 6$ appear $1, 1, 2, 2$ times. These are $\\varphi(1), \\varphi(2), \\varphi(3), \\varphi(6)$, and their sum is $6$.\n",
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