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  "body": "\n## Proof\n\n1. Let $m$ and $n$ be [[coprime|0GFKP8F8AG]] positive integers. For a positive integer $k$, let $U_k$ be the set of integers $x$ with $1 \\le x \\le k$ and $\\gcd(x, k) = 1$. By definition, $\\varphi(k) = \\abs{U_k}$, where $\\varphi$ is [[Euler's totient function|0VV04M41AN]].\n2. Let $k$ be a positive integer. Each integer $x$ is congruent modulo $k$ to exactly one element of $\\{ 1, \\dots, k \\}$. Call it $\\rho_k(x)$.\n3. Proof of step 2: by the [[division algorithm|0CHQKZFK5S]], $x = q k + r$ with $0 \\le r \u003c k$. Take $r$ if $r \u003e 0$ and $k$ if $r = 0$. Two elements of $\\{ 1, \\dots, k \\}$ differ by less than $k$, so at most one of them works.\n4. If $x \\equiv y \\pmod k$, then $\\gcd(x, k) = \\gcd(y, k)$. Indeed, $k$ [[divides|096PW7Q014]] $x - y$, so a divisor of $k$ divides $x$ if and only if it divides $y = x - (x - y)$.\n5. Let $f : \\{ 1, \\dots, m n \\} \\to \\{ 1, \\dots, m \\} \\times \\{ 1, \\dots, n \\}$ be the map $f(x) = (\\rho_m(x), \\rho_n(x))$.\n6. The map $f$ is injective. If $f(x) = f(y)$, then $x \\equiv y \\pmod m$ and $x \\equiv y \\pmod n$. By the uniqueness part of the [[Chinese remainder theorem|0FNN9B4W64]], $x \\equiv y \\pmod{m n}$. Step 2 with $k = m n$ gives $x = y$.\n7. The map $f$ is surjective. Given $(r, s)$, the Chinese remainder theorem gives an integer $z$ with $z \\equiv r \\pmod m$ and $z \\equiv s \\pmod n$. Let $x = \\rho_{m n}(z)$.\n8. Since $m$ and $n$ divide $m n$, and $m n$ divides $x - z$, we get $x \\equiv z \\equiv r \\pmod m$ and $x \\equiv s \\pmod n$. So $f(x) = (r, s)$.\n9. We show that $\\gcd(x, m n) = 1$ if and only if $\\gcd(x, m) = 1$ and $\\gcd(x, n) = 1$.\n10. If $\\gcd(x, m) = d \u003e 1$, then $d$ divides $x$ and $m n$, so $\\gcd(x, m n) \u003e 1$. The same holds for $n$.\n11. Assume $\\gcd(x, m) = \\gcd(x, n) = 1$, and suppose $e = \\gcd(x, m n) \u003e 1$. By the [[existence of a prime divisor|0SPBEDVNW1]], there is a prime $q$ with $q \\mid e$.\n12. Then $q \\mid m n$. By [[Euclid's lemma|051VYKA1EX]], $q \\mid m$ or $q \\mid n$. Since also $q \\mid x$, $q$ is a common divisor greater than $1$ of $x$ and $m$, or of $x$ and $n$. This contradicts step 11.\n13. By steps 4 and 9, for $x \\in \\{ 1, \\dots, m n \\}$, we have $x \\in U_{m n}$ if and only if $\\rho_m(x) \\in U_m$ and $\\rho_n(x) \\in U_n$.\n14. By steps 6, 7 and 13, $f$ restricts to a bijection from $U_{m n}$ onto $U_m \\times U_n$.\n15. By the [[bijection principle|0GV2573FP4]], $\\abs{U_{m n}} = \\abs{U_m} \\cdot \\abs{U_n}$. That is, $\\varphi(m n) = \\varphi(m) \\varphi(n)$.\n\n## Idea\n\nThe Chinese remainder theorem matches each number modulo $m n$ with a pair of numbers modulo $m$ and modulo $n$. A number is coprime to $m n$ exactly when both parts of its pair are coprime to their moduli, so the counts multiply.\n",
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