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  "body": "\n## Proof\n\n1. Let $n$ be a positive integer and let $a$ be an integer. Let $d = \\gcd(a, n)$ be their [[greatest common divisor|02WF878BH1]].\n2. Since $n \\ne 0$, the integers $a$ and $n$ are not both $0$. So $d$ is a positive common divisor of $a$ and $n$.\n3. First assume that $a$ and $n$ are [[coprime|0GFKP8F8AG]], so $d = 1$.\n4. By [[Bézout's identity|0DQVHMYAGA]], there are integers $x$ and $y$ with $a x + n y = 1$.\n5. Then $a x - 1 = n (-y)$, so $n$ [[divides|096PW7Q014]] $a x - 1$. By the definition of [[congruence|0TPW25CBN6]], $a x \\equiv 1 \\pmod n$.\n6. So $x$ is a [[modular multiplicative inverse|00GXTQMY8Z]] of $a$ modulo $n$.\n7. Conversely, assume that $a$ has an inverse $x$ modulo $n$, so $a x \\equiv 1 \\pmod n$.\n8. Then $n$ divides $a x - 1$, so there is an integer $y$ with $a x - 1 = n y$. That is, $a x - n y = 1$.\n9. The integer $d$ divides $a$ and $n$, so it divides $a x - n y = 1$.\n10. Since $d$ is positive by step 2 and divides $1$, we get $d = 1$. So $a$ and $n$ are coprime.\n11. By steps 3 to 10, $a$ has an inverse modulo $n$ if and only if $a$ and $n$ are coprime.\n\n## Idea\n\nAn inverse $x$ of $a$ is the same thing as a way to write $1 = a x + n y$. Bézout's identity gives such a way when $\\gcd(a, n) = 1$. Conversely, any common divisor of $a$ and $n$ divides $a x + n y = 1$.\n",
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