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  "body": "\n## Proof\n\n1. Let $n$ be a positive integer, and let $a, b, c$ be integers with $c a \\equiv c b \\pmod n$. Let $d = \\gcd(c, n)$ be their [[greatest common divisor|02WF878BH1]].\n2. Since $n \\ne 0$, $c$ and $n$ are not both $0$. So $d$ is a positive integer that divides $c$ and $n$.\n3. Write $c = d c'$ and $n = d n'$ with integers $c'$ and $n'$. Since $n \u003e 0$ and $d \u003e 0$, $n' = n / d$ is a positive integer.\n4. By the definition of [[congruence|0TPW25CBN6]], $n$ [[divides|096PW7Q014]] $c a - c b = c (a - b)$. So there is an integer $k$ with $c (a - b) = n k$.\n5. By step 3, this reads $d c' (a - b) = d n' k$. Since $d \\ne 0$, we can divide by $d$ and get $c' (a - b) = n' k$. So $n'$ divides $c' (a - b)$.\n6. By [[Bézout's identity|0DQVHMYAGA]], there are integers $x$ and $y$ with $c x + n y = d$.\n7. By step 3, this reads $d (c' x + n' y) = d$. Dividing by $d$ gives $c' x + n' y = 1$.\n8. Let $e$ be a common divisor of $c'$ and $n'$. By step 7, $e$ divides $1$, so $e = \\pm 1$. Hence $\\gcd(c', n') = 1$, and $n'$ and $c'$ are [[coprime|0GFKP8F8AG]].\n9. By steps 5 and 8 and the [[generalized Euclid's lemma|03BAREW1Z5]], $n'$ divides $a - b$.\n10. Hence $a \\equiv b \\pmod{n / d}$.\n\n## Idea\n\nDivide the relation $n \\mid c (a - b)$ by the common factor $d$ of $c$ and $n$. What is left of $c$ shares no factor with what is left of $n$, so Euclid's lemma removes it.\n",
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