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  "body": "\n## Proof\n\n1. Let $n$ be a positive integer, and let $a, b, c, d$ be integers with $a \\equiv b \\pmod n$ and $c \\equiv d \\pmod n$.\n2. By the definition of [[congruence modulo n|0TPW25CBN6]], $n$ [[divides|096PW7Q014]] $a - b$ and $n$ divides $c - d$. So there are integers $s$ and $t$ with $a - b = n s$ and $c - d = n t$.\n3. Expanding gives $a (c - d) + d (a - b) = a c - a d + a d - b d = a c - b d$.\n4. By steps 2 and 3, $a c - b d = a n t + d n s = n (a t + d s)$.\n5. Since $a t + d s$ is an integer, $n$ divides $a c - b d$.\n6. Hence $a c \\equiv b d \\pmod n$.\n\n## Idea\n\nChange one factor at a time: first replace $c$ by $d$, then $a$ by $b$. Each change alters the product by a multiple of $n$, because the change in the factor is a multiple of $n$.\n",
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        "url": "https://en.wikipedia.org/wiki/Modular_arithmetic"
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