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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]], $n \\mid a - b$ and $n \\mid c - d$.\n3. By the definition of [[divisibility|096PW7Q014]], there are integers $k$ and $l$ with $a - b = n k$ and $c - d = n l$.\n4. Then $(a + c) - (b + d) = (a - b) + (c - d) = n k + n l = n (k + l)$.\n5. Since $k + l$ is an integer, $n \\mid (a + c) - (b + d)$.\n6. Hence $a + c \\equiv b + d \\pmod n$.\n\n## Idea\n\nThe difference of the two sums is the sum of the two differences. Both differences are multiples of $n$, and a sum of multiples of $n$ is again a multiple of $n$.\n",
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