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  "body": "\n## Proof\n\n1. Let $m$ and $n$ be positive integers and let $a$ and $b$ be integers. Let $g = \\gcd(m, n)$ be the [[greatest common divisor|02WF878BH1]] and $M = \\operatorname{lcm}(m, n)$ the [[least common multiple|09S5SN35DJ]].\n2. Since $m$ and $n$ are positive, $g$ is a positive common divisor of $m$ and $n$. Also, $M$ is the smallest positive common multiple of $m$ and $n$.\n3. Necessity. Let $x$ be a solution. By the definition of [[congruence|0TPW25CBN6]], $m$ [[divides|096PW7Q014]] $x - a$ and $n$ divides $x - b$.\n4. Since $g$ divides $m$ and $n$, step 3 shows that $g$ divides $x - a$ and $x - b$. So $g$ divides $(x - b) - (x - a) = a - b$, and $a \\equiv b \\pmod g$.\n5. Sufficiency. Assume $a \\equiv b \\pmod g$, and write $a - b = g t$ with an integer $t$.\n6. By [[Bézout's identity|0DQVHMYAGA]], there are integers $u$ and $v$ with $m u + n v = g$.\n7. Let $x = a - m u t$. Then $x - a = m (- u t)$, so $x \\equiv a \\pmod m$.\n8. By steps 5 and 6, $x - b = (a - b) - m u t = g t - m u t = (g - m u) t = n v t$. So $x \\equiv b \\pmod n$, and $x$ is a solution.\n9. Uniqueness. Let $x$ and $y$ be two solutions, and let $c = x - y$.\n10. The integer $m$ divides $x - a$ and $y - a$, so it divides their difference $c$. In the same way, $n$ divides $c$.\n11. By the [[division algorithm|0CHQKZFK5S]], there are integers $q$ and $r$ with $c = q M + r$ and $0 \\le r \u003c M$.\n12. Since $m$ divides $c$ and $M$, it divides $r = c - q M$. In the same way, $n$ divides $r$. So $r$ is a common multiple of $m$ and $n$.\n13. If $r \u003e 0$, then $r$ is a positive common multiple smaller than $M$, which contradicts step 2. So $r = 0$.\n14. Hence $M$ divides $c = x - y$, that is, $x \\equiv y \\pmod M$.\n15. By steps 4, 8 and 14, the system has a solution if and only if $a \\equiv b \\pmod g$, and then two solutions are congruent modulo $M$.\n\n## Idea\n\nA solution forces $a$ and $b$ to agree modulo every common divisor of the moduli. Conversely, Bézout's identity writes $g$ as $m u + n v$. Multiplying by $t = (a - b) / g$ splits $a - b$ into a multiple of $m$ plus a multiple of $n$. Removing the multiple of $m$ from $a$ lands on $b$ plus a multiple of $n$.\n",
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