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I was wondering if it could be proved without using the Chinese remainder theorem, arithmetically.

Thank you

philok
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    If $2\mid a-b$ and $3\mid a-b$, then $6\mid a-b$ – J. W. Tanner Nov 09 '20 at 11:15
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    Technically you don't need the full power of the Chinese remainder theorem. You can make an argument from a simple lemma involving prime divisors and/or prime factorizations. – Derek Luna Nov 09 '20 at 11:22

2 Answers2

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$2,3$ are relatively prime so $2|(m-n)$, $3|(m-n)$ implies $6|(m-n)$ meaning $m\equiv n \mod 6$.

Derek Luna
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If $c$ is a common multiple of $a$ and $b$, then $c$ is a multiple of their least common multiple $\operatorname{lcm}(a,b)$.

Apply this to $c=m-n$ and $a=2$ and $b=3$.

lhf
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