Let a and b be two distinct prime numbers and x and y are integers. Is the following true?
($x \equiv y \mod a$) and ($x \equiv y \mod b$). So, $a|(x-y)$ and $b|(x-y)$. This means $x-y=ab\phi$ with $\phi \in \mathbb{Z}$.
Can I state the last part or do I need to prove that somehow? It seems logical enough to me but it might be wrong. Your help is appreciated.
Answer
Let $x-y = am = bn$, $m, n\in \mathbb Z$.
$$am = bn$$
Since $a$ and $b$ are coprime and $a \mid bn$, $a\mid n$. Let $n = a\phi$.
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