Let $p$ be a prime number. Let $b \in \mathbb{Z}$ be non-zero. Show that if $p$ does not divide $b$ then for all $a \in \mathbb{Z}$ there is some $m \in \mathbb{Z}$ such that $p|(a-mb)$?

I believe this had to do with a euclidean domain, but I am not positive how to prove.


Since $p$ does not divide $b$ that means they are coprime. Hence by the Euclidean algorithm we can find integers $s,t$ such that:

$sb + tp = 1$

But then scaling by $a$ gives:

$asb + atp = a$

And so $p|(a-(as)b)$, thus $m=as$ works.


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