I am trying to understand why a line in the following proof is true. Suppose $p$ is prime and $p|ab$ where $a$ and $b$ are integers. If $p$ does not divide $a$, then $a$ and $p$ are coprime, and so there exist $x,y\in\mathbb{Z}$ such that $ax+py=1$. Then we have $abx+pby=b$ and $pby=b-abx$. Hence $p|b-abx$.
Now, the next line in the proof says that this implies that $p|b$. Why is this so?
I'd skip the "and" at the end of the proof. Since $p | ab$ we have $ab = pm$ for some $m$. Then $$ b = abx + pby = pmx + pby = p(mx+by) $$ so $p | b$.
I would probably write it like this:
Suppose that $p\mid ab$, and without loss of generality, that $p\nmid a$. Then, by Bézout's lemma, $$(p,a)=1=ua+vp$$ for some $u,v\in\Bbb Z$. Thus, $b=uab+vpb$, and so $p\mid b$.
In your particular case, however, if $p|b-abx$, then by definition $$b-abx=pn,$$ for some $n\in\Bbb Z$ and moreover, $$\frac{b-abx}{p}\in\Bbb Z.$$ Thus we must have that $p\mid b$ and $p\mid abx$. The latter of these statements, though, is guaranteed by hypothesis (remember $p\mid ab$) so we conclude $p\mid b$.
