Number Theory: Problem with proofs There are two propositions in the chapter of Number Theory in my book, the proofs of which I am having trouble to understand.


For Proposition 3
I cannot understand the proof from "Therefore ..." in the third line. I was thinking that maybe "$m-l$ divides $k$" will be "$m-l$ divides $n$". Also, is $k/d$ an integer?
Also, can someone please give an example to clarify this theorem?
For Proposition 4
What does $f(x) \in Z[x]$ mean? Why are the third brackets used? 
I cannot understand how "By proposition 2, $a_ja^j \equiv b_jb^j$ (mod $n$)...".
Can someone please explain? 
 A: For proposition 3, I agree with fleablood's assessment in a comment above that this is probably a typo.
For proposition $4$, note that the thing you're asking about appears in a parenthesis, following the abbreviation "i.e.". That means that even without knowing what "$f\in \Bbb Z[x]$" means, we can infer that it's merely an alternative formulation of whatever came before, namely that $f$ is a polynomial with integral coefficients.
To actually answer the question, given a ring $R$, the notation $R[x]$ means "the ring (or set) of polynomials with coefficients in $R$". So $\Bbb Z[x]$ is the ring of polynomials with integer coefficients. And $f\in \Bbb Z[x]$ means exactly that $f$ is an element of this ring.
As for why we use the brackets there? That's just convention. You could have $\Bbb Z(x)$ as well, but that usually means the ring (or set) of rational functions with integer coefficients.
Finally, the "by proposition 2" thing, note that proposition 2 states that if we have a product of two things, and we change one of the factors to a congruent factor, the product is unchanged modulo $n$. So $$a_ja^j = a_ja^{j-1}\cdot a\equiv a_ja^{j-1}b\pmod n$$
So we can swap one $a$ for a $b$. Now just swap the other $j-1$ $a$'s for $b$'s, one by one as proposition 2 says you're allowed to do, and finally you reach $a_jb^j$.
A: I think the book made a typo and it isn't $m-l$ that divides $k$ (I don't see that as even true) but that $d$ divides $k$.
This follow as $d$ divides $kn$ but is relatively prime to $n$ so $d$ must divide $k$.
And hence, yes, $\frac kd$ is an integer, for which we conclude $m-l \equiv 0 \pmod n$.
Note:  This will not be true if $a, b, n$ will have a common divisor (other than $1$).  Consider $8 \equiv 20 \mod 12$ but $2\equiv 5 \mod 12$ is .... wrong.  (Although $2 \equiv 5 \pmod 3$.....)
--- I empathize.  For a typo that is a doozy to make and utter destroys the intent of the proof.
....
$\mathbb Z[x]$ means the set of all polynomials with integer coefficients.
So $f(x) \in \mathbb Z[x]$ means "Let $f(x)$ be a polynomial with integer coefficients.
