show that the even numbers 2k+2,2k+4,...,4k,4k+2 are congruent mod m to... (first post, hello!)
I'm having a bit of trouble with the following problem:

let k be a positive integer and let $m = 4k + 3$
show that the even numbers $2k+2, 2k+4,..., 4k, 4k+2 $ are congruent mod
  $m$ to the negatives of the odd numbers $2k+1, 2k-1,..., 3, 1$, and deduce
  that $2^{k+1}(k+1)(k+2)...(2k+1) ≡ (-1)^{k+1}(2k+1)(2k-1)...3*1$ (mod
  $m$)

to be entirely honest, I'm not really sure where to start with this and can't seem to get the intuition for it. Can somebody hint me in the right direction? Thanks :)
 A: Hint: remember that $a \equiv b$ (mod $n$) if $n|(a-b)$.
You know that $n = 4k+3$, so what does that tell you about $a$ and $b$? 
(Remember that you also know the form of $a$ and $b$.)
A: First part of the question:
Remember the definition of the negativity in modular arithmetic. You're working in an additive group. Then the negative (inverse) of any element $a$ is the element $b$ which makes $a+b=b+a=i$ where $i$ is the identity of the group, and obviously $i=0$.
Now, as $\forall x\in \mathbb{Z}_{4k+3}, 0\leq x < 4k+3$, we know that any element $2t$'s inverse should be in that range. Then, if we call the inverse of $2t$ as $-(2t)$, then $2t+(-(2t))=0$ in the group. Then, $2t+(-2t) \equiv 0 \pmod{4k+3}$. And as all elements of the set is at the interval  $[0,4k+3)$, then $2t+(-(2t))=4k+3$. And as $2t$ is even and $4k+3$ is odd, then $-(2t)$ should be odd.
I used $2t$ for the representation to make it shorter but in the question they start from $2k+2$ to avoid the problem of inverse of $0$. I have to state that $t<2k+2$ and $t \in \mathbb{Z}$.
I'm gonna think about the second part of the question.
