proof for even and odd integers --number theory First, prove that $r(r + 1)$ is even for any $r ∈ Z$. 
Then, for positive $j ∈ Z$, prove that if $j$ is odd then $8 | (j^2 − 1)$
for the first part can I say if there is an even number being multiplied then we know that $r(r + 1)$ is even?
for the second part: $j$ is odd, it can be written as $2k + 1$ for any integer $k$. 
$j^2 - 1 = (2k + 1)^2 - 1 = 4k^2 + 4k + 1 - 1 = 4k^2 + 4k = 4k(k + 1)$ 
this is where I get stuck with the proof
 A: Clearly 4 divides $j^{2} -1$ and by the first part k(k + 1) is even, so 2 divides k(k + 1).
Hence 8 divides the whole expression.
A: With congruences: any odd integer is congruent to $1,3,5$ or $7$ modulo $8$. The square of any of these is congruent to $1$ modulo $8$.
A: Here is a proof 
that $r(r+1)$ is even
by induction.
Base case:
If $r = 1$,
then $r(r+1) = 2$
which is even.
Induction step:
If $r(r+1)$ is even,
then
$(r+1)(r+2)
=(r+1)r + 2(r+1)
$
and this is even since
both
$r(r+1)$ and $2(r+1)$
are even.
As I discussed earlier,
proving evenness and
oddness can be
surprisingly difficult.
An example is here:
Prove that no positive integer is both even and odd, and that all positive integers are either even or odd
A: The number r(r+1) is the product of two consecutive numbers, one of which must therefore be even and the other odd, so that their product is even.
 For the second part you are basically there, because you know that k(k+1) is even from part 1, and the expression you end up with is also divisible by 4 as well so is overall divisible by 8
