If $8$ does not divide $x^2-1$, then $x$ is even; prove by contrapositive If $8$ does not divide $x^2-1$, then $x$ is even
proof by contrapositive
the contrapositive of this is :
if $x$ is odd, then $8$ divides $x^2-1$
proof by contrapositive:
Assume $x$ is odd
by definition of odd $∃k∈ℤ$ such that $x=2k+1$
Well, $x^2-1$ = $(2k+1)^2-1$
= $4k^2+4k$
= $4k(k+1)$
Therefore, $k(k+1)$ is an even integer
( i also know the definition of division : $∃m∈ℤ$ such that $x^2-1$ = $8m$)
$x^2-1$ = $4 · 2 · k(k+1)/2$ 
Therefore, 
$8(k^2+1)/2$ is divisible by 8

8 does not divide $x^2-1$ is the same as
"if 8 divides $x^2-1$, then x is odd"
which is the same as:
$$8|(x^2-1)\rightarrow 2|(x+1)$$
but
$$x^2-1=(x-1)(x+1)$$
Then since $8=4*2$
Two divides both $x-1$ and $x+1$, and we proven our case
