<s>[INST] 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
 [/INST] Rewrite $$(x^2-1)=(x-1)(x+1)$$ If 8 does not divide $(x^2-1)=(x-1)(x+1)$, then 2 does not devide $x^2-1=(x-1)(x+1)$. But that is the same as saying that neither $x-1$ nor $x+1$ are even. Hence $x$ must be even.
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