Show $8\mid n^2-1$ if $n$ is an odd positive integer. Show that $n^2-1$ is divisible by $8$, if $n$ is an odd positive integer.
Please help me to prove whether this statement is true or false.

This is a nice example of a direct proof. You start with the facts that if $\phi$ is your positive odd integer, then it is in the form $\phi = 2n + 1$ where $n$ is an integer and $\phi^2 - 1 = 8p \quad( p\in\mathbb Z) $
 is true.

Recall that if a number is divisible by 8, then 8 is one of its factors.

This is something like a “case-by-case” proof. You prove the hypothesis for $n \in2\mathbb Z$ and then $n \in 2\mathbb Z + 1$.
Let's start with assuming that $n = 2m + 1$ where $m$ is an integer (of course) to prove for odd numbers. It is clear that,$$\begin{align} \phi^2 - 1 =(2n + 1)^2 - 1 &= 4n^2 + 4n \\ &= 4n(n + 1) \\ &= (8mn + 4n)(2m + 2) \\ &=4n\cdot 2\cdot(2m + 1)\cdot (m + 1) \\ & = 8 \cdot (n(2m + 1)(m +1)) \\ & = 8k  \end{align} $$
$\rm 0.5\times Q.E.D $

Now $n = 2m$ which is gonna be easier. 
$$\begin{align}\phi^2 - 1 =(2n + 1)^2- 1  = 4n^2 + 4n &= 4n(n + 1) \\ &= 8mn(2m +1)\\&= 8(2m^2n + mn) \\ &=8k \end{align}$$

Q.E.D
(yay)
