<s>[INST] 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.
 [/INST] We know that any odd positive integer is of the form $4q+1$ or, $4q+3$ for some integer $q$.
So, we have the following cases:
Case 1 when $n=4q+1$: In this case, we have
$n^2 -1=(4q+1)^2 -1=16q^2 +8q+1-1=16q^2 +8q=8q(2q+1)$
$∴ n^2 -1$ is divisible by 8 [$∵ 8q(2q+1)$ is divisible by 8]
Case 2 when $n=4q+3$: In this case, we have
$n^2 -1=(4q+3)^2 -1= 16q^2 +24q+9-1=16q^2 +24q+8$
⇒ $n^2 -1=8(2q^2 +3q+1)=8(2q+1)(q+1)$
$∴ n^2 -1$ is divisible by 8 [$∵ 8(2q+1)(q+1)$  is divisible by 8]
Hence, $n^2 -1$ is divisible by 8.
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