If n is an odd integer,

let n = 2k+1

then

n^2-1

=(2k+1)^2-1

=4k^2+4k

Divide 4k^2+4k by 8

\dfrac{4k^2+4k}{8} = \dfrac{k^2+k}{2}=\dfrac{k(k+1)}{2}

Since k and k+1 is consecutive two numbers, there must be one is even number, which means k(k+1) is divisible by 2

Therefore, n^2-1 is divisible by 8 if n is an odd number