If $n$ is an odd natural number, then $8$ divides $n^{2}-1$

An easier way to see this is as follows: $n^2-1=(n-1)(n+1)$ where both $n-1$ and $n+1$ are even, and one of them must be divisible by 4.

Being odd means that $n=2m+1$ for some $m$. This gives

$$n^2-1=(2m+1)^2-1=(4m^2+4m+1)-1=4m^2+4m=4m(m+1).$$

Notice that either $m$ is even or $m$ is odd: either way, $m(m+1)$ is even, so can be written as $2k$...

Here are are $8$ proofs $\pm1.\ $ First: $\bmod 8\!:\ {\rm odd}^2 \equiv \{1,\,3,\!\overbrace{5,\,7}^{\large -3,\ -1}\!\!\}^{\large 2}\equiv \{\pm1,\:\!\pm3\}^{\large 2} \equiv 1 $

Alternatively, $\ \ \rm n\ odd\ \Rightarrow\ n = 4k\pm1\ \Rightarrow\ n^2-1 = (4k\pm1)^2-1 = 8k \:\!(2k\pm1)$

Or: $\rm\: n\equiv u = \pm1\pmod 4\:\Rightarrow\: 4\:|\:n\!-\!u,\:2\:|\:n\!+\!u\:\Rightarrow\: 8\:|\:(n\!-\!u)(n\!+\!u) = n^2 - 1$

Or, it's easy by induction: it's true for $\rm\:n = 1,\:$ and if true for all odds below the odd $\rm\:n\!+\!2\:$ then $\rm\:(n\!+\!2)^2\!-1\: =\: n^2\!-\!1 + 4\:\!(n\!+\!1).\:$ But $\rm\:8\:|\:n^2\!-\!1\:$ by induction, $\rm\:8\:|\:4(n\!+\!1)\:$ by $\rm\:n\:$ odd.

Or, notice that $\rm\:mod\ 8,\:$ the function $\rm\:f(n) = (2n\!-\!1)^2\:$ is constant (hence $\rm\:f(n)\equiv f(1)\equiv 1)$ because its first difference is $\equiv 0,\:$ i.e. $\rm\:f(n\!+\!1)-f(n) = (2n\!+\!1)^2-(2n\!-\!1)^2\! = 8n\equiv 0.$

By telescopy the prior proof yields the sum representation below, and a vivid proof.

$$\rm\quad (2n+1)^2 - 1\: =\: \sum_{k\!\:=\!\:1}^n\!\: 8k$$

More generally, it's the special case $\rm\:m = 8,\:\lambda(8)=2\:$ of the Euler-Carmichael theorem $$\rm\ gcd(a,m) = 1\ \Rightarrow\ a^{\lambda(m)}\equiv 1\pmod{m}$$

Since $n$ is odd, $n^2-1 = (n-1)(n+1)$ is the product of two consecutive even numbers, one of which must be divisible by 4.