Please notice that this post is not a duplicate from the ones posted by @Lil or @LizW since I want to use another result.
Proof:
Lets prove the next statement first:
If $n$ is an odd integer, then $n$ is equal to $4k+1$ or $4k+3$
Any integer $n$ divided by $4$ has remainder 0, 1, 2 or 3 using the quotient remainder theorem. Therefore any integer can be written as $4k, 4k+1, 4k+2$ and $4k+3$ for some integer $k$. Clearly, $4k$ and $4k+2$ are even numbers. Therefore, all odd integer numbers are $4k+1$ or $4k+3$.
Using this result, if $a$ and $b$ are odd integers then, $a=4m+1$ and $b=4n+1$ for some integers $m$ and $n$.
By algebra, $a^2=(4m+1)^2$ and $b^2=(4n+1)^2$
Then,
$\begin{align*}a^2-b^2=&(4m+1)^2-(4n+1)^2\\ =&16m^2+8m+1-16n^2-8n-1\\ =&8(2m^2+m-2n^2-n)\end{align*}$
This shows that $8\mid (a^2-b^2)$
Should I prove it for $4k+3$?
Best Answer
Hint: if $a$ is an odd integer, then $a=4m\pm1\,$, so $a^2=16m^2\pm 8m+1=8(2m^2\pm m)+1\,$. Therefore $\,a^2-b^2\,$ is the difference of two numbers that each are a multiple of $\,8\,$ plus $\,1\,$.