Algebraically, prove that $24\mid a^2 - b^2 $ So I have been given this question:

$a$ and $b$ are two odd numbers and are not divisible by $3$. Prove that $24 \mid a^2 – b^2$.

I have successfully completed this using modular arithmetic, but am trying to solve it algebraically as follows:
$$ (2m+1)^2 - (2n+1)^2 $$
by difference of two squares: $ (2m+1+2n+1)(2m+1-2n-1) $, then
$$ (2m+2n+2)(2m-2n) =4(m^2+m-n^2-n)$$
$$ 4(m(m+1)-n(n+1)) $$
Hence it must be divisible by eight, given the $4$ outside the brackets and an even value within the brackets (product of two consecutive integers, one must be even, so both values must be even, and even - even = even).
However this is where I get stuck. To prove divisibility by $24$, one must prove divisibility by $8$ and $3$, and I'm struggling to pull a $3$ out from anywhere here.
Any help very much appreciated :)
 A: You're doing great! Consider the expression $$4(m(m+1)-n(n+1))\tag{1}$$
And notice that both of $m(m+1), n(n+1)$ are even, because two consecutive numbers are being multiplied. Therefore you may write $$8\left(\frac{m(m+1)}{2}-\frac{n(n+1)}{2}\right)\tag{2}$$
Now, you know that $3\not\mid a,b$ so they have remainders either $\pm 1$. Therefore write $a=3k+(-1)^s, b=3l+(-1)^r$, then you have $$a^2-b^2=9k^2+6k(-1)^s+1-9l^2-6l(-1)^r-1\\=3(3k^2+2k(-1)^s-3l^2-2l(-1)^r)$$
A: The problem is that you do not use anywhere that $a$ and $b$ are not divisible by $3$.
A number that is odd and not divisible by three is of the form $6m+1$ or $6m-1$. You could do a similar argument with number of that form instead of $2m+1$.
A priori distinguishing four cases but you can reduce to three easily by symmetry or actually to a unique one using that if it is true for $a$ it is also true for $-a$.
That is, you have on the one hand $a=6m+1$ or $a=6m-1$ and on the other hand $b=6n+1$ or $a=6n-1$.  At first it might look as if one had to consider cases $a= 6m+1$ and $b=6n+1$, $a= 6m+1$ and $b=6n-1$, $a= 6m-1$ and $b=6n+1$, $a= 6m-1$ and $b=6n-1$
However since $a^2 =(-a)^2$ and if $a = 6m-1$ then $-a = 6(-m) + 1$, one can always reduce to the case that $a=6m+1$ and $b=6n+1$.
Now you can give essentially the same argument you had, or even write $a=2m'+1$ and $b=2n'+1$  with $m'=3m$ and $n'=3n$, so that you get with the argument you wrote $4(m'(m'+1) - n'(n'+1))$ and now the expression is divisible by $3$ as well as $m'$ and $n'$ are both divisible by $3$.
