Divisible by 40?

Let

$a=10k+9\phantom{cccccc}b=10j+9\phantom{cccc}k,j\in\mathbb{N}$

Hence

$\begin{aligned} a^2-b^2&=(10k+9)^2-(10j+9)^2\\ &=(10k+9+10j+9)(10k+9-10j-9)\\ &=(10k+10j+18)(10k-10j)\\ &=20(k-j)[5(k+j)+9] \end{aligned}$

Now, $k+j$ and $k-j$ have the same parity so there are two cases

$\begin{cases} k+j\equiv 0\mod 2\Longrightarrow (k-j)[5(k+j)+9]\equiv 0\cdot(5\cdot 0+9)\equiv 0 \mod 2 \Longrightarrow 2|(k-j)[5(k+j)+9]\\ k+j\equiv 1\mod 2\Longrightarrow (k-j)[5(k+j)+9]\equiv 1\cdot(5\cdot 1+9)\equiv 14 \equiv 0 \mod 2 \Longrightarrow 2|(k-j)[5(k+j)+9] \end{cases}$

Thus

$40|20(k-j)[5(k+j)+9]$

$\Longrightarrow \boxed{40|a^2-b^2}$