the largest divisor

$a^4-1=(a^2-1)\times(a^2+1) = (a-1)\times(a+1)\times(a^2+1)$

Since $a$ is odd, $a-1$ and $a+1$ are consecutive even integers, so one of them is divisible by $4$ . $a^2+1$ is also even.

We have 3 even numbers and one of them is divisible by $4$ , so their product will be divisible by $2\times4\times2=16$

Since $a$ is an odd integer which can be written as $2k-1, k\in\mathbb N$ . Now $\begin{aligned} & a^4 -1 = (a+1)(a-1)(a^2+1) \\& a^4-1 = (2k-1+1)(2k-1-1)((2k-1)^2+1) \\& a^4-1 = 2k.(2k-2)(4k^2-4k +2) \\& a^4-1 = 8k(k-1)(2k^2-2k+1)\end{aligned}$ Note for $k>1$ , $k$ and $k-1$ are consecutively co-prime integers then $k(k-1) = 2n\in\mathbb N$ ie the product of two consecutive integers is always even number. $\therefore a^4 -1 = 8.2n.(2k^2-2k+1) = 16n.(2k^2-2k+1)$ Hence, largest integer that divides a evenly is always $16$ .