Let $\displaystyle P$ be $\displaystyle 4^{2n}-1$ is divisible by 15.

Assuming $\displaystyle u_n = 4^{2n}-1$ to be true [1] ($\displaystyle u_n = 15k$),

$\displaystyle u_{n+1} = 4^{2(n+1)} -1 = 4^{2n+2} - 1$

$\displaystyle u_{n+1} = 16 \times 4^{2n} - 1 $

From [1], $\displaystyle u_{n+1} = 16(u_n + 1) -1 $

$\displaystyle u_{n+1} = 16(15k + 1) -1 $

$\displaystyle u_{n+1} = 16(15k)+ 16 -1 = 16(15k) - 15 = 15(16k-1)$

Therefore if $\displaystyle u_n$ is divisible by 15, then so is $\displaystyle u_{n+1}$,

When $\displaystyle n =1 $, $\displaystyle u_n = 4^{2 \times 1} -1 = 15$. Therefore the basis case holds true.

Therefore by mathematical induction,

$\displaystyle \forall n \in \mathbb{Z} , u_n \implies u_{n+1}, u_n \therefore P$