Question

Prove that $2^{4n}-1$ is divisible by $15$.

Answer 1

$2^{4n}-1={\left(2^4\right)}^n-1^n={16}^n-1^n$

$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}{b}^2...........b^{n-1})$

$\Rightarrow {16}^n-1^n=(16-1)({16}^{n-1}+{16}^{n-2}.1+{16}^{n-3}{1}^2......1^{n-1})\Rightarrow {16}^n-1^n=15({16}^{n-1}+{16}^{n-2}.1+{16}^{n-3}{1}^2...........1^{n-1})$

which is clearly divisible by $15$