(i) Show that $3^n− 1$ is a composite integer for $n > 1$.
(ii) Show that $x^n− 1$ is a composite integer for $n > 1$ and $x ≥ 3$.
Hint: $a^n− b^n = (a− b) (a^{n−1}+ a^{n−2}b + a^{n−3}b^2 + ⋯ + ab^{n−2} + b^{n−1})$ for $n > 1$.
(i) We can write, for $n>1$,
$$ 3^n-1 = (3-1)(3^{n−1}+ 3^{n−2} + 3^{n−3} + ⋯ + 3 + 1) $$
This is composite with a factor $(3-1)=2$.
(ii) We can write, for $n>1$,
$$ x^n -1 = (x− 1) (x^{n−1}+ x^{n−2} + x^{n−3}+ ⋯ + x + 1) $$
The condition $x \ge 3$ means $x-1 \ge 2$ and so $(x-1)$ is a factor of 2 or more, and so $x^n-1$ is composite.