Show that if a ≥ 3, then an− 1 has at least 2v(n) divisors. (where v(n) is the number of divisors of n)
Let $\displaystyle d \mid n.$ define $\displaystyle A(d)=\{a^d - 1, \ \frac{a^n - 1}{a^d - 1} \}.$ clearly $\displaystyle A(d)$ is a subset of set of divisors of $\displaystyle a^n -1.$ we need a trivial lemma:
Lemma: if $\displaystyle a \geq 3$ and $\displaystyle k \geq 1, \ \ell \geq 1, \ n \geq 1,$ then: $\displaystyle (a^k - 1)(a^{\ell} - 1) \neq a^n - 1.$
Proof: suppose $\displaystyle (a^k - 1)(a^{\ell} - 1) = a^n - 1.$ then we'll get: $\displaystyle a \mid 2. \ \ \ \Box.$
so, since we've assumed that $\displaystyle a \geq 3,$ the above lemma tells us that for any $\displaystyle d \mid n,$ we have $\displaystyle |A(d)|=2$ and if $\displaystyle d_1 \neq d_2,$ then $\displaystyle A(d_1) \cap A(d_2) = \emptyset. \ \ \ \Box$