Let $a>3$ be an odd integer. Prove that for every positive integer $n$, the number $a^{(2^{n})}-1$ has at least $n+1$ distinct prime divisors.
This problem smells very strongly of induction, but maybe a more complex version than I am trying.
What I have done so far:
Let $a=2k+3$ for positive integers $k$.
Proof by induction.
Base case:
If $n=1$, then $a^{(2^n)}-1$ must have at least one prime divisor (it's greater than 1).
So now assume for $n=k$, that is, $a^{(2^k)}-1$ has at least $k+1$ distinct prime divisors.
So let $a^{(2^k)}-1=p_1 \cdot p_2 \cdot p_3 \cdots p_k \cdot p_{k+1} \cdot m$ for some positive integer $n$ and for distinct primes $p_1, \;p_2, \;p_3, \; \cdots , \;p_{k+1}$.
Now consider $n=k+1$. If $n=k+1$ then the given equation becomes $(p_1 \cdot p_2 \cdot p_3 \cdots p_k \cdot p_{k+1} \cdot m+1)^2-1=(p_1 \cdot p_2 \cdot p_3 \cdots p_k \cdot p_{k+1})(p_1 \cdot p_2 \cdot p_3 \cdots p_k \cdot p_{k+1}+2)$. Will this necessarily have $k+2$ distinct prime divisors, because if so the induction is complete?
Best Answer
$$ a^{2^{n+1}} - 1 = \left( a^{2^n} - 1 \right) \left( a^{2^n} + 1 \right) $$ and $$ \gcd( a^{2^n} - 1 , a^{2^n} + 1 ) = 2. $$