# Prove by induction: $\forall n \in \mathbb{n}, n \geq 1 \Rightarrow 2^{2^{n}}-1$ is divisible by at least n distinct primes [duplicate]

I tried to play around with $$2^{2^{n + 1}} - 1$$, such as splitting it to $$2^{2^{n} \cdot 2} - 1$$, but it didn't helped much. I am kind of stuck and don't know what to do next

• Are you familiar with the difference of squares identity? $a^2-b^2=(a+b)(a-b)$... – abiessu Feb 10 at 21:11
• @abiessu yes, but I don't see how it would help here. – Victor Feb 10 at 21:14
• So, in this case, take $a=2^{2^n}$ and $b=1$, this gives that $2^{2^{n+1}}-1=(2^{2^n}+1)(2^{2^n}-1)$. What prime can divide both portions of this multiplication? – abiessu Feb 10 at 21:18
• @abiessu Thank you so much, I got it. – Victor Feb 10 at 21:20
• @Charlie FYI, the proof to your question is given as "Proof 2" in the question text of Different ways to prove there are infinitely many primes?. – John Omielan Feb 10 at 21:23

As guided in the comments, taking $$2^{2^n}-1$$ as a difference of squares allows for the $$n+1$$ term to be written as
$$2^{2^{n+1}}-1=(2^{2^n}+1)(2^{2^n}-1)$$
On the RHS, any prime which divides both of the factors must divide their difference, which is $$2$$. But both factors are odd, so these factors are relatively prime to one another.