Use recurrence relation Fermat numbers to show that the set of prime numbers is infinite

Question

I'm working on some proofs on a recurrence relation with Fermat numbers.

a). Prove with Fermat numbers $f_n = 2^{2^n} + 1$ that:

$$ f_0 \cdot f_1 \cdot f_2 \cdot f_3 \cdots f_{n-1} = f_n - 2$$

b). Prove that $f_n$ is relatively prime with $f_0, f_1, f_2, ... , f_{n-1}$. In other words:

for any $n$, $f_n$ is not divisible with any factors of its previous Fermat number.

In my attempt on (a) I used the sum of the powers of $2$ :

$$2^0 + 2^1 + 2^2 + 2^3 + ... + 2^n = 2^{n+1} - 1$$

but I didn't come very far.

Some help would be very much appreciated.

Answer 1

a) If $f_0\times f_1\times\cdots\times f_{n-1}=f_n-2$, then\begin{align}f_0\times f_1\times\cdots\times f_{n-1}\times f_n&={f_n}^2-2f_n\\&=\left(2^{2^n}+1\right)^2-2\times\left(2^{2^n}+1\right)\\&=2^{2\times2^n}+2\times2^{2^n}+1-2\times2^{2^n}-2\\&=2^{2^{n+1}}+1-2\\&=f_{n+1}-2.\end{align}

b) If $p$ is a prime number which divides $f_n$ and $f_m$, with $n<m$, then, $p$ divides both $f_0\times f_1\times\cdots\times f_m$ and $f_m$. Therefore, $p$ divides their difference, which is $2$. Since $p$ is prime, $p=2$, but this is impossible, because the Fermat primes are odd. This proves that distinct Fermat numbers are relatively prime.

For each $n\in\mathbb N$, let $p_n$ be the smallest prime factor of $f_n$. Then $(p_n)_{n\in\mathbb N}$ is an injective sequence of prime numbers. Therefore, there are infinitely many prime numbers.