Exercise fermat numbers

Question

Someone can help me with this problem? $F_p=2^{2^p}+1$

I don't know how to do the 1 nor the 4.

For 1), suppose $n=mq$, where $m$ is odd and $\gt 1$. Then $2^n+1=(2^q)^m+1$. Let $x=2^q$. Then $x+1$ divides $x^m+1$. — André Nicolas, Apr 17 '16 at 19:09
It is an immediate consequence of (2) and (3). Let $p_k$ be the smallest prime divisor of $F_k$ (any prime factor will also work). The $p_k$ are distinct, because any two different Fermat numbers are relatively prime. — André Nicolas, Apr 17 '16 at 19:19
Duplicate of Infinitely many primes derived from coprimality of Fermat numbers. — Bill Dubuque, Jun 10 '24 at 14:48

score 1 · Accepted Answer · answered Apr 17 '16 at 19:18

1

Hint:

For 2), use induction on $k$, remembering that $2^{2^{p+k}}=\bigl(2^{2^p}\bigr)^{2^k}$.

For 4): use that each Fermat number has a prime factor.

answered Apr 17 '16 at 19:18

Bernard

Okey, the serie is infinite and every number of the serie if formed by at least one prime divisor who is different from all the primes that form the predecessors of this number, so infinite primes. – joan capell Apr 17 '16 at 20:50

1 Answers1