A Fermat number $F_n$ is of the form $F_n = 2^{2^n} + 1$

Furthermore, $F_n = 2 + F_0F_1F_2......F_{n-1}$ Now I already proved that if $n \neq m$ then $\gcd(F_n,F_m) = 1$

Here is the proof

Without loss of generality , I assume that $m > n$ then we know that $$F_m = 2 + F_0F_1F_2........F_nF_{n+1}.....F_{m-1}$$ I assumed here that $n > 2 $ and $n < m-1$

Now we assume that $\gcd(F_n,F_m) = d$ then $d \mid F_n$ and $d \mid F_m$

Now if $d \mid F_n$ then $d \mid F_0F_1.....F_nF_{n+1}...F_{m-1}$ and so $d$ divides any linear combination of $F_m$ and $F_0F_1.....F_nF_{n+1}...F_{m-1}$, In particular $d \mid Fm- F_0F_1.....F_nF_{n+1}...F_{m-1}=2$ and hence $$d \mid 2$$

Now since all Fermat numbers are odd then $d \neq 2$ and hence $d=1$

Now I want to use this to prove that there exists infinitely many primes.

I know that since Fermat numbers form an infinite sequence of increasing numbers and each Fermat number is relatively prime to all other Fermat numbers then we have infinitely distinct prime divisors for each composite Fermat numbers and we also add to those Fermat prime numbers.

I feel like there is a better mathematical argument that that novel that I wrote above :)

  • $\begingroup$ What is the question? The proof is OK. $\endgroup$ – Winther Apr 23 '15 at 8:16
  • $\begingroup$ Is there a more concrete way to prove that there are infinitely many primes ? using Fermats numbers $\endgroup$ – alkabary Apr 23 '15 at 8:18
  • $\begingroup$ The proof that Fermat numbers are pairwise coprime looks fine. In particular, if you take $p_n = \mbox{minimum prime dividing } F_n$, then $\{ p_n \}_n$ is an infinite set of primes. $\endgroup$ – Crostul Apr 23 '15 at 8:19
  • $\begingroup$ You can make the argument more compact and elegant if you want, but you basically need all the ingredients that is in there now. For the same proof see for example this $\endgroup$ – Winther Apr 23 '15 at 8:20
  • $\begingroup$ A proof "from the book" is given in Aigner, Ziegler. $\endgroup$ – Dietrich Burde Apr 23 '15 at 8:59

I think your proof is essentially fine.

I'm not sure if you need to prove $F_n = 2+F_0F_1F_2\ldots F_{n-1}$, but I'll add that by induction. Base case: $ F_0=3 \\ F_1 = 5 = 2+F_0$

Inductive step: Assume that the stated relation holds for $F_k$, so $F_k-2 = F_0F_1F_2\ldots F_{k-1}$. Then note that

$$ \begin{align} (F_k)(F_k-2) &= (2^{2^k}+1)(2^{2^k}-1) \\ &= (2^{2^k})^2-1\\ &= 2^{2^{k+1}}-1\\ &= F_{k+1}-2\\ \end{align} $$

And so $F_{k+1} = 2+F_k(F_k-2) = 2+ F_0F_1F_2\ldots F_{k-1}F_k$ as required to complete the induction.

To show that for $m>n\ge 0, \gcd(F_m,F_n)=1$, your proof can be shortened, but doesn't change in essence:

$$\begin{align} F_m &= 2+F_0F_1F_2\ldots F_{m-1} \\ &=2+kF_n\\ \end{align}$$

Then for $d=\gcd(F_m,F_n)$ we have $d \mid F_m$ and $d \mid kF_n$, so also $d \mid 2$. We know that $F_m$ is odd $\Rightarrow d \ne 2$, giving $d=1$.

The infinite sequence of Fermat numbers therefore does not share any prime factors between terms, meaning there are an infinite number of primes.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.