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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 :)

Bill Dubuque
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alkabary
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  • What is the question? The proof is OK. – Winther Apr 23 '15 at 08:16
  • Is there a more concrete way to prove that there are infinitely many primes ? using Fermats numbers – alkabary Apr 23 '15 at 08:18
  • 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. – Crostul Apr 23 '15 at 08:19
  • 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 – Winther Apr 23 '15 at 08:20
  • A proof "from the book" is given in Aigner, Ziegler. – Dietrich Burde Apr 23 '15 at 08:59
  • Do you know that the serie of all the reciprocal of prime numbers is divergent? This is another proof. – Piquito Apr 23 '15 at 19:56
  • @LuisGomezSanchez - don't be ridiculous. How on earth can you prove the series of prime number reciprocals is divergent without already knowing that there is an infinitude of prime numbers? – Joffan Apr 23 '15 at 20:59
  • @Joffan - If any ridiculous in this story, not me but Euler. Take it easy and breathe deeply . – Piquito Apr 24 '15 at 01:04
  • @LuisGomezSanchez How do you prove that the sum of reciprocals of prime numbers is divergent using this approach? It is not obvious at all. – Winther Apr 25 '15 at 01:33
  • @Winther No relation between this approach and the famous demonstration of Euler (who incidentally does much more than that with his paper on zeta function) – Piquito Apr 25 '15 at 19:44
  • @LuisGomezSanchez I was trying to understand your claim above: "Do you know that the serie of all the reciprocal of prime numbers is divergent? This is another proof." I don't see how this can be made into another proof of that. – Winther Apr 25 '15 at 19:47
  • @Winther I mean another proof of " infinitely many prime numbers" as Alkabary himself claims in the tittle of his post as one different of Euclides's. It may be that my poor English has led to misunderstandings. – Piquito Apr 26 '15 at 21:06

1 Answers1

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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.

Joffan
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