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Is this proof acceptable?

Theorem (Lucas)

Every prime factor of Fermat number $F _ n = 2 ^ {2 ^ n} + 1$; $(n > 1)$ is of the form $k2 ^{n + 2} + 1$.

Theorem

The set of prime numbers is infinite.

Proof.
Suppose opposite, that there are just finally many prime numbers and we denote the largest prime by $p$. Then $F_p$ must be a composite number because $F_p>p$. By Lucas theorem we know that there is a prime number $q$ of the form $k2 ^{p + 2} + 1$ that divides $F_p$. But $q>p$ , thus we arrived at a contradiction. Hence, the set of prime numbers is infinite.

Bill Dubuque
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Pedja
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  • Note that you can use the Fermat-numbers in another way to show that there are infinite many primes : We have $$F_{n+1}=\prod_{j=0}^n F_n + 2$$ showing that the Fermat numbers are pairwise coprime. Since each Fermat number must have a prime factor , we are done. But of course , the above proof also works. Also, you do not need $F_p>p$ in your proof. Your later $q>p$ is enough. – Peter Jun 10 '24 at 09:31
  • You can also simplify your argument as follows : For every $n>1$ , there is a prime number of the form $k\cdot 2^{n+2}+1$ since every prime factor of $F_n$ has this form. Therefore there is no upper bound for the prime numbers. Hence there are infinite many primes. – Peter Jun 10 '24 at 10:14
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Jun 10 '24 at 13:11
  • Easier with Mersenne vs. Fermat numbers: prime $,q\mid 2^p−1\Rightarrow p\mid q−1\Rightarrow q>p,,$ see the Corollary in the linked dupe. This is well known, e.g. as the 2nd dupe mentions this proof is mentioned in the book Proofs from The Book by Aigner and Ziegler. – Bill Dubuque Jun 10 '24 at 14:17
  • Well , if we argue that this proof is too difficult and the Mersenne variant easier , we can as well argue that Euclid's proof should be used anyway. I think the Lucas theorem is not much less well known then the criterion for the Mersenne factors. – Peter Jun 10 '24 at 16:59

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Yes, your proof is correct, though the premise is a tad silly (in the sense that you're using a fairly sophisticated result to prove a fairly simple result).

DanDan面
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  • I was just playing around. – Pedja Jun 10 '24 at 09:01
  • @Pedja It was amusing and made me chuckle :^) – DanDan面 Jun 10 '24 at 09:03
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    This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – Dominique Jun 10 '24 at 09:33
  • @Dominique How are these types of solution-verification questions supposed to be answered? In my understanding, the question was "Is this proof acceptable?", to which my answer was "Yes, your proof is correct." This did not have enough characters to post as an answer by itself, so I decided to add in something else. – DanDan面 Jun 10 '24 at 09:48
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    What I often read if such questions occur is that "this site is not meant as a proof checking machine". My suggestion would be that such questions are not answered , but handled in the comments. It is no problem if a question remains "unanswered" , better than answers like "this proof is OK." You are right that using this proof is like shooting rockets on ants , but this is not the point. The author wanted to verify this particular proof. – Peter Jun 10 '24 at 10:04