We know that Fermat's numbers are $F_n= 2^{2^n} +1$. My question is: does there exist certain forms of $n$ for which $F_n$ is always composite?
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As far as I know, it's open whether there are infinitely many composite Fermat numbers. – Wojowu Jul 19 '15 at 08:16
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I bet all $n>4$ ))) – Michael Galuza Jul 19 '15 at 08:20
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sir, I asked for certain forms of n. I mean certain types of n or some specific types of n. – Subhash Chand Bhoria Jul 19 '15 at 08:21
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@ michael : can you prove it sir? – Subhash Chand Bhoria Jul 19 '15 at 08:22
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1@SubhashChandBhoria, it's well-known open problem. – Michael Galuza Jul 19 '15 at 08:24
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@ michael : ok sir – Subhash Chand Bhoria Jul 19 '15 at 08:39
2 Answers
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We don't know. According to prothsearch, $F_{3329780}$ is the largest Fermat number that is known to be composite.
wythagoras
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Let $f(1)=1$, and let $f(n+1)=2^{2^{f(n)}}$ for every positive integer $n$.
Let $\Phi$ denote the statement:
if a system $S \subseteq \{x_i \cdot x_j=x_k: i,j,k \in \{1,...,17\}\}
\cup \{2^{2^{x_i}}=x_k: i,k \in \{1,...,17\}\} $ has only finitely many solutions in non-negative
integers $x_1,...,x_{17}$, then each such solution $(x_1,...,x_{17})$ satisfies $x_1,...,x_{17} \leq f(17)$.
Theorem. If the statement $\Phi$ is true, and if $2^{2^n}+1$ is composite for some integer $n \geq f(15)$, then $2^{2^n}+1$ is composite for infinitely many non-negative integers $n$.
Reference:
A. Tyszka, link
A conjecture which implies that there exists a computable upper bound for the
heights of solutions of a Diophantine equation with a finite number of solutions,
Gottfried Helms
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