On Sets ${\mathcal X} \subseteq$ $\mathbb N$ for Which We Know an Algorithm That Computes a Threshold Number $t({\mathcal X}) \in$ $\mathbb N$ Such That ${\mathcal X}$ Is Infinite If and Only If ${\mathcal X}$ Contains an eLement Greater Than $t({\mathcal X})$ [post]

Apoloniusz Tyszka
2018 unpublished
We define computable functions $g,h:$ $\mathbb N$ $\setminus \{0\} \to$ $\mathbb N$ $\setminus \{0\}$. For an integer $n \geqslant 3$, let $\Psi_n$ denote the following statement: if a system ${\mathcal S} \subseteq \Bigl\{x_i!=x_k: (i,k \in \{1,\ldots,n\}) \wedge (i \neq k)\Bigr\} \cup \Bigl\{x_i \cdot x_j=x_k: i,j,k \in \{1,\ldots,n\}\Bigr\}$ has only finitely many solutions in positive integers $x_1,\ldots,x_n$, then each such solution $(x_1,\ldots,x_n)$ satisfies $x_1,\ldots,x_n \leqslant
more » ... ots,x_n \leqslant g(n)$. For a positive integer $n$, let $\Gamma_n$ denote the following statement: if a system $S \subseteq \Bigl\{x_i \cdot x_j=x_k:~i,j,k \in \{1,\ldots,n\}\Bigr\} \cup \Bigl\{2^{\textstyle 2^{\textstyle x_i}}=x_k:~i,k \in \{1,\ldots,n\}\Bigr\}$ has only finitely many solutions in positive integers $x_1,\ldots,x_n$, then each such solution $(x_1,\ldots,x_n)$ satisfies $x_1,\ldots,x_n \leqslant h(n)$. We prove: (1) if the equation $x!+1=y^2$ has only finitely many solutions in positive integers, then the statement $\Psi_6$ guarantees that each such solution $(x,y)$ belongs to the set $\{(4,5),(5,11),(7,71)\}$, (2) the statement $\Psi_9$ proves the following implication: if there exists a positive integer $x$ such that $x^2+1$ is prime and $x^2+1>g(7)$, then there are infinitely many primes of the form $n^2+1$,  (3) the statement $\Psi_9$ proves the following implication: if there exists an integer $x \geqslant g(6)$ such that $x!+1$ is prime, then there are infinitely many primes of the form $n!+1$, (4) the statement $\Psi_{16}$ proves the following implication: if there exists a twin prime greater than $g(14)$, then there are infinitely many twin primes, {\bf (5)}~the statement $\Gamma_{13}$ proves the following implication: if $n \in$ $\mathbb N$ $\setminus \{0\}$ and $2^{\textstyle 2^{\textstyle n}}+1$ is composite and greater than $h(12)$, then $2^{\textstyle 2^{\textstyle n}}+1$ is composite for infinitely many positive integers $n$.
doi:10.20944/preprints201811.0301.v4 fatcat:jgqszi6uivbzdfgd2wbqsdvare