Let  β= (((24!)!)!)!, and let {${\mathcal P}_{n^2+1}$} denote the set of all primes of the form {$n^2+1$}. Let ${\mathcal M}$ denote the set of all positive multiples of elements of the set {${\mathcal P}_{n^2+1} \cap (\beta,\infty)$}. The set {${\mathcal X}=\{0,\ldots,\beta\} \cup {\mathcal M}$} satisfies the following conditions: (1) ${\rm card}({\mathcal X})$ is greater than a huge positive integer and it is conjectured that ${\mathcal X}$ is infinite, (2)} we do not know any algorithm deciding the finiteness of ${\mathcal X}$, (3)~a~known and short algorithm for every {$n \in N$} decides whether or not {$n \in {\mathcal X}$}, (4) a known and short algorithm returns an integer~$n$ such that ${\mathcal X}$ is infinite if and only if ${\mathcal X}$ contains an element greater than $n$. The following problem is open: {\em simply define a set {${\mathcal X} \subseteq N$} such that ${\mathcal X}$ satisfies conditions (1)-(4), and \underline{we do not know any representation of~~${\mathcal X}$~~as a finite union of sets whose definitions are simpler than} \underline{the definition of ${\mathcal X}$}} {\tt (5)}. Let {$f(1)=2$}, {$f(2)=4$}, and let {$f(n+1)=f(n)!$} for every integer {$n \geqslant 2$}. For a positive integer $n$, let {$\Psi_n$} denote the following statement: {\em if a system of equations ${\mathcal S} \subseteq \Bigl\{x_i!=x_k: i,k \in \{1,\ldots,n\}\Bigr\} \cup \Bigl\{x_i \cdot x_j=x_k: i,j,k \in \{1,\ldots,n\}\Bigr\}$ has only finitely many solutions in positive integers \mbox{$x_1,\ldots,x_n$}, then each such solution \mbox{$(x_1,\ldots,x_n)$} satisfies {$x_1,\ldots,x_n \leqslant f(n)$}.} We prove that for every statement $\Psi_n$ the bound {$f(n)$} cannot be decreased. The author's guess is that the statements {$\Psi_1,\ldots,\Psi_9$} are true. We prove that the statement $\Psi_9$ implies that the set ${\mathcal X}$ of all {non-negative} integers $k$ whose number of digits belongs to {${\mathcal P}_{n^2+1}$} satisfies conditions {\tt (1)-(5)}.