1. Composite numbers of the form ${\mathbf{2}}^{{\mathbf{2}}^{\mathit{n}}}+\mathbf{1}$

Let $\mathcal{A}$ denote the following system of equations:

$$\left\{x_i·x_j=x_k:i,j,k\in \{1,\dots ,7\}\right\}\cup \left\{2^{2^{x_j}}=x_k:j,k\in \{1,\dots ,7\}\right\}$$

The following subsystem of $\mathcal{A}$ has exactly one solution in ${(\mathbb{N}\setminus \{0,1\})}^7$, namely $(2,4,16,2,2,2,2)$.

Proof. 

Assume, on the contrary, that Hypothesis 1 holds and $2^{2^{x_1}}+1$ is composite for at most finitely many integers $x_1$ greater than 1. Then, the equation

$$x_2·x_3=2^{2^{x_1}}+1$$

has at most finitely many solutions in ${(\mathbb{N}\setminus \{0,1\})}^3$. By Lemma 1, in positive integers greater than 1, the following subsystem of $\mathcal{A}$ has at most finitely many solutions in ${(\mathbb{N}\setminus \{0,1\})}^7$ and expresses that

$$\left\{\begin{array}{ccc}\hfill x_2·x_3& =& 2^{2^{x_1}}+1\hfill \\ \hfill x_4& =& 2^{2^{x_1}}+1\hfill \\ \hfill x_5& =& 2^{2^{x_1}}\hfill \\ \hfill x_6& =& 2^{2^{2^{2^{x_1}}}}\hfill \\ \hfill x_7& =& 2^{2^{2^{2^{x_1}}+1}}\hfill \end{array}\right.$$

Since $641·6700417=2^{2^5}+1>16$, we get a contradiction.    □

2. The Brocard-Ramanujan equation $\mathit{x}!+\mathbf{1}={\mathit{y}}^{\mathbf{2}}$

Let $\mathcal{B}$ denote the following system of equations:

$$\{x_i·x_j=x_k:i,j,k\in \{1,\dots ,6\}\}\cup \{x_j!=x_k:(j,k\in \{1,\dots ,6\})\wedge (j\ne k)\}$$

The following subsystem of $\mathcal{B}$ has exactly two solutions in positive integers, namely $(1,\dots ,1)$ and $(2,2,4,24,24!,(24!)!)$.

Proof. 

The following system of equations ${\mathcal{B}}_1$ is a subsystem of $\mathcal{B}$. By Lemma 2, in positive integers, the system ${\mathcal{B}}_1$ expresses that $x_1=\dots =x_6=1$ or

$$\left\{\begin{array}{ccc}\hfill x_1!+1& =& x_2^2\hfill \\ \hfill x_3& =& x_1!\hfill \\ \hfill x_4& =& (x_1!)!\hfill \\ \hfill x_5& =& x_1!+1\hfill \\ \hfill x_6& =& (x_1!+1)!\hfill \end{array}\right.$$

If the equation $x_1!+1=x_2^2$ has at most finitely many solutions in positive integers $x_1$ and $x_2$, then ${\mathcal{B}}_1$ has at most finitely many solutions in positive integers $x_1,\dots ,x_6$ and Hypothesis 2 implies that every tuple $(x_1,\dots ,x_6)$ of positive integers that solves ${\mathcal{B}}_1$ satisfies $(x_1!+1)!=x_6⩽(24!)!$. Hence, $x_1\in \{1,\dots ,23\}$. If $x_1\in \{1,\dots ,23\}$, then $x_1!+1$ is a square only for $x_1\in \{4,5,7\}$.    □

3. Erdös’ equation $\mathit{x}(\mathit{x}+\mathbf{1})=\mathit{y}!$

Let $\mathcal{C}$ denote the following system of equations:

$$\{x_i·x_j=x_k:(i,j,k\in \{1,\dots ,6\})\wedge (i\ne j)\}\cup \{x_j!=x_k:(j,k\in \{1,\dots ,6\})\wedge (j\ne k)\}$$

The following subsystem of $\mathcal{C}$ has exactly three solutions in positive integers, namely $(1,\dots ,1)$, $(1,1,2,2,2,2)$, and $(2,2,3,6,720,720!)$.

Proof. 

The following system of equations ${\mathcal{C}}_1$ is a subsystem of $\mathcal{C}$. By Lemma 2, in positive integers, the system ${\mathcal{C}}_1$ expresses that $x_1=\dots =x_6=1$ or

$$\left\{\begin{array}{ccc}\hfill x_1·(x_1+1)& =& x_2!\hfill \\ \hfill x_3& =& x_1·(x_1+1)\hfill \\ \hfill x_4& =& x_1!\hfill \\ \hfill x_5& =& x_1+1\hfill \\ \hfill x_6& =& (x_1+1)!\hfill \end{array}\right.$$

If the equation $x_1(x_1+1)=x_2!$ has at most finitely many solutions in positive integers $x_1$ and $x_2$, then ${\mathcal{C}}_1$ has at most finitely many solutions in positive integers $x_1,\dots ,x_6$ and Hypothesis 3 implies that every tuple $(x_1,\dots ,x_6)$ of positive integers that solves ${\mathcal{C}}_1$ satisfies $x_2!=x_3⩽720!$. Hence, $x_2\in \{1,\dots ,720\}$. If $x_2\in \{1,\dots ,720\}$, then $x_2!$ is a product of two consecutive positive integers only for $x_2\in \{2,3\}$ because the following MuPAD program

for x2 from 1 to 720 do

 x1:=round(sqrt(x2!+(1/4))-(1/2)):

 if x1*(x1+1)=x2! then print(x2) end_if:

 end_for:

returns 2 and 3.    □

4. Hypotheses 2 and 3 cannot be generalized to an arbitrary number of variables

5. Equivalent forms of Hypotheses 1–3

Theorem 5.

The following semi-algorithm ${\mathtt{sem}}_1$ never terminates.If Hypothesis 1 is true, then ${\mathtt{sem}}_1$ endlessly prints consecutive positive integers starting from 1. If Hypothesis 1 is false, then ${\mathtt{sem}}_1$ prints a finite number (including zero) of consecutive positive integers starting from 1.

Theorem 6.

The following semi-algorithm ${\mathtt{sem}}_2$ never terminates.If Hypothesis 2 is true, then ${\mathtt{sem}}_2$ endlessly prints consecutive positive integers starting from 1. If Hypothesis 2 is false, then ${\mathtt{sem}}_2$ prints a finite number (including zero) of consecutive positive integers starting from 1.

Theorem 7.

The following semi-algorithm ${\mathtt{sem}}_3$ never terminates.If Hypothesis 3 is true, then ${\mathtt{sem}}_3$ endlessly prints consecutive positive integers starting from 1. If Hypothesis 3 is false, then ${\mathtt{sem}}_3$ prints a finite number (including zero) of consecutive positive integers starting from 1.

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