Is there an algorithm that decides the solvability of a Diophantine equation with a finite number of solutions?

TL;DR

The paper explores the complexity of determining the solvability of finite Diophantine equations, proposing a conjecture related to bounds on solutions and implications for algorithmic decidability.

Contribution

It introduces a conjecture linking bounds on solutions to Diophantine systems with finite solutions and shows its implications for algorithmic decidability of Diophantine equations.

Findings

Proves that certain bounds imply compositeness of specific numbers.

Shows the conjecture would imply an algorithm for deciding finite Diophantine solvability.

Connects finite-fold Diophantine representations with bounds on functions.

Abstract

For a positive integer n, let {\theta}(n) denote the smallest positive integer b such that for each system S \subseteq {x_i \cdot x_j=x_k, x_i+1=x_k: i,j,k \in {1,...,n}} which has a solution in positive integers x_1,...,x_n and which has only finitely many solutions in positive integers x_1,...,x_n, there exists a solution of S in ([1,b] \cap N)^n. We conjecture that there exists an integer {\delta} \geq 9 such that the inequality {\theta}(n) \leq (2^{2^{n-5}}-1)^{2^{n-5}}+1 holds for every integer n \geq {\delta}. We prove: (1) for every integer n>9, the inequality {\theta}(n)<(2^{2^{n-5}}-1)^{2^{n-5}}+1 implies that 2^{2^{n-5}}+1 is composite, (2) the conjecture implies that there exists an algorithm which takes as input a Diophantine equation D(x_1,...,x_p)=0 and returns the message "Yes" or "No" which correctly determines the solvability of the equation D(x_1,...,x_p)=0 in positive integers, if the solution set is finite, (3) if a function f:N\{0} \to N\{0} has a finite-fold Diophantine representation, then there exists a positive integer m such that f(n)<{\theta}(n) for every integer n>m.

Equations

$$E_n=\{x_i\cdot x_j=x_k,x_i+1=x_k:i,j,k\in \{1,\dots ,n\}\}$$

$$\left\{\begin{array}[]{rcl}x_{1}\cdot x_{1}&=&x_{1}\\ x_{1}+1&=&x_{2}\\ \forall i\in\{2,\ldots,n-1\}~{}x_{i}\cdot x_{i}&=&x_{i+1}\end{array}\right.$$

$$\left\{\begin{array}[]{rcl}\forall i\in\{1,\ldots,n\}~{}x_{i}\cdot x_{i}&=&x_{i+1}\\ x_{1}+1&=&x_{n+2}\\ x_{n+2}+1&=&x_{n+3}\\ x_{n+1}+1&=&x_{n+4}\\ x_{n+3}\cdot x_{n+5}&=&x_{n+4}\end{array}\right.$$

$$x_1^{2^n}+1={\left((x_1+2)-2\right)}^{2^n}+1=$$

$$2^{2^n}+1+(x_1+2)\cdot \sum _{k=1}^{2^n}\left(\genfrac{}{}{0ex}{}{2^n}{k}\right)\cdot (x_1+2{)}^{k-1}\cdot (-2{)}^{2^n-k}=(x_1+2)\cdot x_{n+5}$$

$$2^{2^n}+1=(x_1+2)\cdot \left(x_{n+5}-\sum _{k=1}^{2^n}\left(\genfrac{}{}{0ex}{}{2^n}{k}\right)\cdot (x_1+2{)}^{k-1}\cdot (-2{)}^{2^n-k}\right)$$

$$x_{n+4}=x_{n+1}+1={\left(2^{2^n}-1\right)}^{2^n}+1$$

$$\begin{array}[]{rcl}\forall i\in\{1,\ldots,n+1\}~{}a_{i}&=&\left(2^{\textstyle 2^{n}}-1\right)^{\textstyle 2^{i-1}}\\ a_{n+2}&=&2^{\textstyle 2^{n}}\\ a_{n+3}&=&2^{\textstyle 2^{n}}+1\\ a_{n+4}&=&\left(2^{\textstyle 2^{n}}-1\right)^{\textstyle 2^{n}}+1\\ a_{n+5}&=&1+\sum_{\textstyle k=1}^{\textstyle 2^{n}}\limits\displaystyle{{\textstyle 2^{n}}\choose{\textstyle k}}\cdot\left(2^{\textstyle 2^{n}}+1\right)^{\textstyle k-1}\cdot(-2)^{\textstyle 2^{n}-k}\end{array}$$

$$a_{n+5}=\frac{a_{n+4}}{a_{n+3}}=\frac{{\left(2^{2^n}-1\right)}^{2^n}+1}{2^{2^n}+1}$$

$$\theta (n)⩾{\left(2^{2^{n-5}}-1\right)}^{2^{n-5}}+1$$

$$\theta (9)⩾{\left(2^{2^{9-5}}-1\right)}^{2^{9-5}}+1={\left(2^{16}-1\right)}^{16}+1>{\left(2^{16}-2^{15}\right)}^{16}={\left(2^{15}\right)}^{16}=2^{240}>2^{2^{9-2}}$$

$${\left(2^{2^{6-5}}-1\right)}^{2^{6-5}}+1=10<65536=2^{2^{6-2}}$$

$${\left(2^{2^{7-5}}-1\right)}^{2^{7-5}}+1=50626<4294967296=2^{2^{7-2}}$$

$${\left(2^{2^{8-5}}-1\right)}^{2^{8-5}}+1=17878103347812890626<18446744073709551616=2^{2^{8-2}}$$

$$\theta (n)⩽{\left(2^{2^{n-5}}-1\right)}^{2^{n-5}}+1$$

$$(zx+1)(zy+1)=z^2(xy+1)+1$$

$$zx,zx+1,zy,zy+1,z^2,xy,xy+1,z^2(xy+1),z^2(xy+1)+1$$

$$D({\tilde{x}}_1,\dots ,{\tilde{x}}_p)=0⟺\exists {\tilde{x}}_{p+1},\dots ,{\tilde{x}}_n\in \mathbb{N}\setminus \{0\}({\tilde{x}}_1,\dots ,{\tilde{x}}_p,{\tilde{x}}_{p+1},\dots ,{\tilde{x}}_n)solves\mathcal{T}$$

$$D(x_1,\dots ,x_p)+\tilde{D}(x_1,\dots ,x_p)+1=\tilde{D}(x_1,\dots ,x_p)+1$$

$$\Bigl{(}\Bigl{(}\Bigl{(}\sum_{\textstyle(i_{1},j_{1},\ldots,i_{k},j_{k})\in A}x_{\textstyle i_{1}}^{\textstyle j_{1}}\cdot\ldots\cdot x_{\textstyle i_{k}}^{\textstyle j_{k}}\Bigr{)}+\underbrace{1\Bigr{)}+\ldots\Bigr{)}+1}_{\textstyle a~{}{\rm units}}=$$

$$\Bigl{(}\Bigl{(}\Bigl{(}\sum_{\textstyle(i_{1},j_{1},\ldots,i_{k},j_{k})\in B}x_{\textstyle i_{1}}^{\textstyle j_{1}}\cdot\ldots\cdot x_{\textstyle i_{k}}^{\textstyle j_{k}}\Bigr{)}+\underbrace{1\Bigr{)}+\ldots\Bigr{)}+1}_{\textstyle b~{}{\rm units}}$$

$$(a_1,\dots ,a_n)\in \mathcal{M}⟺\exists x_1,\dots ,x_m\in \mathbb{N}W(a_1,\dots ,a_n,x_1,\dots ,x_m)=0$$

$$(x_1,x_2)\in f⟺\exists x_3,\dots ,x_r\in \mathbb{N}\setminus \{0\}W(x_1,x_2,x_3-1,\dots ,x_r-1)=0$$

$$(x_1,x_2)\in f⟺\exists x_3,\dots ,x_s\in \mathbb{N}\setminus \{0\}\Phi (x_1,x_2,x_3,\dots ,x_s)$$

$$n⩾\left[\frac{n}{2}\right]+\frac{n}{2}>\left[\frac{n}{2}\right]+s+1$$

$$\left\{\begin{array}[]{c}\textrm{all~{}equations~{}occurring~{}in~{}}\Phi(x_{1},x_{2},x_{3},\ldots,x_{s})\\ \begin{array}[]{rcl}\forall i\in\left\{1,\ldots,n-\left[\frac{n}{2}\right]-s-2\right\}~{}u_{i}\cdot u_{i}&=&u_{i}\\ t_{1}\cdot t_{1}&=&t_{1}\\ \forall i\in\left\{1,\ldots,\left[\frac{n}{2}\right]-1\right\}~{}t_{i}+1&=&t_{i+1}\\ t_{2}\cdot t_{\left[\frac{n}{2}\right]}&=&u\\ u+1&=&x_{1}{\rm~{}(if~{}}n{\rm~{}is~{}odd)}\\ t_{1}\cdot u&=&x_{1}{\rm~{}(if~{}}n{\rm~{}is~{}even)}\\ x_{2}+1&=&y\end{array}\end{array}\right.$$

$$f(n)=f(x_1)=x_2<x_2+1=y$$

Full text

Key words and phrases: algorithmic decidability, Diophantine equation with a finite number of solutions, Fermat prime, finite-fold Diophantine representation, smallest solution of a Diophantine equation. 2010 Mathematics Subject Classification: 11U05.

In this article, we propose a conjecture which implies that there exists an algorithm which takes as input a Diophantine equation $D(x_{1},\ldots,x_{p})=0$ and returns the message "Yes" or "No" which correctly determines the solvability of the equation $D(x_{1},\ldots,x_{p})=0$ in positive integers, if the solution set is finite. Let

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For a positive integer $n$, let $\theta(n)$ denote the smallest positive integer $b$ such that for each system $S\subseteq E_{n}$ which has a solution in positive integers $x_{1},\ldots,x_{n}$ and which has only finitely many solutions in positive integers $x_{1},\ldots,x_{n}$, there exists a solution of $S$ in $([1,b]\cap\mathbb{N})^{n}$. We do not know whether or not there exists a computable function $\xi\colon\mathbb{N}\setminus\{0\}\to\mathbb{N}\setminus\{0\}$ which is greater than the function $\theta\colon\mathbb{N}\setminus\{0\}\to\mathbb{N}\setminus\{0\}$.

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has a unique solution in positive integers, namely $\left(1,2,4,16,256,\ldots,2^{\textstyle 2^{n-3}},2^{\textstyle 2^{n-2}}\right)$. ∎

Primes of the form $2^{\textstyle 2^{n}}+1$ are called Fermat primes, as Fermat conjectured that every integer of the form $2^{\textstyle 2^{n}}+1$ is prime ([2, p. 1]). Fermat correctly remarked that $2^{\textstyle 2^{0}}+1=3$, $2^{\textstyle 2^{1}}+1=5$, $2^{\textstyle 2^{2}}+1=17$, $2^{\textstyle 2^{3}}+1=257$, and $2^{\textstyle 2^{4}}+1=65537$ are all prime ([2, p. 1]). Open Problem. Are there infinitely many composite numbers of the form $2^{\textstyle 2^{n}}+1$? ([2, p. 159]) Most mathematicians believe that $2^{\textstyle 2^{n}}+1$ is composite for every integer $n\geqslant 5$.

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has a unique solution $\left(a_{1},\ldots,a_{n+5}\right)$ in non-negative integers. The numbers $a_{1},\ldots,a_{n+5}$ are positive and ${\rm max}\left(a_{1},\ldots,a_{n+5}\right)=a_{n+4}=\left(2^{\textstyle 2^{n}}-1\right)^{\textstyle 2^{n}}+1$.

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Hence,

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Therefore, $x_{1}+2$ divides $2^{\textstyle 2^{n}}+1$. Since $x_{1}+2\geqslant 2$ and $2^{\textstyle 2^{n}}+1$ is prime, we get $x_{1}=2^{\textstyle 2^{n}}-1$. Hence, $x_{n+2}=2^{\textstyle 2^{n}}$ and $x_{n+3}=2^{\textstyle 2^{n}}+1$. Next, $x_{n+1}=x_{1}^{\textstyle 2^{n}}=\left(2^{\textstyle 2^{n}}-1\right)^{\textstyle 2^{n}}$ and

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The following positive integers

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give the solution which is unique in non-negative integers. The number $a_{n+5}$ is positive because

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∎

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In particular,

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The numbers $2^{\textstyle 2^{n-5}}+1$ are prime when $n\in\{6,7,8\}$, but

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Conjecture. (cf. [7, p. 710]) There exists an integer $\delta\geqslant 9$ such that the inequality

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holds for every integer $n\geqslant\delta$.

Let $\alpha$, $\beta$, and $\gamma$ denote variables.

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∎

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Condition 2.* If positive integers $\tilde{x}_{1},\ldots,\tilde{x}_{p}$ satisfy $D(\tilde{x}_{1},\ldots,\tilde{x}_{p})=0$, then there exists a unique tuple $(\tilde{x}_{p+1},\ldots,\tilde{x}_{n})\in(\mathbb{N}\setminus\{0\})^{n-p}$ such that the tuple $(\tilde{x}_{1},\ldots,\tilde{x}_{p},\tilde{x}_{p+1},\ldots,\tilde{x}_{n})$ solves ${\cal T}$. Conditions 1 and 2 imply that the equation $D(x_{1},\ldots,x_{p})=0$ and the system ${\cal T}$ have the same number of solutions in positive integers.*

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There exist positive integers $a$ and $b$ and finite non-empty lists $A$ and $B$ such that the above equation is equivalent to

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and all the numbers $k,i_{1},j_{1},\ldots,i_{k},j_{k}$ belong to $\mathbb{N}\setminus\{0\}$. Next, we apply Corollary 5. ∎

The Davis-Putnam-Robinson-Matiyasevich theorem states that every recursively enumerable set ${\cal M}\subseteq{\mathbb{N}}^{n}$ has a Diophantine representation, that is

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for some polynomial $W$ with integer coefficients, see [3]. The polynomial $W$ can be computed, if we know the Turing machine $M$ such that, for all $(a_{1},\ldots,a_{n})\in{\mathbb{N}}^{n}$, $M$ halts on $(a_{1},\ldots,a_{n})$ if and only if $(a_{1},\ldots,a_{n})\in{\cal M}$, see [3]. The representation (R) is said to be single-fold, if for every $a_{1},\ldots,a_{n}\in\mathbb{N}$ the equation $W(a_{1},\ldots,a_{n},x_{1},\ldots,x_{m})=0$ has at most one solution $(x_{1},\ldots,x_{m})\in{\mathbb{N}}^{m}$. The representation (R) is said to be finite-fold, if for every $a_{1},\ldots,a_{n}\in\mathbb{N}$ the equation $W(a_{1},\ldots,a_{n},x_{1},\ldots,x_{m})=0$ has only finitely many solutions $(x_{1},\ldots,x_{m})\in{\mathbb{N}}^{m}$. Yu. Matiyasevich conjectured that each recursively enumerable set ${\cal M}\subseteq{\mathbb{N}}^{n}$ has a single-fold (finite-fold) Diophantine representation, see [1, pp. 341–342] and [4, p. 42]. Currently, he seems agnostic on his conjectures, see [5, p. 749]. In [8, p. 581], the author explains why Matiyasevich’s conjectures although widely known are less widely accepted.

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and for each positive integers $x_{1},x_{2}$ at most finitely many tuples $(x_{3},\ldots,x_{r})$ of positive integers satisfy $W(x_{1},x_{2},x_{3}-1,\ldots,x_{r}-1)=0$. By Lemma 2, there exists an integer $s\geqslant 3$ such that for every positive integers $x_{1},x_{2}$,

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where $\Phi(x_{1},x_{2},x_{3},\ldots,x_{s})$ is a conjunction of formulae of the forms $x_{i}+1=x_{k}$ and $x_{i}\cdot x_{j}=x_{k}$, the indices $i,j,k$ belong to $\{1,\ldots,s\}$, and for each positive integers $x_{1},x_{2}$ at most finitely many tuples $(x_{3},\ldots,x_{s})$ of positive integers satisfy $\Phi(x_{1},x_{2},x_{3},\ldots,x_{s})$. Let $[\cdot]$ denote the integer part function, and let an integer $n$ be greater than $m=2s+2$. Then,

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and $n-\left[\frac{n}{2}\right]-s-2\geqslant 0$. Let $T_{n}$ denote the following system with $n$ variables:

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By the equivalence (E), the system $T_{n}$ is solvable in positive integers, $2\cdot\left[\frac{n}{2}\right]=u$, $n=x_{1}$, and

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The system $T_{n}$ consists of equations of the forms $\alpha+1=\gamma$ and $\alpha\cdot\beta=\gamma$. Since $T_{n}$ has only finitely many solutions in positive integers, $y\leqslant\theta(n)$. Hence, $f(n)<\theta(n)$. ∎