<p>From this 2017 paper on <a href="https://www.degruyter.com/view/journals/comp/7/1/article-p17.xml?language=en" rel="nofollow noreferrer">upper bounds for solutions to diophantine equations</a>:</p>

<blockquote>
  <p>Conjecture 1. If a system of equations S ⊆ Bn has exactly one solution
  in positive integers x1, . . . , xn , then x1, . . . , xn ≤ g(2n). In
  other words, ξ (n) ≤ g(2n) for every positive integer n.</p>
</blockquote>

<p>And following this:</p>

<blockquote>
  <p>Theorem 2. Conjecture 1 implies that there exists an algorithm which
  takes as input a Diophantine equation D(x1, . . . , xp) = 0 and
  returns a positive integer d with the following property: for every
  positive integers a1, . . . , ap, if the tuple (a1, . . . , ap) solely
  solves the equation D(x1, . . . , xp) = 0 in positive integers, then
  a1, . . . , ap ≤ d.</p>
</blockquote>

<p>This means that given a diophantine equation, we can first compute <span class="math-container">$d$</span>, and then check the finitely many tuples less than or equal to <span class="math-container">$d$</span> to find a solution. If we don't find a solution, this means the diophantine equation does not have exactly one solution - it has none or more than one.</p>

<p>Now from <a href="http://www.math.umd.edu/~laskow/Pubs/713/Diophantine.pdf" rel="nofollow noreferrer">Hilbert's 10th problem is unsolvable</a>:</p>

<blockquote>
  <p>Let #(P) be the number of solutions of the Diophantinequation P = 0.
  Thus O ≤ #(P) ≤ N_0. Hilbert's tenth problem seeks an algorithm for
  deciding of a given P whether or not #(P) = 0. But there are many
  related questions: Is there an algorithm for testing whether #(P) = N,
  or #(P) = 1, or #(P) is even? I was able to show easily (beginning
  with the unsolvability of Hilbert's tenth problem) that all of these
  problems are unsolvable. In fact if A = {0,1,2,3,.. N_0} and B subset of
  A, B not null set, B not equal to A, then one can readily show that
  there is no algorithm for determining whether or not #(P) is an
  element of B</p>
</blockquote>

<p>Does that mean conjecture 1 and hence theorem 2 (and to be frank atleast half of the rest of the paper) is false? Or am I missing something?</p>

CC BY-SA 4.0
2020-04-18T16:27:01.767Z
If you check all tuples below $d$ and you *do* find a solution, this does not imply that there exists a unique solution. It only implies that there exists at least one solution. So, this does not actually contradict Davis’s claim. Having said that, Tyszka is known for inventing various conjectures of dubious plausibility, some of which have been disproved.
Emil Jeřábek
CC BY-SA 4.0
2020-04-18T16:31:21.453Z
@EmilJeřábek Thanks, that answers it. Feel free to post it as an answer, others I could delete the question if that's better
ghosts_in_the_code
<nt.number-theory><upper-bounds><decidability><comp-number-theory><undecidability>
Is this a proof that diophantine equation solutions can't be bounded by power towers?
ghosts_in_the_code
# Is this a proof that diophantine equation solutions can't be bounded by power towers?

(Asked by: username_0 on 2020-04-18)

From this 2017 paper on [upper bounds for solutions to diophantine equations](https://www.degruyter.com/view/journals/comp/7/1/article-p17.xml?language=en):

> Conjecture 1. If a system of equations S ⊆ Bn has exactly one solution
> in positive integers x1, . . . , xn , then x1, . . . , xn ≤ g(2n). In
> other words, ξ (n) ≤ g(2n) for every positive integer n.

And following this:

> Theorem 2. Conjecture 1 implies that there exists an algorithm which
> takes as input a Diophantine equation D(x1, . . . , xp) = 0 and
> returns a positive integer d with the following property: for every
> positive integers a1, . . . , ap, if the tuple (a1, . . . , ap) solely
> solves the equation D(x1, . . . , xp) = 0 in positive integers, then
> a1, . . . , ap ≤ d.

This means that given a diophantine equation, we can first compute $d$, and then check the finitely many tuples less than or equal to $d$ to find a solution. If we don't find a solution, this means the diophantine equation does not have exactly one solution - it has none or more than one.

Now from [Hilbert's 10th problem is unsolvable](http://www.math.umd.edu/~laskow/Pubs/713/Diophantine.pdf):

> Let #(P) be the number of solutions of the Diophantinequation P = 0.
> Thus O ≤ #(P) ≤ N\_0. Hilbert's tenth problem seeks an algorithm for
> deciding of a given P whether or not #(P) = 0. But there are many
> related questions: Is there an algorithm for testing whether #(P) = N,
> or #(P) = 1, or #(P) is even? I was able to show easily (beginning
> with the unsolvability of Hilbert's tenth problem) that all of these
> problems are unsolvable. In fact if A = {0,1,2,3,.. N\_0} and B subset of
> A, B not null set, B not equal to A, then one can readily show that
> there is no algorithm for determining whether or not #(P) is an
> element of B

Does that mean conjecture 1 and hence theorem 2 (and to be frank atleast half of the rest of the paper) is false? Or am I missing something?

**Tags:** nt.number-theory, upper-bounds, decidability, comp-number-theory, undecidability

> **Comments:**
> **username_1**: If you check all tuples below $d$ and you *do* find a solution, this does not imply that there exists a unique solution. It only implies that there exists at least one solution. So, this does not actually contradict Davis’s claim. Having said that, Tyszka is known for inventing various conjectures of dubious plausibility, some of which have been disproved.  
> **username_0**: @EmilJeřábek Thanks, that answers it. Feel free to post it as an answer, others I could delete the question if that's better  

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## 0 answers

