Hilbert’s Tenth Problem was whether there is an algorithm which will answer whether any Diophantine equation has solutions (where we want integer solutions). Hilbert’s Tenth has a negative solution by Matiyasevich’s theorem which showed that the question is equivalent to the Halting problem.

The equivalent problem is now known to also have a negative solution for many rings other than Z by essentially generalizing Matiyasevich’s theorem to other rings. Early work allowed one to replace Z by any quadratric extension of Z, and later work showed the same result ring of integers of any algebraic number field whose Galois group over the rationals is abelian.

However, the case of the rationals is still open. The naive thing to do would be to try to find a Diophantine equation over Q whose solutions are exactly the integer. To see why this would suffice, note that given two Diophantine equations over Q one can make a single Diophantine equation whose solutions is their intersection, so one could just then combine this equation with Matiyasevich’s equations. However, attempts to construct such a Diophantine equation failed. There is a conjecture of Barry Mazur that implies that in fact no such equation exist. Mazur conjectured the following:

**Conjecture** (Mazur): Given a variety over the rationals, the topological closure over the reals of the set of solutions has only finitely many connected components.

This question concern’s what happens if one tries to make such an equation in the most naive way possible. Say for example, that one knows that every positive integer is the sum of four perfect squares. So one could try looking at the Diophantine equation $$x -a_1^2 +a_2^2 + a_3^2 +a_4^2=0. ,, (1)$$

One might hope that maybe when one looks at Equation 1 that only some positive rational values of $x$ allow a solution in the $a_i$. But alas, the homogenous nature of $a_1^2 +a_2^2 + a_3^2 +a_4^2$ implies that every positive rational number leads to solutions. And the same would hold true if one tried to use any homogenous way of representing any positive integer (such as as a sum of cubes, or anything else arising from Waring’s problem). But we can without too much work modify this equation a little bit, and show that the set $A_0$= {$x|x$ x is rational and $x =0, 1$ or $x geq 2$ }is Diophantine over Q. So now, we can define for $i geq 1$

$$A_i = {x|x in A_{i-1}, exists a_1, a_2, a_3, a_4 in A_{i-1}, s.t. x -a_1^2 +a_2^2 + a_3^2 +a_4^2=0 }.$$

Notice that if $r in A_i$, and $r$ is not an integer, then $r > 2^{2^i}$. But for any $i$, $A_i$ will contain non-integer values. One could try the same sort of approach using instead of Equation 1, some other quadratic form which represents all positive integers, and one would get roughly the same growth. If one instead used for example that every positive integer is expressible as the sum of 16 perfect fourth powers, and called that $B_i$ then one could still construct an explicit constant $C$ such that for $i>1$, the smallest non-integer rational in $B_i$ is no more than $2^{C^i}$, and one could do the same for any other Waring type relation.

What we would like to do is essentially the same approach as with the $A_i$ or $B_i$ but where the iteration of constructing more and more complicated Diophantine equations, which correspond to a series of Diophantine sets $S_i$ recursively by insisting that all variables in our corresponding equation live in $S_{i-1}$. More explicitly:

Given a set of rational numbers $R$, let us write $f(R)$ to be the infimum of all non-integer elements of $R$.

Let $P(x,y_1, y_2, cdots y_k)$ be a polynomial with integer coefficients where for any sufficiently large integer $x geq M$ (for some integer M), the Diophantine relationship satisfied $P(x,y_1, y_2, cdots y_k)=0$ has a non-negative integer solution $y_i$. Then we can define $S_0(P)$ to be the set of numbers which are $1, 2, 3 cdots max{2,M}$, or are rationals $r$ such that $P(r, y_1, y_2, cdots y_k)=0$ has a solution in rational $y_j$. We can then define $S_i$ recursively, by at each stage insisting that the $y_j$ be in $S_{i-1}$.

The question then is:

**Question 1:** Can we construct a $P$ such that there for any positive real numbers $beta$ and $alpha$, for sufficiently large $i$, $f(S_i(P)) > beta^{alpha^i}$?

Essentially this is asking if we can by repeatedly applying the same Diophantine relations that we know are true for integers, can we very quickly build up Diophantine relations over Q which avoid many small non-integer values.

If $P$ is restricted to being just a set of homogenous relations like those from Waring’s problem, we cannot get the fast growth we want for Question 1. We also cannot do so even if one is non-homogenous if one is allowed to have floating isolated powers. For example, a form like $x=a^2 + b^2c + bc^2 + bcd^4 +e^3$ would not beat what we want because of the $a^2$ and $e^3$ terms.

Note also that if Z or N is Diophantine over Q, then we can trivially get a solution to Question 1 by using the relevant relation, since then we can construct a set of $S_i$ such that $f(S_i) = infty$ for all $i$. So if the answer to Question 1 is “No” this would be likely very difficult to prove. However, my suspicion is that the answer is in fact yes, although it may take more than simply a series of relations of the form $x=$ blah, and may require more sophisticated interaction.