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"An unusual cubic representation problem" by Bremner & Macleod (2014) describes an infinite set $S$ of even positive integers, such that for each $n\in S$, the equation $$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}=n\tag{1}$$ has positive integer solutions $(a>0,b>0,c>0).$ In particular, for each $n\in S$, there is a smallest such solution $(a,b,c)$, whose maximum element ($\max\{a,b,c\}$) has a number of digits we denote by $M(n)$. The function $M()$ is unbounded but not monotonic, with occasional very large increases; e.g. (from results in the paper), $$M(4)=81\\ M(136)=26942\\ M(178)=398605460\\ M(198)=726373\\ M(896)\gt 2187147111901. $$

Now, Equation (1) is shown to correspond to an elliptic curve $$ y^2 = x^3 + (4n^2 + 12n − 3)x^2 + 32(n + 3)x\tag{2}$$ in which the coefficients are increasing functions of $n$, and [this Quora posting] asserts that "The negative solution of Hilbert’s 10th problem means that the growth of the solutions as the coefficients get larger is an uncomputable function", and that "the correspondence 4→ 80-digit numbers, 178→ hundreds-of-millions-digit numbers and 896→ trillions of digits gives us a glimpse into the first few tiny steps of that monstrous, uncomputable function."

Question: Is $M()$ an uncomputable function? If so, how does this follow from the negative solution of Hilbert’s 10th problem, or how is it otherwise proved?

I'm aware that certain one-parameter families of Diophantine equations (e.g. this one) supply a negative solution to Hilbert's 10th problem, but is the present case such a family?

r.e.s.
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  • Clarify. You just said the algorithm enumerating and testing the $(a,b,c) \le k$ halts for some $k=M(n)$, but your linked article discusses the opposite (for $n$ odd). – reuns Sep 11 '17 at 02:22
  • @reuns : He takes $S$ to be the domain of $M$. The process of iterating $m$ to find a solution is shown to halt for elements of $S$. – Eric Towers Sep 11 '17 at 03:06
  • but then $M$ is computable by definition of $S$. – reuns Sep 11 '17 at 03:07
  • @reuns - Function $M()$, defined for the infinitely many even $n\in S$, relates to pages 36-40, esp. Theorem 5.3, of the Bremner & Macleod article. It seems to me (but I'm not 100% sure) that the procedures they use to calculate $M(n)$ for some $n$, do in fact provide an algorithm for all $n\in S$, so I suspect that the cited Quora posting is simply incorrect in claiming $M()$ to be uncomputable. I'm uncertain because the mathematics involved is somewhat above my pay grade :). – r.e.s. Sep 11 '17 at 03:28
  • The problem is for the $n$ such that there is no solution for $(a,b,c)$, in that case you also need an algorithm to detect those and exactly those $n$. $M$ is non-computable iff such an algorithm doesn't exist (you know the halting problem ?) – reuns Sep 11 '17 at 03:30
  • @reuns -- Yes, I'm supposing that $S$ itself is likewise computable. – r.e.s. Sep 11 '17 at 03:32
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    Note that (1) has been discussed before, here and on MO. https://math.stackexchange.com/questions/402537/find-integer-in-the-form-fracabc-fracbca-fraccab and https://mathoverflow.net/questions/227713/estimating-the-size-of-solutions-of-a-diophantine-equation and elsewhere. – Gerry Myerson Sep 11 '17 at 03:47
  • Additional sources of which I was not aware include https://mathoverflow.net/q/278747/20307 and the links in comments there. – r.e.s. Jun 20 '22 at 05:34

1 Answers1

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The negative solution to Hilbert's 10th problem says there is no algorithm that decides solvability over the integers for all possible Diophantine equations. This negative result is silent on particular Diophantine equations, and this is a particular one.

Edit: I may have been a bit hasty in reading the method description prior to table 2. The method is given for rank one curves. The method may or may not work, or work with modification, for other ranks of curves.

At this point, the following shows that $M_1(N)$, the version of $M$ restricted to rank one curves, is computable. The argument does not eliminate an obstruction to computability for other ranks.

The described method to construct table 2 (in section 4) of the Bremner & Macleod (2014) paper shows that $M$ is not uncomputable. We see it has been sequentially computed for $S \cap [1,200]$. Continuing the computation in the paper, we could extend $M$ as far as desired (at relatively rapidly increasing computational cost). (They clearly computed $M$ for some arguments greater that $200$. For instance, they know that $M(896) > 2.187 \times 10^{12}$ (penultimate paragraph of p. 40).)

Aside: The method they used is essentially the method of using enumerators for decidability and semidecidability. In other words, it was running $|m|$ sequentially through the natural numbers (the parameter $m$ is described in section 6 of the paper) until two inequalities (the paper's (4.1) and (4.2)) are satisfied, then computed the maximum of the number of digits of $a$, $b$, or $c$ for the corresponding solution. This is entirely do-able for a Turing Machine and the results in the sections prior to section 6 show that such a TM computation will terminate for each element of $S$ (although it could take a very long time to do so). Thus, $M$ is computable.

Eric Towers
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  • Thanks for this answer, esp. the Aside. (I think you should restate the part that says the existence of Table 2 shows $M$ to be computable, because, as you know, the computability of $M(n)$ for some $n$ says nothing about the computability of the function $M()$.) – r.e.s. Sep 11 '17 at 03:45
  • @r.e.s. : Okay. I've reduced the vigor of the table 2 sentence... – Eric Towers Sep 11 '17 at 07:06
  • Something still bothers me ... Suppose $S$ is the set of all $n\in \mathbb{N}$ such that (1) has solutions as required. The paper mentions a parameterization giving a generator point ($P$) on the elliptic curve as a function of parameter $t$, such that by considering successive multiples $mP$, one will eventually find a point that satisfies 4.1 or 4.2. So that's enough to show $S$ is infinite, etc. But aside from that one particular parameterization, how can one determine whether $n\in S$ for all $n\in \mathbb{N}$? – r.e.s. Sep 11 '17 at 14:24
  • @r.e.s. : Hmm... You might be on to something. I had glossed that the explanation of their method prior to table 2 was only for rank one curves. I don't imagine there is an insurmountable obstruction for other ranks, but I haven't studied that aspect of the problem. I'm adding a little text to this effect. – Eric Towers Sep 11 '17 at 15:44