0

I have recently been doing a side project and researching if there is a deep and beautiful mathematical structure in the process of recursion itself. My idea was: once we treat algorithms themselves as mathematical objects, what can we learn about both algorithms and their mathematical recursive structure?

My (wild?) guess is that the continued fraction of e plays an important role in what I have called "mathematical recursive structure".

My yes/no question is:

(1) "for the general class of Diophantine equations with positive coefficients, is it possible to write an algorithm which produces the solution of any Diophantine equation in this class?"

I believe the answer is: no, in the general case no such algorithm can exist, and the reason for this depends essentially on questions related to continuity. I arrived at this idea from considering the class of polynomials with rational coefficients and demanding only rational number solutions. I can further explain my reason for believing this but for brevity am omitting the details.

But what about the Chinese Remainder Theorem? My idea is: what if represent rational numbers as repeating decimal binary expansions, and really ask ourselves the question:

"Did Qiaochu Yuan answer a 4th question in a post where I asked two questions and he gave three answers over thirteen years ago?"

The post in question is [1], I will briefly quote the relevant passage now:

The answer to question 3 is the following.

Lemma: If $n, m$ are relatively prime odd numbers, then $\text{ord}_{mn}(2) = \text{lcm}(\text{ord}_n(2), \text{ord}_m(2))$.

Proof. $\text{ord}_{mn}(2)$ is the order of the element $2$ in the multiplicative group of $\mathbb{Z}/mn\mathbb{Z}$, which we will denote $U(mn)$. By the Chinese remainder theorem, $U(mn)$ is isomorphic to the direct product $U(m) \times U(n)$, so the order of $2$ in $U(mn)$ must be the $\text{lcm}$ of the orders of $2$ in $U(m)$ and $U(n)$.

Given that I am myself ultimately still an amateur mathematician (I am a computer programmer), I am starting to reach the limits of being able to have all the heavy machinery to answer the types of questions I am encountering.

Question: is there a strong reason to believe the Chinese Remainder Theorem may play an important role in answering the question (1) which I have posed.

Thanks!

[1] On the binary expansion of the reciprocals of prime numbers

Martin Brandenburg
  • 181,922
Matt Calhoun
  • 4,414
  • 1
    A hurdle: "an algorithm which produces the solution" is infeasible for many Diophantine equation (systems), especially the ones having infinitely many solutions. Every computably enumerable set is the solution set of a Diophantine system and the "generic behaviour" of such sets is that they're infinite. Maybe requiring positive coefficients changes this, but no one seems to have published this claim. – Eric Towers Oct 04 '24 at 04:18
  • @EricTowers thats fascinating, and good to know. Thanks! – Matt Calhoun Oct 04 '24 at 22:24
  • I just wanted to leave a quick note that the reason for posing the question about positive coefficients is this makes the problem decidable with the PIT algorithm. whats fascinating to me is: if its decidable but no algorithm can produce the solution, then how can we verify that the problem is decidable in the first place? I know we can I just mean thats very interesting. – Matt Calhoun Oct 04 '24 at 22:47
  • 1
    So there are three broad classes of proofs that one should attempt on a problem class: existence of solutions, uniqueness of solutions, and computation of solutions. Just because an existence proof exists does not mean a computation method exists. I.e., decidability only requires one bit of output but the computation may require infinite output. – Eric Towers Oct 05 '24 at 00:49

0 Answers0