<p>The starting point of this question is the observation that the smallest positive integers $a,b,c$ satisfying</p>

<p>$$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 4$$</p>

<p>are <a href="https://plus.google.com/+johncbaez999/posts/Pr8LgYYxvbM" rel="noreferrer">absurdly high</a>. This leads to the following general question: Is the set $C\subseteq {\mathbb N}$ defined by $$ C = \left\{n\in\mathbb{N}\setminus\{0\}: \big(\exists a,b,c \in\mathbb{N}\setminus\{0\}\big):\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = n\right\}$$
computable?</p>

CC BY-SA 3.0
2017-10-26T15:39:56.097Z
I'm not an expert, but if you simplify the equation you get a cubic diophantine equation in three variables, and I think it's an open problem if universality can be achieved in such a setting. In every case, quadratic diophantine equations in two variables ($ax^2 + by + c = 0$) are already NP-complete (see [NP-complete decision problems for quadratic polynomials](https://dl.acm.org/citation.cfm?id=803627)) ... so I'm not surprised if switching to cubic+three vars leads to very hard instances (outside NP).
Marzio De Biasi
CC BY-SA 3.0
2017-10-26T16:05:23.300Z
Cross-posted from https://mathoverflow.net/questions/278747/is-this-set-computable
Emil Jeřábek
<computability>
Computability of a "weird" set
Dominic van der Zypen
# Computability of a "weird" set

(Asked by: username_0 on 2017-10-26)

The starting point of this question is the observation that the smallest positive integers $a,b,c$ satisfying

$$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 4$$

are [absurdly high](https://plus.google.com/+johncbaez999/posts/Pr8LgYYxvbM). This leads to the following general question: Is the set $C\subseteq {\mathbb N}$ defined by $$ C = \left\{n\in\mathbb{N}\setminus\{0\}: \big(\exists a,b,c \in\mathbb{N}\setminus\{0\}\big):\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = n\right\}$$
computable?

**Tags:** computability

> **Comments:**
> **username_1**: I'm not an expert, but if you simplify the equation you get a cubic diophantine equation in three variables, and I think it's an open problem if universality can be achieved in such a setting. In every case, quadratic diophantine equations in two variables ($ax^2 + by + c = 0$) are already NP-complete (see [NP-complete decision problems for quadratic polynomials](https://dl.acm.org/citation.cfm?id=803627)) ... so I'm not surprised if switching to cubic+three vars leads to very hard instances (outside NP).  
> **username_2**: Cross-posted from https://mathoverflow.net/questions/278747/is-this-set-computable  

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## 0 answers

