The set ( C ) is computable.

To determine if a given ( n \in \mathbb{N} \setminus {0} ) is in ( C ), we need to check if there exist positive integers ( a, b, c ) satisfying ( \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = n ). Through transformation, this equation can be rephrased in terms of ( x = b+c ), ( y = a+c ), ( z = a+b ), leading to the Diophantine equation:

[ (x +...

Short answer: We can semi-decide membership in C (i.e., C is computably enumerable), but whether C is decidable is currently unknown. Under standard number-theoretic conjectures (e.g., Birch–Swinnerton-Dyer plus finiteness of the Tate–Shafarevich group), C would be decidable.

Details:

1. Turning the condition into a polynomial Diophantine equation

Let E(a,b,c) = a/(b+c) + b/(a+c) + c/(a+b). The...

Let

S(a ,b ,c ) = a/(b+c) + b/(a+c) + c/(a+b)

and

C = { n ≥ 1 : ∃ a ,b ,c ≥ 1  S(a ,b ,c ) = n } .

1. The problem is equivalent to a single polynomial equation

   F(a ,b ,c ,n) = 0 ,

   where F is obtained by clearing the denominators:

   a(a+b)(a+c) + b(a+b)(b+c) + c(a+c)(b+c)
   

   – n (a+b)(a+c)(b+c) = 0 .

   Expanding, F is an integral polynomial of total degre...

First, rewrite the defining condition

 n ∈ C  ⟺ ∃ a,b,c∈ℕ₊ : a/(b+c) + b/(a+c) + c/(a+b) = n.

Clearing denominators gives a single homogeneous cubic Diophantine equation

 F(a,b,c,n) = a(a+c)(a+b) + b(b+c)(b+a) + c(c+a)(c+b)         – n·(a+b)(b+c)(c+a) = 0.

Hence

(i) C is clearly recursively enumerable: on input n, dovetail all triples (a,b,c), check F(a,b,c,n)=0, and halt if ...

Let

        a     b     c

E(a,b,c) = ─── + ─── + ─── . b+c a+c a+b

For every positive integer n let

     F n (X,Y,Z)  =  n(Y+Z)(X+Z)(X+Y)
                    – X(X+Z)(X+Y)
                    – Y(Y+Z)(X+Y)
                    – Z(Y+Z)(X+Z)            (1)

and put

     C = { n ≥ 1 :  F n (a,b,c)=0 has a solution
                    a,b,c ∈ ℕ,  a,b...

This is an excellent question that connects a seemingly simple number theory problem to the foundations of computability theory.

Here is a breakdown of the answer, structured into two parts:

1. Showing that the set is computably enumerable (also known as recursively enumerable or semidecidable).
2. Arguing that the set is not computable (or decidable).

Summary

The set ...

Computability of a "Weird" Set

This question explores the computability of a set defined by a specific Diophantine equation. Let me analyze this thoroughly.

Understanding the Problem

The set consists of positive integers such that there exist positive integers , , and satisfying:

We're told that the smallest solution...

<think> I need to determine if the set C is computable. C is defined as the set of natural numbers n such that there exist positive integers a, b, c satisfying the equation:

[\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = n]

The question mentions that for n=4, the smallest solutions are very large, which is interesting but might not directly help with computability.

First, I should understan...

This is an interesting question about the computability of a set defined by a Diophantine-like condition. Let me analyze this carefully.

Analysis of the Set C

The set C consists of all positive integers n such that the equation has a solution in positive integers a, b, c.

Key Observations

1. Clearing denominators: Multiplying thr...