Is There a Single-Variable Solution to This Diophantine Equation?

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Discussion Overview

The discussion revolves around the possibility of finding a single-variable solution to the Diophantine equation a(3b+1)=c. Participants explore whether solutions can be expressed in terms of one variable, specifically looking for forms like a(n), b(n), and c(n).

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant presents the equation a(3b+1)=c and seeks a solution in terms of single-variable functions a(n), b(n), and c(n).
  • Another participant asks for clarification on what is meant by "a(n), b(n) (and so, c(n))".
  • A participant suggests that a solution in one variable exists, providing an example of a known solution for the equation x²+y²=z².
  • One participant claims there are infinitely many solutions and proposes a general form involving functions F(n) and G(n) to express a, b, and c.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the existence or form of a single-variable solution, with multiple viewpoints and examples presented without resolution.

Contextual Notes

Some assumptions about the nature of the functions F(n) and G(n) are not explicitly stated, and the discussion does not clarify the conditions under which the proposed solutions hold.

danieldf
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a(3b+1)=c
The solutions are obvious. I want to know if there is a solution a(n), b(n) (and so, c(n)) or at least a solution a(b).
Can someone help with that..?
 
Last edited:
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Sorry.. I didn't explain before.
 
danieldf said:
a(3b+1)=c
The solutions are obvious. I want to know if there is a solution a(n), b(n) (and so, c(n)) or at least a solution a(b).
Can someone help with that..?
What do you mean by "a(n), b(n) (and so, c(n))" ?
 
I mean a solutions in one variable. Like (2n, n²-1,n²+1) is a solutioin for x²+y²=z²
 
danieldf said:
I mean a solutions in one variable. Like (2n, n²-1,n²+1) is a solutioin for x²+y²=z²
There are an infinite number. The only comprehensive one of all possibilities though is
a = F(n), b = G(n), so c= F(n)*(3G(n)+1). Simple as that.
 
Last edited:

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