Proving Non-Equality of Cubes of Natural Numbers

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Discussion Overview

The discussion centers around the proposition that the sum of the cubes of two natural numbers cannot equal the cube of a third natural number, specifically exploring the implications of Fermat's Last Theorem (FLT) in this context. Participants engage in a mix of theoretical reasoning and attempts to understand the proof of FLT, particularly for the case when n=3, as well as related concepts in number theory.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asks for help in proving that \(a^3 + b^3 \neq c^3\) for \(a, b, c \in \mathbb{N}\).
  • Some participants reference Fermat's Last Theorem, stating that there are no solutions to the equation \(x^n + y^n = z^n\) for \(n > 2\).
  • There is a discussion about the difficulty of proving FLT in general, with some suggesting that proving it for specific cases, like \(n=3\), is simpler.
  • One participant expresses doubt about the simplicity of proving FLT for \(n=3\), while another asserts that it is indeed manageable.
  • Several participants discuss the historical context of FLT, including Andrew Wiles' proof and the complexity involved in it.
  • There are references to the mathematical background required for understanding proofs related to FLT, including group theory and Galois theory.
  • One participant suggests that any square number can be expressed as a sum of consecutive odd numbers and questions if cubes can be expressed similarly.
  • Another participant provides reasoning for the infinitude of solutions for \(n=2\) and discusses methods for generating Pythagorean triples.
  • Some participants challenge the generality of certain proofs and reasoning presented in the thread.

Areas of Agreement / Disagreement

Participants express a mix of agreement and disagreement regarding the simplicity of proving FLT for specific cases. While some believe that proving it for \(n=3\) is straightforward, others caution against underestimating the complexity involved. The discussion remains unresolved on the broader implications of these proofs and the nature of Fermat's original claim.

Contextual Notes

Participants note the limitations of their discussions, including the need for deeper mathematical knowledge and the challenges of proving statements for all natural numbers. There is also an acknowledgment that the proofs for specific cases do not necessarily extend to the general case.

  • #31
mathbalarka said:
It's quite trivial to prove that.

Absolutely!My understanding

Let $$b^2$$ be an odd square number.There exists a pythagorean triple including $$b^2$$ because the odd number $$b^2$$ can form the last number of the string of consecutive odd numbers of another square number $$a^2$$.So there are PPTs as long as there are odd numbers and their squares,so they are infinite...
 
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  • #32
i have a little different approach to prove there are infinite triples ,(though its unnecessary)
let x,z be two variables ,
let x^2+z^2=(x+y)^2
then z^2=y^2+2xy =y(y+2x)
and if we consider y to be perfect square we can adjust x such that y+2x is perfect square
since there are infinite perfect squares there and infinite x's and thus we can find infinite z's thus there are infinite solutions to x^2+y^2=z^2

- - - Updated - - -

i tried applying same method to n=3 but i found myself cycling around ... :mad:
 

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