What is the Connection Between the ABC Conjecture and Fermat's Last Theorem?

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Discussion Overview

The discussion revolves around the connection between the ABC conjecture and Fermat's Last Theorem, exploring the implications of the ABC conjecture on the validity of Fermat's Last Theorem and related number theory concepts. Participants engage with mathematical reasoning, conjectures, and examples related to integer solutions and prime numbers.

Discussion Character

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

Main Points Raised

  • Some participants suggest that a proof of the ABC conjecture could lead to a concise proof of Fermat's Last Theorem and other conjectures in number theory.
  • Questions are raised about how the proposed equations demonstrate that x, y, and z cannot be integers when p > 2, with some arguing that a, b, and c can take any values.
  • One participant asserts that p, x, y, a, b, and c are always positive integers when p is a prime number, while noting exceptions when p is 1.
  • Another participant challenges the assertion that z and c cannot be positive integers for p > 2, stating that proof is required for such claims.
  • Concerns are expressed about the placement of the discussion in the forum, with one participant confused by the initial mathematical expressions presented.
  • Examples are provided to illustrate the relationships between the variables and the conditions under which certain equations hold, particularly for p = 2.
  • Some participants emphasize that Fermat's theorem applies to all integers, not just primes, and call for clarification on the specific theorem being referenced.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the ABC conjecture for Fermat's Last Theorem, the nature of integer solutions for various values of p, and the definitions of the variables involved. The discussion remains unresolved with multiple competing perspectives presented.

Contextual Notes

Limitations include unclear assumptions regarding the definitions of variables and the conditions under which certain statements are made. The mathematical steps and reasoning presented are not fully resolved, leading to ongoing debate.

PFanalog57
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5^1 = 1*0 + 5

5^2 = 2*10 + 5

5^3 = 3*40 + 5

5^p = p*a + 5

x^p = p*a + x



x^p = p*a + x

y^p = p*b + y

z^p = p*c + z



...x^p + y^p = z^p



p*a + x + p*b + y = p*c + z

p*[a + b - c] = z - [x + y]


p = [z - (x + y)]/[a + b - c]



http://www.maa.org/mathland/mathtrek_12_8.html


Astonishingly, a proof of the ABC conjecture would provide a way of establishing Fermat's last theorem in less than a page of mathematical reasoning. Indeed, many famous conjectures and theorems in number theory would follow immediately from the ABC conjecture, sometimes in just a few lines.



 
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how does this prove that x,y and z can't be integers when p>2? Also a,b and as far as I know can be anything you want.
 
FulhamFan3 said:
how does this prove that x,y and z can't be integers when p>2? Also a,b and as far as I know can be anything you want.


p, x, y, a,b,c, are always integers > 0, when p is a prime number. When p is 1, a = 0, b = 0, and c = 0.


z and c cannot be a +integer for p > 2.


[x^p - x]/p = a

[y^p - y]/p = b

[z^p - z]/p = c


for example:


[3^2 - 3]/2 = 3

[3^3 - 3]/3 = 8

[3^5 - 3]/5 = 48

[+integer^p - +integer]/p = another +integer.

It works for all prime numbers.


3^2 + 4^2 = 5^2

2*3 + 3 = 3^2

2*6 + 4 = 4^2



2*3 + 4*3 + 3 + 4 = 6*3 + 7 = 5^2
 
Last edited:
Russel Rierson,

I don't know why this is here and no in the number theory section of the math board.

But as long as it's here, you lost me on your ifrst line: 5^1 = 1*0 + 5
 
Russell E. Rierson said:
p, x, y, a,b,c, are always integers > 0, when p is a prime number. When p is 1, a = 0, b = 0, and c = 0.


z and c cannot be a +integer for p > 2.

z and c can be integers when p>2. you also have to prove that statement you can't just say it.

Also fermat's theorem as far as I know isn't just for primes, it's for all numbers. Unless this is some other theorem by fermat. In which case you need to state it so we know what your talking about.
 
Last edited:
FulhamFan3 said:
z and c can be integers when p>2. you also have to prove that statement you can't just say it.

Also fermat's theorem as far as I know isn't just for primes, it's for all numbers. Unless this is some other theorem by fermat. In which case you need to state it so we know what your talking about.


If p = 2 :

x^2 = 2*[1 + 2 + 3+...+ x-1] + x

2^2 = 2*[1] + 2

3^2 = 2*[1 + 2] + 3

4^2 = 2*[1 + 2 + 3] + 4

5^2 = 2*[1 + 2 + 3 + 4] + 5

etc...

This congruence does not hold for p > 2
 
5^2 = 2*[1 + 2 + 3 + 4] + 5


5^2 = 2*[1 + 2] + 2*[3 + 4] + 5



5^2 = 2*[1 + 2] + 2*[3 + (4 - 1)] + 5 + 2*1



5^2 = 2*[1 + 2] + 2*[3 + 3] + 7



5^2 = 2*[1 + 2] + 3 + 2*[ 1 + 2 + 3] + 4



5^2 = 3^2 + 4^2
 

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