Proving No Integers Exist: a^n + b^n

  • Thread starter Thread starter halvizo1031
  • Start date Start date
  • Tags Tags
    Integers
Click For Summary

Homework Help Overview

The problem involves proving that there are no integers \(a\), \(b\), and \(n\) greater than 1 such that \((a^n - b^n)\) divides \((a^n + b^n)\). The subject area pertains to number theory, specifically exploring properties of divisibility.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the possibility of using contradiction as a method, with one suggesting that examining the greatest common divisor (gcd) of \((a^n - b^n)\) and \((a^n + b^n)\) could be a viable approach. Questions arise regarding the meaning of the divisibility notation used in the problem statement.

Discussion Status

Some participants have offered insights into the gcd approach and its implications, while others are seeking clarification on the notation used in the problem. There is an ongoing exploration of the concepts involved without a clear consensus on the next steps.

Contextual Notes

Participants note that they are working within the constraints of number theory and express varying levels of confidence in their understanding of the problem. There is an acknowledgment of the need for further clarification on specific terms used in the discussion.

halvizo1031
Messages
77
Reaction score
0

Homework Statement



Prove that they are no integers a,b,n>1 such that (a^n - b^n) | (a^n + b^n).

Homework Equations





The Attempt at a Solution


Do I solve this by contradiction? If so, how do I start it?
 
Physics news on Phys.org
I think it is by contradiction (i suppose you could show the gcd of [itex](a^n - b^n , a^n + b^n )[/itex] is 1 or 2 ) viz,
let d be the gcd clearly then , d must divide the sum (and the difference) of the two , [tex]d | a^n + b^n + a^n - b^n[/tex]
[tex]d | 2a^n[/tex]
which implies, [itex]d|2,[/itex] [itex]d|a^n[/itex] this last result shows d is either 1 or 2 , thus if the gcd of the two is 2 or 1, no integers a,b, c >1 can exist to satisfy the requirement (there are no numbers a,b, c> 1 that can divide in that manner) though I am only into number theory as a hobby, i might not be quite on the mark, ;-) ,
good luck
 
Last edited:
Mechdude said:
I think it is by contradiction (i suppose you could show the gcd of [itex](a^n - b^n , a^n + b^n )[/itex] is 1 or 2 ) viz,
let d be the gcd clearly then , d must divide the sum (and the difference) of the two , [tex]d | a^n + b^n + a^n - b^n[/tex]
[tex]d | 2a^n[/tex]
which implies, [itex]d|2,[/itex] [itex]d|a^n[/itex] this last result shows d is either 1 or 2 , thus if the gcd of the two is 2 or 1, no integers a,b, c >1 can exist to satisfy the requirement (there are no numbers a,b, c> 1 that can divide in that manner) though I am only into number theory as a hobby, i might not be quite on the mark, ;-) ,
good luck


I consulted with my study partners and we agree that what you have is correct. But we didn't get exactly what you got so we had to make some corrections. thanks for your help.
 
(a^n - b^n) | (a^n + b^n)

what does that mean?? does the bar symbol mean "given"
 
frustr8photon said:
(a^n - b^n) | (a^n + b^n)

what does that mean?? does the bar symbol mean "given"

The bar symbol means "divides" in the sense of integer divisibility.
 
Last edited:

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
20
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
Replies
12
Views
2K
  • · Replies 16 ·
Replies
16
Views
5K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K