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

[halvizo1031]

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?

 

[Mechdude]

I think it is by contradiction (i suppose you could show the gcd of $(a^n - b^n , a^n + b^n )$ is 1 or 2 ) viz,
let d be the gcd clearly then , d must divide the sum (and the difference) of the two , $$d | a^n + b^n + a^n - b^n$$
$$d | 2a^n$$
which implies, $d|2,$ $d|a^n$ 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

 

[halvizo1031]

I think it is by contradiction (i suppose you could show the gcd of $(a^n - b^n , a^n + b^n )$ is 1 or 2 ) viz,
let d be the gcd clearly then , d must divide the sum (and the difference) of the two , $$d | a^n + b^n + a^n - b^n$$
$$d | 2a^n$$
which implies, $d|2,$ $d|a^n$ 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.

 

[frustr8photon]

(a^n - b^n) | (a^n + b^n)

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

 

[Dick]

(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.