Divisibility question

[joeblow]

How can I show that (a^n -b^n) doesn't divide (a^n + b ^n) for all integers a,b, and n?

I have that if (a^n -b^n) did divide (a^n + b ^n), then (a^n +b^n) = q (a^n -b^n) which implies b^n = -q*b^n (mod a^n). Then 1 = -q (mod a^n), meaning gcd(b, a^n) = 1. I am unsure of what more I can deduce. Any help is appreciated.

[Petek]

I think you need to assume that a, b and n are all > 1. Otherwise there are simple counterexamples.

[joeblow]

Yes, of course.

[CRGreathouse]

a^n - b^n | a^n + b^n
implies
a^n - b^n | 2b^n
and so, if a >= b,
ka^n = (2 - k)b^n
for some positive k. It must be the case that k = 1 (why?), so
a^n = b^n
and hence
a = b.

[Petek]

a^n - b^n | a^n + b^n
implies
a^n - b^n | 2b^n
and so, if a >= b,
ka^n = (2 - k)b^n
for some positive k. It must be the case that k = 1 (why?), so
a^n = b^n
and hence
a = b.

Shouldn't the equation

ka^n = (2 - k)b^n

instead be

ka^n = (2 + k)b^n?

[Petek]

I think I have a solution. Here's a hint: Prove that you can reduce the problem to the case in which a and b are relatively prime.

HTH

Petek

[robert Ihnot]

We assume a and b have no common factor, because if they did, it could be removed.

To avoid the case of a^n-b^n = 0, we need only consider a prime p, such that p divides a^n-b^n. We eliminate the trivial cases where n=1, or where a,b or n =0. Then a^n \equiv b^n Mod p ; a^n + b^n \equiv 2b^n Mod p

Thus p divides 2 or p divides b^n. but 2 can be a factor of b since then it would also divide a, and consequently we can chose p >2. But then it follows both a and b are divisible by p contrary to assumption.

[Kittel Knight]

How can I show that (a^n -b^n) doesn't divide (a^n + b ^n) for all integers a,b, and n?

I have that if (a^n -b^n) did divide (a^n + b ^n), then (a^n +b^n) = q (a^n -b^n) which implies b^n = -q*b^n (mod a^n). Then 1 = -q (mod a^n), meaning gcd(b, a^n) = 1. I am unsure of what more I can deduce. Any help is appreciated.

That is easy.
a=3,b=2,n=2 implies
a^n+b^n=13 which is not divisible by a^n-b^n=5.
So, if it is not valid for 3,2,and 2, then it is not valid for all integers.

[Borek]

a^n - b^n | a^n + b^n
implies
a^n - b^n | 2b^n

Hm, if a^n+b^n=2b^n doesn't it mean a^n-b^n=0?

[Petek]

We assume a and b have no common factor, because if they did, it could be removed.

To avoid the case of a^n-b^n = 0, we need only consider a prime p, such that p divides a^n-b^n. We eliminate the trivial cases where n=1, or where a,b or n =0. Then a^n \equiv b^n Mod p ; a^n + b^n \equiv 2b^n Mod p

Thus p divides 2 or p divides b^n. but 2 can be a factor of b since then it would also divide a, and consequently we can chose p >2. But then it follows both a and b are divisible by p contrary to assumption.

This proof seems to be correct (with "but 2 can be a factor of b" an obvious typo for "but 2 cannot be a factor of b"), and is simpler than the one I had in mind. Nice!

Petek

[robert Ihnot]

This proof seems to be correct (with "but 2 can be a factor of b" an obvious typo for "but 2 cannot be a factor of b"), and is simpler than the one I had in mind. Nice!

Petek

Yes, thank you: "But 2 CANNOT be a factor of b since then it would also divide a...."

[Office_Shredder]

Hm, if a^n+b^n=2b^n doesn't it mean a^n-b^n=0?

an-bn|an+bn

Then

an - bn|an+bn-(an-bn)
Which gives

an-bn|2bn

Nothing here says they're equal though

[Borek]

Nothing here says they're equal though

Thanks :smile: