If $m>n$ prove that $a^{2^n}+1$ is a divisor of $a^{2^m}-1$

Question

We have to prove that $a^{2^n} +1\mid a^{2^m}-1$ when $m>n$.

What I've done: I considered the two cases when $a= 2k$ and when $a= 2k+1$.

Do you mean $a^{2^{n+1}}$ or $a^{2^{n}+1}$ or $a^{2^{n}}+1$? — Andrew Chin, Dec 04 '19 at 15:31
@AndrewChin The first two interpretations seem to make the problem too easy (though the first hits issues if $m=n+1$) so I would guess the third was intended comparing $a^{2^{n}}+1$ and $a^{2^{m}}-1$ — Henry, Dec 04 '19 at 15:37
See my post in https://math.stackexchange.com/questions/123524/fermat-numbers-are-coprime — lab bhattacharjee, Dec 04 '19 at 15:40

score 0 · Answer 1 · answered Dec 04 '19 at 15:44

0

Hint: $x^2 - 1 = (x-1)(x+1)$.

A bigger hint:

$x^{2^n} - 1 = ( x^{2^{n-1}} -1 ) ( x^{2^{n-1}} + 1)$

answered Dec 04 '19 at 15:44

Calvin Lin

Please give me solution. I am totally clueless about this question – Nilabh Kumar Dec 04 '19 at 17:30
What have you tried? Can you prove it for $m=n+1$ using the first hint? How about for $m=n+2$ using the second hint too? – Calvin Lin Dec 04 '19 at 17:34
How can you say m=n+1. Or m=n+2. Sir please give me the proper solution . I will be very thankful to you – Nilabh Kumar Dec 05 '19 at 13:26

1 Answers1