Show that if $ \ m>n \ $ then $ \ 2^{2^{ \large n}} +1 \ $ divides $ \ 2^{2^{\large m}}-1 \ $. 

Also show that $\, \gcd(2^{2^{\large n}}+1,2^{2^{\large m}}+1)=1$

**Answer:**

Since $ \ m>n \ $ , we have $ \ m=n+p \ $, where $\ p \in \mathbb{N} $ 

Thus,

$ 2^{2^m}-1 \\ = 2^{2^{n+p}}-1 \\ = \left(\left(2^{2^n} \right)^{2^{p-1}} \right)^2-1 \\ = \left( \left(2^{2^n} \right)^{2^{p-1}}-1 \right) \left( \left(2^{2^n} \right)^{2^{p-1}} +1 \right) \\ = \left( 2^{2^n}-1 \right) {\color{Blue} {\left(2^{2^n}+1 \right)}} ........... \left( \left(2^{2^n} \right)^{2^{p-1}} +1 \right) \ \ (Decomposing \ \ the \ \ first \ \ term) 
\ $ 

Thus from the last line we can see that $ \ 2^{2^n}+1 \ $ divides $ \ 2^{2^m}-1 \ $.

**Kindly check my calculation.**

Next part,

Since $ \ 2^{2^n}+1 \ $ divides $ \ 2^{2^m}-1 \ $ , we conclude 

$ 2^{2^n}+1 \ $ does not divide $ \ 2^{2^m}+1 \ $.

Hence we have

$\gcd(2^{2^{ \large n}} +1, 2^{2^{ \large m}} +1)=1 \ $

**Please check my calculation in both parts and confirm my work.**