How to show that: $ \gcd \Big(2^{2^a}+1 , 2^{2^b}+1\Big)=1$ How to show that:
$$ \gcd     \Big(2^{2^a}+1 , 2^{2^b}+1 \Big)=1$$
 A: Assume WLOG that $b>a$.
Note that $(2^{2^a}+1)|((2^{2^a})^{2^{b-a}}-1)=(2^{2^b}-1)$. (This follows from the fact that $x^2-1|x^{2^k}-1$ and $x+1|x^2-1$ for integer x and any positive integer k)
Thus, it follows that $2^{2^b}-1=k(2^{2^a}+1)$. Now add two to both sides to get:
$$2^{2^b}+1=k(2^{2^a}+1)+2$$
 Thus, $\gcd(2^{2^a}+1,2^{2^b}+1)|2$. Since both numbers are odd it follows that $\gcd(2^{2^a}+1,2^{2^b}+1)=1$
A: Hint:
Call $2^{2^n}+1=F_n$, and then show that $F_n = F_{n-1}...F_1 + 2$.
Now any number greater than $2$ that divides $F_m$ for $n>m$ will divide only part of the right side, and therefore not the left.
A: We know $(x-y)\mid(x^c-y^c)$ where $x,y,c$ are natural numbers.
Putting $x=a^{2^n},b=-1, c=2^{m-n}$
$$\{a^{2^n}-(-1)\}\mid \{(a^{2^n})^{2^{m-n}}-(-1)^{2^{m-n}}\}$$ if $m>n$ where $a,m,n$ are natural numbers.
$$\implies (a^{2^n}+1)\mid (a^{2^m}-1)$$   if $m>n$
Now, $$(a^{2^m}-1,a^{2^m}+1)=(a^{2^m}-1,a^{2^m}+1-(a^{2^m}-1))=(a^{2^m}-1,2)$$ is $2$ if $a$ is odd, is $1$ if $a$ is even
So, $$(a^{2^n}+1,a^{2^m}+1)$$ will be $1$ or $2$ accordingly. 
