How to show that: $ \gcd \Big(2^{2^a}+1 , 2^{2^b}+1\Big)=1$

Question

How to show that: $$ \gcd \Big(2^{2^a}+1 , 2^{2^b}+1 \Big)=1$$

Answer 1

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.

Answer 2

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.

Answer 3

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 $\gcd(2^{2^a}+1,2^{2^b}+1)=1$