Prove that $\gcd(2^{2^{22}}+1,2^{2^{222}}+1)=1$

Question

The great common divisor (gcd) of $2^{2^{22}}+1$ and $2^{{2}^{222}}+1$ is

My work,

\begin{align} F_{n}-2&= 2^{2^{n}}+1-2 \\ &=(2^{2^{n-1}}+1)(2^{2^{n-2}}+1)(2^{2^{n-2}}-1)\\ &=(2^{2^{m}}+1)(2^{2^{m}}-1)(2^{2^{m-n-1}}+1)\ \ \ [\text{ where } m \geq n] \end{align}

See https://math.stackexchange.com/questions/3687760/prove-that-gcd-leftna1-nb1-right-divides-n-gcda-b1/3687796#3687796 — ShBh, Jul 19 '20 at 14:04
In your edit it appears that you go from "$gcd(n^{a}+1,n^{b}+1)$ divides $n^{gcd(a,b)}+1$" to "$gcd(n^{a}+1,n^{b}+1)$ equals $n^{gcd(a,b)}+1$" without justification. — halrankard, Jul 19 '20 at 14:45

score 2 · Answer 1 · answered Jul 19 '20 at 15:14

2

This is actually a general result that $\gcd(F_n,F_m)=1$ whenever $m\neq n$

Proof: Let $\mathrm{WLOG}$ $m>n$. Then we have, $$F_m-2=(2^{2^{m-1}}-1)(2^{2^{m-1}}+1)=(2^{2^{m-1}}-1)F_{m-1}$$ Proceeding this way we get, $$F_m-2=F_{m-1}F_{m-2}\cdots F_0=MF_n$$ for some $M\in\mathbb{N}$. Therefore $\gcd(F_m,F_n)\mid2$. Since $F_m,F_n$ are odd we have $\gcd(F_m,F_n)=1$. In particular $\gcd(2^{2^{22}}+1,2^{2^{222}}+1)=1$.

Done!

answered Jul 19 '20 at 15:14

ShBh

6,054

Why am I not getting ans using that formula? – Jul 19 '20 at 15:17
You have some error in computation. Check once again. – ShBh Jul 19 '20 at 15:18
1

$$F_{n}-2= 2^{2^{n}}+1-2 =(2^{2^{n-1}}+1)(2^{2^{n-2}}+1)(2^{2^{n-2}}-1)=\color{red}{ (2^{2^{m}}+1)(2^{2^{m}}-1)(2^{2^{m-n-1}}+1)}$$ The red part deduction is incorrect. – ShBh Jul 19 '20 at 15:23
thanks, got it! – Jul 19 '20 at 15:42

Prove that $\gcd(2^{2^{22}}+1,2^{2^{222}}+1)=1$

1 Answers1