Fermat numbers are coprime

Question

So today in my final for number theory I had to prove that the Fermat numbers ($F_n=2^{2^n}+1$) are coprime.

I know that the standard proof uses the following: $F_n=F_1\cdots F_{n-1}+2$ and then the $\gcd$ divides $2$, and it cannot be two, and hence the numbers are coprime.

However, he asked us to use the hint: "Let $l$ be a prime dividing $F_n$. What can you say about the order of $2$ in $(\mathbb{Z}/l\mathbb{Z})^\times$?"

I have been thinking about it and I cannot figure out how to use his hint.

Any ideas?

Answer 1

If $l$ is a prime that divides $F_n$, then $2^{2^n}\equiv -1 \pmod{l}$, and therefore $2^{2^{n+1}}\equiv 1 \pmod{l}$. So $2$ has order $2^{n+1}$ modulo $l$. Equivalently, but in more group-theoretic language, (the equivalence class of) $2$ has order $2^{n+1}$ in $(\mathbb{Z}/l\mathbb{Z})^\times$. (The order of $2$ must divide $2^{n+1}$, but cannot be $\le 2^n$.)

If $p$ is a prime that divides $2^{2^k}+1$, then by the same reasoning $2$ has order $2^{k+1}$ modulo $p$. If $k<n$, that means that $p$ cannot be equal to $l$. So we have proved that no prime that divides $F_n$ can divide any $F_k$ with $k<n$. This shows that $F_n$ and $F_k$ are relatively prime.

Remark: We do not really need to identify the order of $2$ explicitly. For if $l$ is a prime that divides $F_n$, then $2^{2^n}\equiv -1 \pmod{l}$. But if $p$ is a prime that divides $F_k$ for some $k<n$, then $2^{2^k}\equiv -1\pmod{p}$, and therefore $2^{2^{k+1}}\equiv 1\pmod{p}$. Since $k+1 \le n$, this is impossible.

Answer 2

Let $A_r=a^{2^r}+1,$

$A_{n+t}=a^{2^{n+t}}+1=(a^{2^n})^{2^t}+1=(A_n-1)^{2^t}+1\equiv2\pmod{A_n}$ for integer $t\ge1$

$\implies(A_{n+t},A_n)=(2,A_n)$

If $a$ is even, $A_r$ will be odd if $r\ge1$

$\implies(2,A_n)=1$ for $n\ge1$