Extension on my one of previous questions about each element in a sequence being coprime.

Question

So my previous question states that: We are given the sequence $_{n}= 6^{({2}^n)} + 1$. We must prove that the elements of this sequence are pairwise co-prime, i.e prove that if m ≠ n then $(_{m},_{n}) = 1$.

I previously proved that $k_n | k_{n+1} - 2$

Then using answers I got on my previous question I was able to show that if $p | k_n$ then it follows that $p | k_{n+1} -2$. Using this I could say that that if $p|k_{n+1}$ then $p|(k_{n+1} - (k_{n+1} -2)$. This means the only prime that could divide $k_n$ and $k_{n+1}$ is $2$, however all the terms in the sequence are odd so therefore $gcd(k_n,k_{n+1})=1$.

This therefore proves that every pair of numbers in the sequence are coprime however I'm struggling to extend it to show every element is coprime so I need to show $gcd(k_m,k_n)=1$.

All help is very much appreciated as always cheers.

Answer 1

I don't see why the same approach doesn't work. If $m>n$

$$k_m-2=6^{2^m}-1=\big(6^{2^n}\big)^{2^{m-n}}-1\equiv(-1)^{2^{m-n}}-1\equiv1-1\equiv0\pmod{k_n}$$

So, indeed $k_n\mid k_m-2$ ...

Edit: Further elaboration, as $k_n=6^{2^n}+1$ we must have

$$6^{2^n}+1\equiv 0\pmod{k_n}\Rightarrow 6^{2^n}\equiv -1\pmod{k_n}$$

also, note that $m-n>0$ so $2^{m-n}$ is even, so $(-1)^{2^{m-n}}=1$.

Answer 2

${\bf Hint}\rm\quad\ \ \gcd(a+1,\,\ a^{\large 2K}\!+1)\ =\ gcd(a+1,\ \ \color{#0a0}2)\,\ $ by the Euclidean algorithm

${\bf Proof}\rm\ \ mod\,\ a+1\!:\,\ \color{#c00}a^{\large 2K}\!+1 \equiv (\color{#c00}{-1})^{\large 2K}\!\!+1\equiv\color{#0a0} 2,\ \ {\rm by}\ \ \color{#c00}{a\equiv {-}1}\quad {\bf QED}$

Specializing $\rm\,\ a=6^{\Large 2^{M}}\!,\ \ 2K = 2^{\large \,N-M} \Rightarrow\, a^{\Large 2K}\! = 6^{\Large 2^{N}},\,\ N>M,\, $ yields the claim.