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We are given the sequence $k$n = 6$^{{({2}^n)}}$ + 1. We must prove that the elements of this sequence are pairwise co-prime, i.e prove that if m $\neq$ n then $gcd$($k$m,$k$n) = $1$.

I have proved that $k$n | ($k$n+1 - $2$) however I can't seem to extend this proof in order to prove every element is co prime.

All help would be greatly appreciated, cheers.

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  • $\begingroup$ What you have already shows you that $k_n$ and $k_{n+1}$ are coprime: the only possible common divisor would be $2$, but they're all odd. Could you do something similar for $k_{n}$ and $k_{n+2}$? Could something like this generalise? $\endgroup$ – user3482749 Jan 31 at 16:41
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Claim: $$\boxed {5\prod_{i=0}^nk_i = k_{n+1}-2}$$

Pf: Consider the product $$P_n=\prod_{i=0}^nk_i$$

Since $5=6^{(2^0)}-1$ we note that $$5P_n=\left(6^{(2^0)}-1\right)\times \left(6^{(2^0)}+1\right)\times \prod_{i=1}^nk_i =\left(6^{(2^1)}-1\right)\times \left(6^{(2^1)}+1\right)\times \prod_{i=2}^nk_i=$$ $$=\left(6^{(2^2)}-1\right)\times \left(6^{(2^2)}+1\right)\times \prod_{i=3}^nk_i$$

Continuing in this way we see that $$5P_n=6^{(2^{n+1})}-1=k_{n+1}-2$$ as desired.

It follows that any common divisor of two of the $k_i$ would have to be a divisor of $2$. As all the $k_i$ are odd, we are done.

Note: since the point was raised in the comments, let me elaborate on the final paragraph. Suppose $i<j$. We wish to prove that $\gcd(k_i,k_j)=1$. But $i<j\implies i≤j-1\implies k_i\,|\,P_{j-1}$ Thus, $k_{i}\,|\,5P_{j-1}=k_j-2$ Thus any common divisor of $k_i,k_j$ would have to divide $2$.

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  • $\begingroup$ I did have to think a bit more about the last line to see it myself. I do see it now though. If the $k_n$s were not pairwise coprime then there exists an $n$ such that $k_{n+1}$ shares a common factor with $k_l$ for some $l<n+1$. If $k_{n+1}$ shares any common factors with any $k_l$; $l<n+1$, then $k_{n+1}$ shares a common factor with $5P_n$. But $5P_n +2 = k_{n+1}$, so the only factor $k_{n+1}$ could share with $5P_n$ is 2, and as $k_{n+1}$ is odd, it does not even share a factor of 2. Good answer! $\endgroup$ – Mike Jan 31 at 18:17
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    $\begingroup$ @Mike Yes, that's right. If $i<j$ then $k_i\,|\,5P_{j-1}=k_j-2$ $\endgroup$ – lulu Jan 31 at 18:18
  • $\begingroup$ @lulu is there any way I could then use this result to then prove that there are infinitely many primes combined with the fundamental theorem of arithmetic? $\endgroup$ – user637295 Jan 31 at 18:55
  • $\begingroup$ @LittleRichard Sure...let $p_i$ denote the least prime dividing $k_i$. Then all of the $\{p_i\}$ are distinct. $\endgroup$ – lulu Jan 31 at 18:58
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If $p|k_n$, then $p|(k_{n+1}-2)$. If also $p|k_{n+1}$, then $$p|\big(k_{n+1}-(k_{n+1}-2)\big)$$ or $p|2$. Hence the only prime that could divide both $k_n$ and $k_{n+1}$ is $2$. However all the terms are odd, so $\gcd(k_n,k_{n+1})=1$.

Now, you need a similar relationship between $k_n$ and $k_{n+m}$.

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  • $\begingroup$ So would I just run through the same argument but instead of using k$_{n+1}$Just use k$_m$ ? $\endgroup$ – user637295 Jan 31 at 19:07
  • $\begingroup$ You will first need to prove an analogue of $k_n|(k_{n+1}-2)$, but with $k_{n+m}$ instead of $k_{n+1}$. $\endgroup$ – vadim123 Jan 31 at 19:53
  • $\begingroup$ so would I literally just use $k_{n+m}$ in replace of $k_{n+1}$ ? $\endgroup$ – user637295 Feb 1 at 17:05
  • $\begingroup$ I'm not 100% sure how I can prove the k$_{n+m}$ case, is there any way you could help? Cheers $\endgroup$ – user637295 Feb 3 at 17:39
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note that the expression can be mod $$6(2^m)+1,\quad m<n$$ this gives $$6(2^n \bmod 6(2^m)+1)+1$$ if the mod is $2^m$ then $2^n$ is different from a value we know divides by it ( the number itself) by a multiple of the moduli itself. That implies $$2^n=(6(2^m)+1)k+2^m$$ dividing boths sides by $2^m$ gives $$2^{n-m}=6k+k=7k$$, since 7 divides no power of 2 it then follows the mod can't hold, which makes the division by divisors of $6(2^m)+1$ also not work unless they can divide a power of 2 ( which being odd they can't)

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