Proving that the elements of a sequence will always be co-prime to each other.

Question

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.

Answer 1

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 $k_{n+m}$.

Answer 2

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 $2$.

Answer 3

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)