Mixing
Proving that the elements of a sequence will always be co-prime to each other.
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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.
sequences-and-series number-theory elementary-number-theory proof-writing coprime
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add a comment |
$begingroup$
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.
sequences-and-series number-theory elementary-number-theory proof-writing coprime
$endgroup$
$begingroup$
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?
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– user3482749
Jan 31 at 16:41
add a comment |
$begingroup$
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.
sequences-and-series number-theory elementary-number-theory proof-writing coprime
$endgroup$
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.
sequences-and-series number-theory elementary-number-theory proof-writing coprime
sequences-and-series number-theory elementary-number-theory proof-writing coprime
asked Jan 31 at 16:35
user637295
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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
add a comment |
$begingroup$
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
$begingroup$
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
$begingroup$
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
add a comment |
3 Answers
3
active
oldest
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Claim: $$boxed {5prod_{i=0}^nk_i = k_{n+1}-2}$$
Pf: Consider the product $$P_n=prod_{i=0}^nk_i$$
Since $5=6^{(2^0)}-1$ we note that $$5P_n=left(6^{(2^0)}-1right)times left(6^{(2^0)}+1right)times prod_{i=1}^nk_i =left(6^{(2^1)}-1right)times left(6^{(2^1)}+1right)times prod_{i=2}^nk_i=$$ $$=left(6^{(2^2)}-1right)times left(6^{(2^2)}+1right)times prod_{i=3}^nk_i$$
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 $i<jimplies i≤j-1implies k_i,|,P_{j-1}$ Thus, $k_{i},|,5P_{j-1}=k_j-2$ Thus any common divisor of $k_i,k_j$ would have to divide $2$.
answered Jan 31 at 16:55
lulululu
43.6k25081
$endgroup$
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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!
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– Mike
Jan 31 at 18:17
1
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@Mike Yes, that's right. If $i<j$ then $k_i,|,5P_{j-1}=k_j-2$
$endgroup$
– lulu
Jan 31 at 18:18
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@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
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@LittleRichard Sure...let $p_i$ denote the least prime dividing $k_i$. Then all of the ${p_i}$ are distinct.
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– lulu
Jan 31 at 18:58
add a comment |
$begingroup$
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 $k_{n+m}$.
answered Jan 31 at 16:42
vadim123vadim123
76.5k897191
$endgroup$
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So would I just run through the same argument but instead of using k$_{n+1}$Just use k$_m$ ?
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– user637295
Jan 31 at 19:07
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You will first need to prove an analogue of $k_n|(k_{n+1}-2)$, but with $k_{n+m}$ instead of $k_{n+1}$.
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– vadim123
Jan 31 at 19:53
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so would I literally just use $k_{n+m}$ in replace of $k_{n+1}$ ?
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– user637295
Feb 1 at 17:05
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I'm not 100% sure how I can prove the k$_{n+m}$ case, is there any way you could help? Cheers
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– user637295
Feb 3 at 17:39
add a comment |
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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)
answered Feb 23 at 16:40
Roddy MacPheeRoddy MacPhee
722118
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3 Answers
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3 Answers
3
active
oldest
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active
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$begingroup$
Claim: $$boxed {5prod_{i=0}^nk_i = k_{n+1}-2}$$
Pf: Consider the product $$P_n=prod_{i=0}^nk_i$$
Since $5=6^{(2^0)}-1$ we note that $$5P_n=left(6^{(2^0)}-1right)times left(6^{(2^0)}+1right)times prod_{i=1}^nk_i =left(6^{(2^1)}-1right)times left(6^{(2^1)}+1right)times prod_{i=2}^nk_i=$$ $$=left(6^{(2^2)}-1right)times left(6^{(2^2)}+1right)times prod_{i=3}^nk_i$$
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 $i<jimplies i≤j-1implies k_i,|,P_{j-1}$ Thus, $k_{i},|,5P_{j-1}=k_j-2$ Thus any common divisor of $k_i,k_j$ would have to divide $2$.
answered Jan 31 at 16:55
lulululu
43.6k25081
$endgroup$
$begingroup$
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
1
$begingroup$
@Mike Yes, that's right. If $i<j$ then $k_i,|,5P_{j-1}=k_j-2$
$endgroup$
– lulu
Jan 31 at 18:18
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@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
add a comment |
$begingroup$
Claim: $$boxed {5prod_{i=0}^nk_i = k_{n+1}-2}$$
Pf: Consider the product $$P_n=prod_{i=0}^nk_i$$
Since $5=6^{(2^0)}-1$ we note that $$5P_n=left(6^{(2^0)}-1right)times left(6^{(2^0)}+1right)times prod_{i=1}^nk_i =left(6^{(2^1)}-1right)times left(6^{(2^1)}+1right)times prod_{i=2}^nk_i=$$ $$=left(6^{(2^2)}-1right)times left(6^{(2^2)}+1right)times prod_{i=3}^nk_i$$
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 $i<jimplies i≤j-1implies k_i,|,P_{j-1}$ Thus, $k_{i},|,5P_{j-1}=k_j-2$ Thus any common divisor of $k_i,k_j$ would have to divide $2$.
answered Jan 31 at 16:55
lulululu
43.6k25081
$endgroup$
$begingroup$
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
1
$begingroup$
@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
add a comment |
$begingroup$
Claim: $$boxed {5prod_{i=0}^nk_i = k_{n+1}-2}$$
Pf: Consider the product $$P_n=prod_{i=0}^nk_i$$
Since $5=6^{(2^0)}-1$ we note that $$5P_n=left(6^{(2^0)}-1right)times left(6^{(2^0)}+1right)times prod_{i=1}^nk_i =left(6^{(2^1)}-1right)times left(6^{(2^1)}+1right)times prod_{i=2}^nk_i=$$ $$=left(6^{(2^2)}-1right)times left(6^{(2^2)}+1right)times prod_{i=3}^nk_i$$
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 $i<jimplies i≤j-1implies k_i,|,P_{j-1}$ Thus, $k_{i},|,5P_{j-1}=k_j-2$ Thus any common divisor of $k_i,k_j$ would have to divide $2$.
answered Jan 31 at 16:55
lulululu
43.6k25081
$endgroup$
Claim: $$boxed {5prod_{i=0}^nk_i = k_{n+1}-2}$$
Pf: Consider the product $$P_n=prod_{i=0}^nk_i$$
Since $5=6^{(2^0)}-1$ we note that $$5P_n=left(6^{(2^0)}-1right)times left(6^{(2^0)}+1right)times prod_{i=1}^nk_i =left(6^{(2^1)}-1right)times left(6^{(2^1)}+1right)times prod_{i=2}^nk_i=$$ $$=left(6^{(2^2)}-1right)times left(6^{(2^2)}+1right)times prod_{i=3}^nk_i$$
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 $i<jimplies i≤j-1implies k_i,|,P_{j-1}$ Thus, $k_{i},|,5P_{j-1}=k_j-2$ Thus any common divisor of $k_i,k_j$ would have to divide $2$.
answered Jan 31 at 16:55
lulululu
43.6k25081
edited Jan 31 at 20:09
answered Jan 31 at 16:55
lulululu
43.6k25081
answered Jan 31 at 16:55
lulululu
43.6k25081
answered Jan 31 at 16:55
lulululu
43.6k25081
43.6k25081
$begingroup$
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
1
$begingroup$
@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
add a comment |
$begingroup$
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
1
$begingroup$
@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
$begingroup$
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
$begingroup$
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
1
1
$begingroup$
@Mike Yes, that's right. If $i<j$ then $k_i,|,5P_{j-1}=k_j-2$
$endgroup$
– lulu
Jan 31 at 18:18
$begingroup$
@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
$begingroup$
@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
$begingroup$
@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
add a comment |
$begingroup$
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 $k_{n+m}$.
answered Jan 31 at 16:42
vadim123vadim123
76.5k897191
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $k_{n+m}$.
answered Jan 31 at 16:42
vadim123vadim123
76.5k897191
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $k_{n+m}$.
answered Jan 31 at 16:42
vadim123vadim123
76.5k897191
$endgroup$
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 $k_{n+m}$.
answered Jan 31 at 16:42
vadim123vadim123
76.5k897191
answered Jan 31 at 16:42
vadim123vadim123
76.5k897191
answered Jan 31 at 16:42
vadim123vadim123
76.5k897191
answered Jan 31 at 16:42
vadim123vadim123
76.5k897191
76.5k897191
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
add a comment |
$begingroup$
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)
answered Feb 23 at 16:40
Roddy MacPheeRoddy MacPhee
722118
$endgroup$
add a comment |
$begingroup$
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)
answered Feb 23 at 16:40
Roddy MacPheeRoddy MacPhee
722118
$endgroup$
add a comment |
$begingroup$
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)
answered Feb 23 at 16:40
Roddy MacPheeRoddy MacPhee
722118
$endgroup$
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)
answered Feb 23 at 16:40
Roddy MacPheeRoddy MacPhee
722118
edited Feb 23 at 16:46
answered Feb 23 at 16:40
Roddy MacPheeRoddy MacPhee
722118
answered Feb 23 at 16:40
Roddy MacPheeRoddy MacPhee
722118
answered Feb 23 at 16:40
Roddy MacPheeRoddy MacPhee
722118
722118
add a comment |
add a comment |
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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