Mixing
Prove a sequence is a divisibility sequence [duplicate]
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This question already has an answer here:
How to prove $,x^a-1 mid x^b-1 iff amid b$
5 answers
Sequence: $a_n=2^n-1$
I want to prove that $a_m|a_n$ whenever $m|n$.
I started by approaching with induction, base/trivial cases being $m=n$, $n=0$, and $m=1$, but I'm not sure where to go from there or if there is a more straightforward method.
Note:
$m|n$ means for some $x$, $mx = n$
$a_m|a_n$ means for some $y$, $a_my=a_n$
Other similar questions appeal to laws beyond basic arithmetic and the definitions of divisibility and the sequence.
sequences-and-series elementary-number-theory divisibility
asked Jan 30 at 19:54
raznbagelraznbagel
206
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marked as duplicate by Bill Dubuque divisibility
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Jan 30 at 20:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
How to prove $,x^a-1 mid x^b-1 iff amid b$
5 answers
Sequence: $a_n=2^n-1$
I want to prove that $a_m|a_n$ whenever $m|n$.
I started by approaching with induction, base/trivial cases being $m=n$, $n=0$, and $m=1$, but I'm not sure where to go from there or if there is a more straightforward method.
Note:
$m|n$ means for some $x$, $mx = n$
$a_m|a_n$ means for some $y$, $a_my=a_n$
Other similar questions appeal to laws beyond basic arithmetic and the definitions of divisibility and the sequence.
sequences-and-series elementary-number-theory divisibility
asked Jan 30 at 19:54
raznbagelraznbagel
206
$endgroup$
marked as duplicate by Bill Dubuque divisibility
Users with the divisibility badge can single-handedly close divisibility questions as duplicates and reopen them as needed.
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Jan 30 at 20:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Use the fact that $a^n-b^n=(a-b)(a^{n-1} + a^{n-2}b + ldots + b^{n-1})$
$endgroup$
– Ashish K
Jan 30 at 20:00
add a comment |
$begingroup$
This question already has an answer here:
How to prove $,x^a-1 mid x^b-1 iff amid b$
5 answers
Sequence: $a_n=2^n-1$
I want to prove that $a_m|a_n$ whenever $m|n$.
I started by approaching with induction, base/trivial cases being $m=n$, $n=0$, and $m=1$, but I'm not sure where to go from there or if there is a more straightforward method.
Note:
$m|n$ means for some $x$, $mx = n$
$a_m|a_n$ means for some $y$, $a_my=a_n$
Other similar questions appeal to laws beyond basic arithmetic and the definitions of divisibility and the sequence.
sequences-and-series elementary-number-theory divisibility
asked Jan 30 at 19:54
raznbagelraznbagel
206
$endgroup$
This question already has an answer here:
How to prove $,x^a-1 mid x^b-1 iff amid b$
5 answers
Sequence: $a_n=2^n-1$
I want to prove that $a_m|a_n$ whenever $m|n$.
I started by approaching with induction, base/trivial cases being $m=n$, $n=0$, and $m=1$, but I'm not sure where to go from there or if there is a more straightforward method.
Note:
$m|n$ means for some $x$, $mx = n$
$a_m|a_n$ means for some $y$, $a_my=a_n$
Other similar questions appeal to laws beyond basic arithmetic and the definitions of divisibility and the sequence.
This question already has an answer here:
How to prove $,x^a-1 mid x^b-1 iff amid b$
5 answers
sequences-and-series elementary-number-theory divisibility
sequences-and-series elementary-number-theory divisibility
asked Jan 30 at 19:54
raznbagelraznbagel
206
asked Jan 30 at 19:54
raznbagelraznbagel
206
edited Jan 30 at 20:18
raznbagel
asked Jan 30 at 19:54
raznbagelraznbagel
206
asked Jan 30 at 19:54
raznbagelraznbagel
206
asked Jan 30 at 19:54
raznbagelraznbagel
206
206
marked as duplicate by Bill Dubuque divisibility
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Jan 30 at 20:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Bill Dubuque divisibility
Users with the divisibility badge can single-handedly close divisibility questions as duplicates and reopen them as needed.
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Jan 30 at 20:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Use the fact that $a^n-b^n=(a-b)(a^{n-1} + a^{n-2}b + ldots + b^{n-1})$
$endgroup$
– Ashish K
Jan 30 at 20:00
add a comment |
$begingroup$
Use the fact that $a^n-b^n=(a-b)(a^{n-1} + a^{n-2}b + ldots + b^{n-1})$
$endgroup$
– Ashish K
Jan 30 at 20:00
$begingroup$
Use the fact that $a^n-b^n=(a-b)(a^{n-1} + a^{n-2}b + ldots + b^{n-1})$
$endgroup$
– Ashish K
Jan 30 at 20:00
$begingroup$
Use the fact that $a^n-b^n=(a-b)(a^{n-1} + a^{n-2}b + ldots + b^{n-1})$
$endgroup$
– Ashish K
Jan 30 at 20:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $m | n$ then
$n = km$
so
$begin{array}\
2^n-1
&=2^{km}-1\
&=(2^{m})^k-1\
&=(2^m-1)sum_{j=0}^{k-1}2^{jm}\
end{array}
$
answered Jan 30 at 20:00
marty cohenmarty cohen
74.9k549130
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $m | n$ then
$n = km$
so
$begin{array}\
2^n-1
&=2^{km}-1\
&=(2^{m})^k-1\
&=(2^m-1)sum_{j=0}^{k-1}2^{jm}\
end{array}
$
answered Jan 30 at 20:00
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
If $m | n$ then
$n = km$
so
$begin{array}\
2^n-1
&=2^{km}-1\
&=(2^{m})^k-1\
&=(2^m-1)sum_{j=0}^{k-1}2^{jm}\
end{array}
$
answered Jan 30 at 20:00
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
If $m | n$ then
$n = km$
so
$begin{array}\
2^n-1
&=2^{km}-1\
&=(2^{m})^k-1\
&=(2^m-1)sum_{j=0}^{k-1}2^{jm}\
end{array}
$
answered Jan 30 at 20:00
marty cohenmarty cohen
74.9k549130
$endgroup$
If $m | n$ then
$n = km$
so
$begin{array}\
2^n-1
&=2^{km}-1\
&=(2^{m})^k-1\
&=(2^m-1)sum_{j=0}^{k-1}2^{jm}\
end{array}
$
answered Jan 30 at 20:00
marty cohenmarty cohen
74.9k549130
answered Jan 30 at 20:00
marty cohenmarty cohen
74.9k549130
answered Jan 30 at 20:00
marty cohenmarty cohen
74.9k549130
answered Jan 30 at 20:00
marty cohenmarty cohen
74.9k549130
74.9k549130
add a comment |
add a comment |
$begingroup$
Use the fact that $a^n-b^n=(a-b)(a^{n-1} + a^{n-2}b + ldots + b^{n-1})$
$endgroup$
– Ashish K
Jan 30 at 20:00