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Prove that $n$ is not divider of number $2^n-1$, if $n>1$ [duplicate]
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For $n geq 2$, show that $n nmid 2^{n}-1$
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Show that $n$ does not divide $2^n - 1$ where $n$ is an integer greater than $1$? [duplicate]
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Prove that $n$ is not divider of number $2^n-1$, if $n>1$
If we think opposite, then $n$ is divider of number $2^{n}-1$, since $2^{n}-1$ is odd number then $nnot=2k$, where $kin mathbb Z$, now let try with $n=2k+1$ but then $2^{2k} equiv 1 mod (2k+1)$, then $2^{2k+1} equiv 2 (mod 2k+1)$, so $nnot | 2^n-1$ for any number n>1, so only option is that n=1, is this ok?
discrete-mathematics modular-arithmetic divisibility
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Marko Škorić
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marked as duplicate by lulu, Servaes, Bill Dubuque divisibility
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This question already has an answer here:
For $n geq 2$, show that $n nmid 2^{n}-1$
2 answers
Show that $n$ does not divide $2^n - 1$ where $n$ is an integer greater than $1$? [duplicate]
1 answer
Prove that $n$ is not divider of number $2^n-1$, if $n>1$
If we think opposite, then $n$ is divider of number $2^{n}-1$, since $2^{n}-1$ is odd number then $nnot=2k$, where $kin mathbb Z$, now let try with $n=2k+1$ but then $2^{2k} equiv 1 mod (2k+1)$, then $2^{2k+1} equiv 2 (mod 2k+1)$, so $nnot | 2^n-1$ for any number n>1, so only option is that n=1, is this ok?
discrete-mathematics modular-arithmetic divisibility
asked yesterday
Marko Škorić
58110
marked as duplicate by lulu, Servaes, Bill Dubuque divisibility
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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.
Your argument only works in the case where $2k+1$ is prime. For example, $2^{14}equiv 4pmod {15}$.
– lulu
yesterday
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This question already has an answer here:
For $n geq 2$, show that $n nmid 2^{n}-1$
2 answers
Show that $n$ does not divide $2^n - 1$ where $n$ is an integer greater than $1$? [duplicate]
1 answer
Prove that $n$ is not divider of number $2^n-1$, if $n>1$
If we think opposite, then $n$ is divider of number $2^{n}-1$, since $2^{n}-1$ is odd number then $nnot=2k$, where $kin mathbb Z$, now let try with $n=2k+1$ but then $2^{2k} equiv 1 mod (2k+1)$, then $2^{2k+1} equiv 2 (mod 2k+1)$, so $nnot | 2^n-1$ for any number n>1, so only option is that n=1, is this ok?
discrete-mathematics modular-arithmetic divisibility
asked yesterday
Marko Škorić
58110
This question already has an answer here:
For $n geq 2$, show that $n nmid 2^{n}-1$
2 answers
Show that $n$ does not divide $2^n - 1$ where $n$ is an integer greater than $1$? [duplicate]
1 answer
Prove that $n$ is not divider of number $2^n-1$, if $n>1$
If we think opposite, then $n$ is divider of number $2^{n}-1$, since $2^{n}-1$ is odd number then $nnot=2k$, where $kin mathbb Z$, now let try with $n=2k+1$ but then $2^{2k} equiv 1 mod (2k+1)$, then $2^{2k+1} equiv 2 (mod 2k+1)$, so $nnot | 2^n-1$ for any number n>1, so only option is that n=1, is this ok?
This question already has an answer here:
For $n geq 2$, show that $n nmid 2^{n}-1$
2 answers
Show that $n$ does not divide $2^n - 1$ where $n$ is an integer greater than $1$? [duplicate]
1 answer
discrete-mathematics modular-arithmetic divisibility
discrete-mathematics modular-arithmetic divisibility
asked yesterday
Marko Škorić
58110
asked yesterday
Marko Škorić
58110
asked yesterday
Marko Škorić
58110
asked yesterday
Marko Škorić
58110
asked yesterday
Marko Škorić
58110
58110
marked as duplicate by lulu, Servaes, Bill Dubuque divisibility
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marked as duplicate by lulu, Servaes, Bill Dubuque divisibility
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yesterday
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.
Your argument only works in the case where $2k+1$ is prime. For example, $2^{14}equiv 4pmod {15}$.
– lulu
yesterday
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Your argument only works in the case where $2k+1$ is prime. For example, $2^{14}equiv 4pmod {15}$.
– lulu
yesterday
Your argument only works in the case where $2k+1$ is prime. For example, $2^{14}equiv 4pmod {15}$.
– lulu
yesterday
Your argument only works in the case where $2k+1$ is prime. For example, $2^{14}equiv 4pmod {15}$.
– lulu
yesterday
add a comment |
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Your argument only works in the case where $2k+1$ is prime. For example, $2^{14}equiv 4pmod {15}$.
– lulu
yesterday