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?






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marked as duplicate by lulu, Servaes, Bill Dubuque divisibility
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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















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  • 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?






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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













up vote
0
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up vote
0
down vote

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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







share|cite|improve this question














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






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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 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


















  • 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















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