Find all positive integer $m$ such $2^{m}+1mid5^m-1$ [closed]












5












$begingroup$



Find all positive integer $m$ such
$$2^{m}+1mid5^m-1,.$$




It seem there no solution, I think it might be necessary to use quadratic reciprocity knowledge to solve this problem.



Let $M=2^m+1$, so we have
$$5^{m}equiv 1pmod{M}. $$
If $m$ is odd, then we have
$$left(5^{frac{m+1}{2}}right)^2=5pmod {M}Longrightarrow left(dfrac{5}{M}right)=1,$$
so
$$left(dfrac{M}{5}right)(-1)^{frac{(5-1)(M-1)}{4}}=left(dfrac{M}{5}right)(-1)^M=-left(dfrac{M}{5}right).$$






number-theory elementary-number-theory modular-arithmetic divisibility diophantine-equations







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closed as off-topic by davidlowryduda♦ Dec 15 '18 at 16:47



  • This question does not appear to be about math within the scope defined in the help center.

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    What did you try?
    $endgroup$
    – Robert Z
    Dec 13 '18 at 8:15










  • $begingroup$
    @RobertZ I have post my some try,Following step I can't do it
    $endgroup$
    – function sug
    Dec 13 '18 at 8:22








  • 1




    $begingroup$
    Are we assuming $(2^m+1)mid(5^m-1)$?
    $endgroup$
    – manooooh
    Dec 13 '18 at 8:26






  • 3




    $begingroup$
    @manooooh,yes. it's right
    $endgroup$
    – function sug
    Dec 13 '18 at 8:27






  • 1




    $begingroup$
    If $m$ is odd then $2^m +1$ is divisible by 3 but $5^m - 1$ is not. No need for quadratic reciprocity
    $endgroup$
    – user420261
    Dec 13 '18 at 8:34
















5












$begingroup$



Find all positive integer $m$ such
$$2^{m}+1mid5^m-1,.$$




It seem there no solution, I think it might be necessary to use quadratic reciprocity knowledge to solve this problem.



Let $M=2^m+1$, so we have
$$5^{m}equiv 1pmod{M}. $$
If $m$ is odd, then we have
$$left(5^{frac{m+1}{2}}right)^2=5pmod {M}Longrightarrow left(dfrac{5}{M}right)=1,$$
so
$$left(dfrac{M}{5}right)(-1)^{frac{(5-1)(M-1)}{4}}=left(dfrac{M}{5}right)(-1)^M=-left(dfrac{M}{5}right).$$






number-theory elementary-number-theory modular-arithmetic divisibility diophantine-equations







share|cite|improve this question











$endgroup$



closed as off-topic by davidlowryduda♦ Dec 15 '18 at 16:47



  • This question does not appear to be about math within the scope defined in the help center.

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    What did you try?
    $endgroup$
    – Robert Z
    Dec 13 '18 at 8:15










  • $begingroup$
    @RobertZ I have post my some try,Following step I can't do it
    $endgroup$
    – function sug
    Dec 13 '18 at 8:22








  • 1




    $begingroup$
    Are we assuming $(2^m+1)mid(5^m-1)$?
    $endgroup$
    – manooooh
    Dec 13 '18 at 8:26






  • 3




    $begingroup$
    @manooooh,yes. it's right
    $endgroup$
    – function sug
    Dec 13 '18 at 8:27






  • 1




    $begingroup$
    If $m$ is odd then $2^m +1$ is divisible by 3 but $5^m - 1$ is not. No need for quadratic reciprocity
    $endgroup$
    – user420261
    Dec 13 '18 at 8:34














5












5








5


7



$begingroup$



Find all positive integer $m$ such
$$2^{m}+1mid5^m-1,.$$




It seem there no solution, I think it might be necessary to use quadratic reciprocity knowledge to solve this problem.



Let $M=2^m+1$, so we have
$$5^{m}equiv 1pmod{M}. $$
If $m$ is odd, then we have
$$left(5^{frac{m+1}{2}}right)^2=5pmod {M}Longrightarrow left(dfrac{5}{M}right)=1,$$
so
$$left(dfrac{M}{5}right)(-1)^{frac{(5-1)(M-1)}{4}}=left(dfrac{M}{5}right)(-1)^M=-left(dfrac{M}{5}right).$$






number-theory elementary-number-theory modular-arithmetic divisibility diophantine-equations







share|cite|improve this question











$endgroup$





Find all positive integer $m$ such
$$2^{m}+1mid5^m-1,.$$




It seem there no solution, I think it might be necessary to use quadratic reciprocity knowledge to solve this problem.



Let $M=2^m+1$, so we have
$$5^{m}equiv 1pmod{M}. $$
If $m$ is odd, then we have
$$left(5^{frac{m+1}{2}}right)^2=5pmod {M}Longrightarrow left(dfrac{5}{M}right)=1,$$
so
$$left(dfrac{M}{5}right)(-1)^{frac{(5-1)(M-1)}{4}}=left(dfrac{M}{5}right)(-1)^M=-left(dfrac{M}{5}right).$$






number-theory elementary-number-theory modular-arithmetic divisibility diophantine-equations




number-theory elementary-number-theory modular-arithmetic divisibility diophantine-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 21:25









Batominovski

33.1k33293




33.1k33293










asked Dec 13 '18 at 8:13









function sugfunction sug

3411438




3411438




closed as off-topic by davidlowryduda♦ Dec 15 '18 at 16:47



  • This question does not appear to be about math within the scope defined in the help center.

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by davidlowryduda♦ Dec 15 '18 at 16:47



  • This question does not appear to be about math within the scope defined in the help center.

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    What did you try?
    $endgroup$
    – Robert Z
    Dec 13 '18 at 8:15










  • $begingroup$
    @RobertZ I have post my some try,Following step I can't do it
    $endgroup$
    – function sug
    Dec 13 '18 at 8:22








  • 1




    $begingroup$
    Are we assuming $(2^m+1)mid(5^m-1)$?
    $endgroup$
    – manooooh
    Dec 13 '18 at 8:26






  • 3




    $begingroup$
    @manooooh,yes. it's right
    $endgroup$
    – function sug
    Dec 13 '18 at 8:27






  • 1




    $begingroup$
    If $m$ is odd then $2^m +1$ is divisible by 3 but $5^m - 1$ is not. No need for quadratic reciprocity
    $endgroup$
    – user420261
    Dec 13 '18 at 8:34














  • 1




    $begingroup$
    What did you try?
    $endgroup$
    – Robert Z
    Dec 13 '18 at 8:15










  • $begingroup$
    @RobertZ I have post my some try,Following step I can't do it
    $endgroup$
    – function sug
    Dec 13 '18 at 8:22








  • 1




    $begingroup$
    Are we assuming $(2^m+1)mid(5^m-1)$?
    $endgroup$
    – manooooh
    Dec 13 '18 at 8:26






  • 3




    $begingroup$
    @manooooh,yes. it's right
    $endgroup$
    – function sug
    Dec 13 '18 at 8:27






  • 1




    $begingroup$
    If $m$ is odd then $2^m +1$ is divisible by 3 but $5^m - 1$ is not. No need for quadratic reciprocity
    $endgroup$
    – user420261
    Dec 13 '18 at 8:34








1




1




$begingroup$
What did you try?
$endgroup$
– Robert Z
Dec 13 '18 at 8:15




$begingroup$
What did you try?
$endgroup$
– Robert Z
Dec 13 '18 at 8:15












$begingroup$
@RobertZ I have post my some try,Following step I can't do it
$endgroup$
– function sug
Dec 13 '18 at 8:22






$begingroup$
@RobertZ I have post my some try,Following step I can't do it
$endgroup$
– function sug
Dec 13 '18 at 8:22






1




1




$begingroup$
Are we assuming $(2^m+1)mid(5^m-1)$?
$endgroup$
– manooooh
Dec 13 '18 at 8:26




$begingroup$
Are we assuming $(2^m+1)mid(5^m-1)$?
$endgroup$
– manooooh
Dec 13 '18 at 8:26




3




3




$begingroup$
@manooooh,yes. it's right
$endgroup$
– function sug
Dec 13 '18 at 8:27




$begingroup$
@manooooh,yes. it's right
$endgroup$
– function sug
Dec 13 '18 at 8:27




1




1




$begingroup$
If $m$ is odd then $2^m +1$ is divisible by 3 but $5^m - 1$ is not. No need for quadratic reciprocity
$endgroup$
– user420261
Dec 13 '18 at 8:34




$begingroup$
If $m$ is odd then $2^m +1$ is divisible by 3 but $5^m - 1$ is not. No need for quadratic reciprocity
$endgroup$
– user420261
Dec 13 '18 at 8:34










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