Find all positive integer $m$ such $2^{m}+1mid5^m-1$ [closed]
$begingroup$
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
edited Dec 14 '18 at 21:25
Batominovski
33.1k33293
asked Dec 13 '18 at 8:13
function sugfunction sug
3411438
$endgroup$
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.
|
show 5 more comments
$begingroup$
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
edited Dec 14 '18 at 21:25
Batominovski
33.1k33293
asked Dec 13 '18 at 8:13
function sugfunction sug
3411438
$endgroup$
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
|
show 5 more comments
$begingroup$
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
edited Dec 14 '18 at 21:25
Batominovski
33.1k33293
asked Dec 13 '18 at 8:13
function sugfunction sug
3411438
$endgroup$
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
edited Dec 14 '18 at 21:25
Batominovski
33.1k33293
asked Dec 13 '18 at 8:13
function sugfunction sug
3411438
edited Dec 14 '18 at 21:25
Batominovski
33.1k33293
asked Dec 13 '18 at 8:13
function sugfunction sug
3411438
edited Dec 14 '18 at 21:25
Batominovski
33.1k33293
edited Dec 14 '18 at 21:25
Batominovski
33.1k33293
edited Dec 14 '18 at 21:25
Batominovski
33.1k33293
33.1k33293
asked Dec 13 '18 at 8:13
function sugfunction sug
3411438
asked Dec 13 '18 at 8:13
function sugfunction sug
3411438
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
|
show 5 more comments
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
|
show 5 more comments
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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