Prove for any postive $k$,such $2^m-2^n|k^m-k^n$ The 2019 Stack Overflow Developer Survey Results Are Inif for any postive integer $(b-k)|(a-k^n)$ why $a=b^n?$Find all odd positive integers $n$ greater than $1$ such that for any coprime divisors …Prove that there are no positive integers $a$ and $b$, such that $b^4 + b + 1 = a^4$Find all integer solutions to $a+b+c|ab+bc+ca|abc$May be this Conjecture is hold? with $n$ consecutive postive integers problemproof - infinite pair integers $a$ and $b$ such that $a + b = 100$ and $gcd(a, b) = 5$Find all numbers such that “Product of all divisors=cube of number”.$x^a+b-1$ is divisible by $x^a+mx^b+1$ for every positive integers $x$.Find all positive integers $a$ and $b$ such that $(1 + a)(8 + b)(a + b) = 27ab$.Prove $n$ having to be an exponent of 2 for $b^n + 1 =$ a prime number
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Prove for any postive $k$,such $2^m-2^n|k^m-k^n$
The 2019 Stack Overflow Developer Survey Results Are Inif for any postive integer $(b-k)|(a-k^n)$ why $a=b^n?$Find all odd positive integers $n$ greater than $1$ such that for any coprime divisors …Prove that there are no positive integers $a$ and $b$, such that $b^4 + b + 1 = a^4$Find all integer solutions to $a+b+c|ab+bc+ca|abc$May be this Conjecture is hold? with $n$ consecutive postive integers problemproof - infinite pair integers $a$ and $b$ such that $a + b = 100$ and $gcd(a, b) = 5$Find all numbers such that “Product of all divisors=cube of number”.$x^a+b-1$ is divisible by $x^a+mx^b+1$ for every positive integers $x$.Find all positive integers $a$ and $b$ such that $(1 + a)(8 + b)(a + b) = 27ab$.Prove $n$ having to be an exponent of 2 for $b^n + 1 =$ a prime number
$begingroup$
Let positive integers such $m>n$,and if $$2^m-2^n|3^m-3^n$$
then for any positive integers $k$ have
$$2^m-2^n|k^m-k^n$$
I have looked at the simpler case $k=1,2,3$.I was able to solve this,How to prove $kge 4$. Any progress on the problem is welcome.
number-theory
asked Jan 20 '16 at 12:06
function sugfunction sug
2681439
$endgroup$
|
show 2 more comments
$begingroup$
Let positive integers such $m>n$,and if $$2^m-2^n|3^m-3^n$$
then for any positive integers $k$ have
$$2^m-2^n|k^m-k^n$$
I have looked at the simpler case $k=1,2,3$.I was able to solve this,How to prove $kge 4$. Any progress on the problem is welcome.
number-theory
asked Jan 20 '16 at 12:06
function sugfunction sug
2681439
$endgroup$
$begingroup$
Surely this does not hold for $k=1$? Shouldn't this be for $kge2$?
$endgroup$
– Mufasa
Jan 20 '16 at 12:17
2
$begingroup$
@Mufasa Is not zero divisible by every number?
$endgroup$
– Chad Shin
Jan 20 '16 at 12:20
$begingroup$
@ChadShin - Oops! Sorry I misread this as the other way around - thanks for clarifying
$endgroup$
– Mufasa
Jan 20 '16 at 12:21
$begingroup$
Could anyone give me $(m,n)$ such that $m-n$ is not 2 or 1?
$endgroup$
– Chad Shin
Jan 20 '16 at 12:41
2
$begingroup$
FWIW I did a small Mathematica search. If $1le nle 100$ and $n<mle n+100$, then the only pairs that work are $$(1,2),(1,3),(1,5),(2,4),(2,6),(2,8),(2,14),(3,5),(3,7),(3,9),(3,15),(4,8),(4,16).$$ If that list is complete proving the claim is easy, because no prime factor $>13$ occurs as a factor of $2^m-2^n$ in those cases.
$endgroup$
– Jyrki Lahtonen
Jan 20 '16 at 13:39
|
show 2 more comments
$begingroup$
Let positive integers such $m>n$,and if $$2^m-2^n|3^m-3^n$$
then for any positive integers $k$ have
$$2^m-2^n|k^m-k^n$$
I have looked at the simpler case $k=1,2,3$.I was able to solve this,How to prove $kge 4$. Any progress on the problem is welcome.
number-theory
asked Jan 20 '16 at 12:06
function sugfunction sug
2681439
$endgroup$
Let positive integers such $m>n$,and if $$2^m-2^n|3^m-3^n$$
then for any positive integers $k$ have
$$2^m-2^n|k^m-k^n$$
I have looked at the simpler case $k=1,2,3$.I was able to solve this,How to prove $kge 4$. Any progress on the problem is welcome.
number-theory
number-theory
asked Jan 20 '16 at 12:06
function sugfunction sug
2681439
asked Jan 20 '16 at 12:06
function sugfunction sug
2681439
edited Jan 21 '16 at 10:13
function sug
asked Jan 20 '16 at 12:06
function sugfunction sug
2681439
asked Jan 20 '16 at 12:06
function sugfunction sug
2681439
asked Jan 20 '16 at 12:06
function sugfunction sug
2681439
2681439
$begingroup$
Surely this does not hold for $k=1$? Shouldn't this be for $kge2$?
$endgroup$
– Mufasa
Jan 20 '16 at 12:17
2
$begingroup$
@Mufasa Is not zero divisible by every number?
$endgroup$
– Chad Shin
Jan 20 '16 at 12:20
$begingroup$
@ChadShin - Oops! Sorry I misread this as the other way around - thanks for clarifying
$endgroup$
– Mufasa
Jan 20 '16 at 12:21
$begingroup$
Could anyone give me $(m,n)$ such that $m-n$ is not 2 or 1?
$endgroup$
– Chad Shin
Jan 20 '16 at 12:41
2
$begingroup$
FWIW I did a small Mathematica search. If $1le nle 100$ and $n<mle n+100$, then the only pairs that work are $$(1,2),(1,3),(1,5),(2,4),(2,6),(2,8),(2,14),(3,5),(3,7),(3,9),(3,15),(4,8),(4,16).$$ If that list is complete proving the claim is easy, because no prime factor $>13$ occurs as a factor of $2^m-2^n$ in those cases.
$endgroup$
– Jyrki Lahtonen
Jan 20 '16 at 13:39
|
show 2 more comments
$begingroup$
Surely this does not hold for $k=1$? Shouldn't this be for $kge2$?
$endgroup$
– Mufasa
Jan 20 '16 at 12:17
2
$begingroup$
@Mufasa Is not zero divisible by every number?
$endgroup$
– Chad Shin
Jan 20 '16 at 12:20
$begingroup$
@ChadShin - Oops! Sorry I misread this as the other way around - thanks for clarifying
$endgroup$
– Mufasa
Jan 20 '16 at 12:21
$begingroup$
Could anyone give me $(m,n)$ such that $m-n$ is not 2 or 1?
$endgroup$
– Chad Shin
Jan 20 '16 at 12:41
2
$begingroup$
FWIW I did a small Mathematica search. If $1le nle 100$ and $n<mle n+100$, then the only pairs that work are $$(1,2),(1,3),(1,5),(2,4),(2,6),(2,8),(2,14),(3,5),(3,7),(3,9),(3,15),(4,8),(4,16).$$ If that list is complete proving the claim is easy, because no prime factor $>13$ occurs as a factor of $2^m-2^n$ in those cases.
$endgroup$
– Jyrki Lahtonen
Jan 20 '16 at 13:39
$begingroup$
Surely this does not hold for $k=1$? Shouldn't this be for $kge2$?
$endgroup$
– Mufasa
Jan 20 '16 at 12:17
$begingroup$
Surely this does not hold for $k=1$? Shouldn't this be for $kge2$?
$endgroup$
– Mufasa
Jan 20 '16 at 12:17
2
2
$begingroup$
@Mufasa Is not zero divisible by every number?
$endgroup$
– Chad Shin
Jan 20 '16 at 12:20
$begingroup$
@Mufasa Is not zero divisible by every number?
$endgroup$
– Chad Shin
Jan 20 '16 at 12:20
$begingroup$
@ChadShin - Oops! Sorry I misread this as the other way around - thanks for clarifying
$endgroup$
– Mufasa
Jan 20 '16 at 12:21
$begingroup$
@ChadShin - Oops! Sorry I misread this as the other way around - thanks for clarifying
$endgroup$
– Mufasa
Jan 20 '16 at 12:21
$begingroup$
Could anyone give me $(m,n)$ such that $m-n$ is not 2 or 1?
$endgroup$
– Chad Shin
Jan 20 '16 at 12:41
$begingroup$
Could anyone give me $(m,n)$ such that $m-n$ is not 2 or 1?
$endgroup$
– Chad Shin
Jan 20 '16 at 12:41
2
2
$begingroup$
FWIW I did a small Mathematica search. If $1le nle 100$ and $n<mle n+100$, then the only pairs that work are $$(1,2),(1,3),(1,5),(2,4),(2,6),(2,8),(2,14),(3,5),(3,7),(3,9),(3,15),(4,8),(4,16).$$ If that list is complete proving the claim is easy, because no prime factor $>13$ occurs as a factor of $2^m-2^n$ in those cases.
$endgroup$
– Jyrki Lahtonen
Jan 20 '16 at 13:39
$begingroup$
FWIW I did a small Mathematica search. If $1le nle 100$ and $n<mle n+100$, then the only pairs that work are $$(1,2),(1,3),(1,5),(2,4),(2,6),(2,8),(2,14),(3,5),(3,7),(3,9),(3,15),(4,8),(4,16).$$ If that list is complete proving the claim is easy, because no prime factor $>13$ occurs as a factor of $2^m-2^n$ in those cases.
$endgroup$
– Jyrki Lahtonen
Jan 20 '16 at 13:39
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
There are only finitely many of $(m,n)$'s for which $2^m-2^n|k^m-k^n$ holds for all $k$ (as observed also by Jyrki in comments).
This is proven in paper by Sun Qi and Zhang Ming-Zhi Pairs where $2^a - 2^b$ divides $n^a - n^b$ for all $n$, where they show that only $(m,n)$ for which this holds with $0 leq n < m$ are:
$$
(1,0),(2,1),(3,1),(4,2),(5,3),(5,1),(6, 2), (7, 3),
(8, 4), (8, 2), (9, 3), (14, 2), (15, 3), (16, 4).
$$
This is also referred in $B47$ in Richard Guy's book Unsolved Problems in Number Theory.
answered Mar 23 at 18:15
SilSil
5,65821745
$endgroup$
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are only finitely many of $(m,n)$'s for which $2^m-2^n|k^m-k^n$ holds for all $k$ (as observed also by Jyrki in comments).
This is proven in paper by Sun Qi and Zhang Ming-Zhi Pairs where $2^a - 2^b$ divides $n^a - n^b$ for all $n$, where they show that only $(m,n)$ for which this holds with $0 leq n < m$ are:
$$
(1,0),(2,1),(3,1),(4,2),(5,3),(5,1),(6, 2), (7, 3),
(8, 4), (8, 2), (9, 3), (14, 2), (15, 3), (16, 4).
$$
This is also referred in $B47$ in Richard Guy's book Unsolved Problems in Number Theory.
answered Mar 23 at 18:15
SilSil
5,65821745
$endgroup$
add a comment |
$begingroup$
There are only finitely many of $(m,n)$'s for which $2^m-2^n|k^m-k^n$ holds for all $k$ (as observed also by Jyrki in comments).
This is proven in paper by Sun Qi and Zhang Ming-Zhi Pairs where $2^a - 2^b$ divides $n^a - n^b$ for all $n$, where they show that only $(m,n)$ for which this holds with $0 leq n < m$ are:
$$
(1,0),(2,1),(3,1),(4,2),(5,3),(5,1),(6, 2), (7, 3),
(8, 4), (8, 2), (9, 3), (14, 2), (15, 3), (16, 4).
$$
This is also referred in $B47$ in Richard Guy's book Unsolved Problems in Number Theory.
answered Mar 23 at 18:15
SilSil
5,65821745
$endgroup$
add a comment |
$begingroup$
There are only finitely many of $(m,n)$'s for which $2^m-2^n|k^m-k^n$ holds for all $k$ (as observed also by Jyrki in comments).
This is proven in paper by Sun Qi and Zhang Ming-Zhi Pairs where $2^a - 2^b$ divides $n^a - n^b$ for all $n$, where they show that only $(m,n)$ for which this holds with $0 leq n < m$ are:
$$
(1,0),(2,1),(3,1),(4,2),(5,3),(5,1),(6, 2), (7, 3),
(8, 4), (8, 2), (9, 3), (14, 2), (15, 3), (16, 4).
$$
This is also referred in $B47$ in Richard Guy's book Unsolved Problems in Number Theory.
answered Mar 23 at 18:15
SilSil
5,65821745
$endgroup$
There are only finitely many of $(m,n)$'s for which $2^m-2^n|k^m-k^n$ holds for all $k$ (as observed also by Jyrki in comments).
This is proven in paper by Sun Qi and Zhang Ming-Zhi Pairs where $2^a - 2^b$ divides $n^a - n^b$ for all $n$, where they show that only $(m,n)$ for which this holds with $0 leq n < m$ are:
$$
(1,0),(2,1),(3,1),(4,2),(5,3),(5,1),(6, 2), (7, 3),
(8, 4), (8, 2), (9, 3), (14, 2), (15, 3), (16, 4).
$$
This is also referred in $B47$ in Richard Guy's book Unsolved Problems in Number Theory.
answered Mar 23 at 18:15
SilSil
5,65821745
answered Mar 23 at 18:15
SilSil
5,65821745
answered Mar 23 at 18:15
SilSil
5,65821745
answered Mar 23 at 18:15
SilSil
5,65821745
5,65821745
add a comment |
add a comment |
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$begingroup$
Surely this does not hold for $k=1$? Shouldn't this be for $kge2$?
$endgroup$
– Mufasa
Jan 20 '16 at 12:17
2
$begingroup$
@Mufasa Is not zero divisible by every number?
$endgroup$
– Chad Shin
Jan 20 '16 at 12:20
$begingroup$
@ChadShin - Oops! Sorry I misread this as the other way around - thanks for clarifying
$endgroup$
– Mufasa
Jan 20 '16 at 12:21
$begingroup$
Could anyone give me $(m,n)$ such that $m-n$ is not 2 or 1?
$endgroup$
– Chad Shin
Jan 20 '16 at 12:41
2
$begingroup$
FWIW I did a small Mathematica search. If $1le nle 100$ and $n<mle n+100$, then the only pairs that work are $$(1,2),(1,3),(1,5),(2,4),(2,6),(2,8),(2,14),(3,5),(3,7),(3,9),(3,15),(4,8),(4,16).$$ If that list is complete proving the claim is easy, because no prime factor $>13$ occurs as a factor of $2^m-2^n$ in those cases.
$endgroup$
– Jyrki Lahtonen
Jan 20 '16 at 13:39