Has $ 2^n-1equiv 2^41+1mod n $ a solution?Suppose $p$ is an odd prime. Show that $x^4 equiv-1$ (mod $p$) has a solution if and only if $p equiv1$ (mod $8$).Number of quadractic residues $mod p$ and $mod n$.$x^16-16 equiv 0 mod p$ has a solution for each primeIf $pequiv 2$ mod $3$, $x^3equiv a$ mod $p$ has only one solution modulo $p$.Show that if $n equiv 3 (mod 4)$, then n has a prime factor $p equiv 3 (modtext 4)$Show that if $m equiv 3 mod 4$, then $m$ has a prime factor $p equiv 3 mod 4$.Number of solutions of $x^e equiv c mod p$Can the smallest solution of $varphi(k)=n$ be an even positive integer?Can I find all solutions of $2^n-1equiv kmod n$?Is there a solution for $2^n-1equiv 2^16+1mod n$ or $2^n-1equiv 2^26+1mod n$?
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Has $ 2^n-1equiv 2^41+1mod n $ a solution?
Suppose $p$ is an odd prime. Show that $x^4 equiv-1$ (mod $p$) has a solution if and only if $p equiv1$ (mod $8$).Number of quadractic residues $mod p$ and $mod n$.$x^16-16 equiv 0 mod p$ has a solution for each primeIf $pequiv 2$ mod $3$, $x^3equiv a$ mod $p$ has only one solution modulo $p$.Show that if $n equiv 3 (mod 4)$, then n has a prime factor $p equiv 3 (modtext 4)$Show that if $m equiv 3 mod 4$, then $m$ has a prime factor $p equiv 3 mod 4$.Number of solutions of $x^e equiv c mod p$Can the smallest solution of $varphi(k)=n$ be an even positive integer?Can I find all solutions of $2^n-1equiv kmod n$?Is there a solution for $2^n-1equiv 2^16+1mod n$ or $2^n-1equiv 2^26+1mod n$?
$begingroup$
Has $$2^n-1equiv 2^41+1mod n$$ a solution with a positive integer $ n>1 $ ?
Motivation : The equation $$2^n-1equiv kmod n$$ has always a solution, if $ k-1 $ has an odd prime factor (this odd prime factor is then a solution) and for $ k=2^m+1 $ , I know a solution for $$m=1,2,3,cdots,40$$ Hence, this is the smallest number for which I know no solution. Upto $ n=10^9 $, there is no solution.
number-theory elementary-number-theory modular-arithmetic
asked Mar 29 at 10:05
PeterPeter
49.1k1240138
$endgroup$
|
show 3 more comments
$begingroup$
Has $$2^n-1equiv 2^41+1mod n$$ a solution with a positive integer $ n>1 $ ?
Motivation : The equation $$2^n-1equiv kmod n$$ has always a solution, if $ k-1 $ has an odd prime factor (this odd prime factor is then a solution) and for $ k=2^m+1 $ , I know a solution for $$m=1,2,3,cdots,40$$ Hence, this is the smallest number for which I know no solution. Upto $ n=10^9 $, there is no solution.
number-theory elementary-number-theory modular-arithmetic
asked Mar 29 at 10:05
PeterPeter
49.1k1240138
$endgroup$
1
$begingroup$
Could you please say something more about the cases $m<41$? How did you find solutions in those cases?
$endgroup$
– Leo163
Mar 29 at 12:00
1
$begingroup$
Do you know $n$ modulo $3$ or $4$? Can you rule out the case $11mid n$?
$endgroup$
– W-t-P
Mar 29 at 19:54
$begingroup$
n is odd. not prime. etc.
$endgroup$
– Roddy MacPhee
Mar 30 at 15:06
$begingroup$
it's also not 0 mod 3.
$endgroup$
– Roddy MacPhee
Mar 30 at 15:55
1
$begingroup$
I have done some computations, and off they are correct, there is no solution with $n<10^10$. Out of curiosity, what are solutions for $m=2$ and $m=12$?
$endgroup$
– Julián Aguirre
Mar 30 at 16:50
|
show 3 more comments
$begingroup$
Has $$2^n-1equiv 2^41+1mod n$$ a solution with a positive integer $ n>1 $ ?
Motivation : The equation $$2^n-1equiv kmod n$$ has always a solution, if $ k-1 $ has an odd prime factor (this odd prime factor is then a solution) and for $ k=2^m+1 $ , I know a solution for $$m=1,2,3,cdots,40$$ Hence, this is the smallest number for which I know no solution. Upto $ n=10^9 $, there is no solution.
number-theory elementary-number-theory modular-arithmetic
asked Mar 29 at 10:05
PeterPeter
49.1k1240138
$endgroup$
Has $$2^n-1equiv 2^41+1mod n$$ a solution with a positive integer $ n>1 $ ?
Motivation : The equation $$2^n-1equiv kmod n$$ has always a solution, if $ k-1 $ has an odd prime factor (this odd prime factor is then a solution) and for $ k=2^m+1 $ , I know a solution for $$m=1,2,3,cdots,40$$ Hence, this is the smallest number for which I know no solution. Upto $ n=10^9 $, there is no solution.
number-theory elementary-number-theory modular-arithmetic
number-theory elementary-number-theory modular-arithmetic
asked Mar 29 at 10:05
PeterPeter
49.1k1240138
asked Mar 29 at 10:05
PeterPeter
49.1k1240138
edited Mar 29 at 10:18
Peter
asked Mar 29 at 10:05
PeterPeter
49.1k1240138
asked Mar 29 at 10:05
PeterPeter
49.1k1240138
asked Mar 29 at 10:05
PeterPeter
49.1k1240138
49.1k1240138
1
$begingroup$
Could you please say something more about the cases $m<41$? How did you find solutions in those cases?
$endgroup$
– Leo163
Mar 29 at 12:00
1
$begingroup$
Do you know $n$ modulo $3$ or $4$? Can you rule out the case $11mid n$?
$endgroup$
– W-t-P
Mar 29 at 19:54
$begingroup$
n is odd. not prime. etc.
$endgroup$
– Roddy MacPhee
Mar 30 at 15:06
$begingroup$
it's also not 0 mod 3.
$endgroup$
– Roddy MacPhee
Mar 30 at 15:55
1
$begingroup$
I have done some computations, and off they are correct, there is no solution with $n<10^10$. Out of curiosity, what are solutions for $m=2$ and $m=12$?
$endgroup$
– Julián Aguirre
Mar 30 at 16:50
|
show 3 more comments
1
$begingroup$
Could you please say something more about the cases $m<41$? How did you find solutions in those cases?
$endgroup$
– Leo163
Mar 29 at 12:00
1
$begingroup$
Do you know $n$ modulo $3$ or $4$? Can you rule out the case $11mid n$?
$endgroup$
– W-t-P
Mar 29 at 19:54
$begingroup$
n is odd. not prime. etc.
$endgroup$
– Roddy MacPhee
Mar 30 at 15:06
$begingroup$
it's also not 0 mod 3.
$endgroup$
– Roddy MacPhee
Mar 30 at 15:55
1
$begingroup$
I have done some computations, and off they are correct, there is no solution with $n<10^10$. Out of curiosity, what are solutions for $m=2$ and $m=12$?
$endgroup$
– Julián Aguirre
Mar 30 at 16:50
1
1
$begingroup$
Could you please say something more about the cases $m<41$? How did you find solutions in those cases?
$endgroup$
– Leo163
Mar 29 at 12:00
$begingroup$
Could you please say something more about the cases $m<41$? How did you find solutions in those cases?
$endgroup$
– Leo163
Mar 29 at 12:00
1
1
$begingroup$
Do you know $n$ modulo $3$ or $4$? Can you rule out the case $11mid n$?
$endgroup$
– W-t-P
Mar 29 at 19:54
$begingroup$
Do you know $n$ modulo $3$ or $4$? Can you rule out the case $11mid n$?
$endgroup$
– W-t-P
Mar 29 at 19:54
$begingroup$
n is odd. not prime. etc.
$endgroup$
– Roddy MacPhee
Mar 30 at 15:06
$begingroup$
n is odd. not prime. etc.
$endgroup$
– Roddy MacPhee
Mar 30 at 15:06
$begingroup$
it's also not 0 mod 3.
$endgroup$
– Roddy MacPhee
Mar 30 at 15:55
$begingroup$
it's also not 0 mod 3.
$endgroup$
– Roddy MacPhee
Mar 30 at 15:55
1
1
$begingroup$
I have done some computations, and off they are correct, there is no solution with $n<10^10$. Out of curiosity, what are solutions for $m=2$ and $m=12$?
$endgroup$
– Julián Aguirre
Mar 30 at 16:50
$begingroup$
I have done some computations, and off they are correct, there is no solution with $n<10^10$. Out of curiosity, what are solutions for $m=2$ and $m=12$?
$endgroup$
– Julián Aguirre
Mar 30 at 16:50
|
show 3 more comments
1 Answer
1
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oldest
votes
$begingroup$
Yes, $n=24189255799819$ is a solution (may be not the smallest one). Tried to search for $n=pq$ with $p,q$ prime and $p$ small, by factoring $2^p-2k$, and got it already at $p=11$.
answered Mar 31 at 20:02
metamorphymetamorphy
3,8721721
$endgroup$
add a comment |
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$begingroup$
Yes, $n=24189255799819$ is a solution (may be not the smallest one). Tried to search for $n=pq$ with $p,q$ prime and $p$ small, by factoring $2^p-2k$, and got it already at $p=11$.
answered Mar 31 at 20:02
metamorphymetamorphy
3,8721721
$endgroup$
add a comment |
$begingroup$
Yes, $n=24189255799819$ is a solution (may be not the smallest one). Tried to search for $n=pq$ with $p,q$ prime and $p$ small, by factoring $2^p-2k$, and got it already at $p=11$.
answered Mar 31 at 20:02
metamorphymetamorphy
3,8721721
$endgroup$
add a comment |
$begingroup$
Yes, $n=24189255799819$ is a solution (may be not the smallest one). Tried to search for $n=pq$ with $p,q$ prime and $p$ small, by factoring $2^p-2k$, and got it already at $p=11$.
answered Mar 31 at 20:02
metamorphymetamorphy
3,8721721
$endgroup$
Yes, $n=24189255799819$ is a solution (may be not the smallest one). Tried to search for $n=pq$ with $p,q$ prime and $p$ small, by factoring $2^p-2k$, and got it already at $p=11$.
answered Mar 31 at 20:02
metamorphymetamorphy
3,8721721
edited Mar 31 at 20:18
answered Mar 31 at 20:02
metamorphymetamorphy
3,8721721
answered Mar 31 at 20:02
metamorphymetamorphy
3,8721721
answered Mar 31 at 20:02
metamorphymetamorphy
3,8721721
3,8721721
add a comment |
add a comment |
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1
$begingroup$
Could you please say something more about the cases $m<41$? How did you find solutions in those cases?
$endgroup$
– Leo163
Mar 29 at 12:00
1
$begingroup$
Do you know $n$ modulo $3$ or $4$? Can you rule out the case $11mid n$?
$endgroup$
– W-t-P
Mar 29 at 19:54
$begingroup$
n is odd. not prime. etc.
$endgroup$
– Roddy MacPhee
Mar 30 at 15:06
$begingroup$
it's also not 0 mod 3.
$endgroup$
– Roddy MacPhee
Mar 30 at 15:55
1
$begingroup$
I have done some computations, and off they are correct, there is no solution with $n<10^10$. Out of curiosity, what are solutions for $m=2$ and $m=12$?
$endgroup$
– Julián Aguirre
Mar 30 at 16:50