Mixing
Determine when $frac{n!-1}{3n+1}$ is an integer
0
$begingroup$
I want to find $n$ so that $frac{n!-1}{3n+1}$ is an integer. Assume that $3n+1$ is prime (there are only finitely many solutions where $3n+1$ is composite).
number-theory elementary-number-theory
asked Jan 11 at 8:10
acreativenameacreativename
5816
$endgroup$
|
show 8 more comments
0
$begingroup$
I want to find $n$ so that $frac{n!-1}{3n+1}$ is an integer. Assume that $3n+1$ is prime (there are only finitely many solutions where $3n+1$ is composite).
number-theory elementary-number-theory
asked Jan 11 at 8:10
acreativenameacreativename
5816
$endgroup$
2
$begingroup$
How do yo know that there are finitely many solutions with $3n+1$ composite and why this makes the problem uninteresting in this case? Why do you think there are infinitely many solutions with $3n+1$ prime? What is the origin of the problem?
$endgroup$
– W-t-P
Jan 11 at 8:23
1
$begingroup$
There was a stack exchange question similar to this one except that it read $frac{n!-1}{2n+1}$ which can be tackled by modifying wilson's theorem. Note that if 3n+1 is composite there is a non-trivial divisor of 3n+1 at most $(3n+1)^{0.5}$ which is less than $n$ for sufficiently large n but $n! - 1$ is co prime to all numbers less than or equal to $n$
$endgroup$
– acreativename
Jan 11 at 8:28
1
$begingroup$
If $3n+1$ is composite and divides $n!-1$.
$endgroup$
– acreativename
Jan 11 at 8:40
1
$begingroup$
I think the problem with $2n+1$ is much more tractable. Computations show that there are 19 primes $p$ up to $10^6$ with $((p-1)/3)!equiv 1pmod p$, the largest of them being $p=909,151$.
$endgroup$
– W-t-P
Jan 11 at 8:55
1
$begingroup$
@acreativename Are you content with the first few positive integers $n$ satisfying your condition, or do you want to classify those integers ?
$endgroup$
– Peter
Jan 11 at 10:43
|
show 8 more comments
0
0
0
1
$begingroup$
I want to find $n$ so that $frac{n!-1}{3n+1}$ is an integer. Assume that $3n+1$ is prime (there are only finitely many solutions where $3n+1$ is composite).
number-theory elementary-number-theory
asked Jan 11 at 8:10
acreativenameacreativename
5816
$endgroup$
I want to find $n$ so that $frac{n!-1}{3n+1}$ is an integer. Assume that $3n+1$ is prime (there are only finitely many solutions where $3n+1$ is composite).
number-theory elementary-number-theory
number-theory elementary-number-theory
asked Jan 11 at 8:10
acreativenameacreativename
5816
asked Jan 11 at 8:10
acreativenameacreativename
5816
asked Jan 11 at 8:10
acreativenameacreativename
5816
asked Jan 11 at 8:10
acreativenameacreativename
5816
asked Jan 11 at 8:10
acreativenameacreativename
5816
5816
2
$begingroup$
How do yo know that there are finitely many solutions with $3n+1$ composite and why this makes the problem uninteresting in this case? Why do you think there are infinitely many solutions with $3n+1$ prime? What is the origin of the problem?
$endgroup$
– W-t-P
Jan 11 at 8:23
1
$begingroup$
There was a stack exchange question similar to this one except that it read $frac{n!-1}{2n+1}$ which can be tackled by modifying wilson's theorem. Note that if 3n+1 is composite there is a non-trivial divisor of 3n+1 at most $(3n+1)^{0.5}$ which is less than $n$ for sufficiently large n but $n! - 1$ is co prime to all numbers less than or equal to $n$
$endgroup$
– acreativename
Jan 11 at 8:28
1
$begingroup$
If $3n+1$ is composite and divides $n!-1$.
$endgroup$
– acreativename
Jan 11 at 8:40
1
$begingroup$
I think the problem with $2n+1$ is much more tractable. Computations show that there are 19 primes $p$ up to $10^6$ with $((p-1)/3)!equiv 1pmod p$, the largest of them being $p=909,151$.
$endgroup$
– W-t-P
Jan 11 at 8:55
1
$begingroup$
@acreativename Are you content with the first few positive integers $n$ satisfying your condition, or do you want to classify those integers ?
$endgroup$
– Peter
Jan 11 at 10:43
|
show 8 more comments
2
$begingroup$
How do yo know that there are finitely many solutions with $3n+1$ composite and why this makes the problem uninteresting in this case? Why do you think there are infinitely many solutions with $3n+1$ prime? What is the origin of the problem?
$endgroup$
– W-t-P
Jan 11 at 8:23
1
$begingroup$
There was a stack exchange question similar to this one except that it read $frac{n!-1}{2n+1}$ which can be tackled by modifying wilson's theorem. Note that if 3n+1 is composite there is a non-trivial divisor of 3n+1 at most $(3n+1)^{0.5}$ which is less than $n$ for sufficiently large n but $n! - 1$ is co prime to all numbers less than or equal to $n$
$endgroup$
– acreativename
Jan 11 at 8:28
1
$begingroup$
If $3n+1$ is composite and divides $n!-1$.
$endgroup$
– acreativename
Jan 11 at 8:40
1
$begingroup$
I think the problem with $2n+1$ is much more tractable. Computations show that there are 19 primes $p$ up to $10^6$ with $((p-1)/3)!equiv 1pmod p$, the largest of them being $p=909,151$.
$endgroup$
– W-t-P
Jan 11 at 8:55
1
$begingroup$
@acreativename Are you content with the first few positive integers $n$ satisfying your condition, or do you want to classify those integers ?
$endgroup$
– Peter
Jan 11 at 10:43
2
2
$begingroup$
How do yo know that there are finitely many solutions with $3n+1$ composite and why this makes the problem uninteresting in this case? Why do you think there are infinitely many solutions with $3n+1$ prime? What is the origin of the problem?
$endgroup$
– W-t-P
Jan 11 at 8:23
$begingroup$
How do yo know that there are finitely many solutions with $3n+1$ composite and why this makes the problem uninteresting in this case? Why do you think there are infinitely many solutions with $3n+1$ prime? What is the origin of the problem?
$endgroup$
– W-t-P
Jan 11 at 8:23
1
1
$begingroup$
There was a stack exchange question similar to this one except that it read $frac{n!-1}{2n+1}$ which can be tackled by modifying wilson's theorem. Note that if 3n+1 is composite there is a non-trivial divisor of 3n+1 at most $(3n+1)^{0.5}$ which is less than $n$ for sufficiently large n but $n! - 1$ is co prime to all numbers less than or equal to $n$
$endgroup$
– acreativename
Jan 11 at 8:28
$begingroup$
There was a stack exchange question similar to this one except that it read $frac{n!-1}{2n+1}$ which can be tackled by modifying wilson's theorem. Note that if 3n+1 is composite there is a non-trivial divisor of 3n+1 at most $(3n+1)^{0.5}$ which is less than $n$ for sufficiently large n but $n! - 1$ is co prime to all numbers less than or equal to $n$
$endgroup$
– acreativename
Jan 11 at 8:28
1
1
$begingroup$
If $3n+1$ is composite and divides $n!-1$.
$endgroup$
– acreativename
Jan 11 at 8:40
$begingroup$
If $3n+1$ is composite and divides $n!-1$.
$endgroup$
– acreativename
Jan 11 at 8:40
1
1
$begingroup$
I think the problem with $2n+1$ is much more tractable. Computations show that there are 19 primes $p$ up to $10^6$ with $((p-1)/3)!equiv 1pmod p$, the largest of them being $p=909,151$.
$endgroup$
– W-t-P
Jan 11 at 8:55
$begingroup$
I think the problem with $2n+1$ is much more tractable. Computations show that there are 19 primes $p$ up to $10^6$ with $((p-1)/3)!equiv 1pmod p$, the largest of them being $p=909,151$.
$endgroup$
– W-t-P
Jan 11 at 8:55
1
1
$begingroup$
@acreativename Are you content with the first few positive integers $n$ satisfying your condition, or do you want to classify those integers ?
$endgroup$
– Peter
Jan 11 at 10:43
$begingroup$
@acreativename Are you content with the first few positive integers $n$ satisfying your condition, or do you want to classify those integers ?
$endgroup$
– Peter
Jan 11 at 10:43
|
show 8 more comments
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2
$begingroup$
How do yo know that there are finitely many solutions with $3n+1$ composite and why this makes the problem uninteresting in this case? Why do you think there are infinitely many solutions with $3n+1$ prime? What is the origin of the problem?
$endgroup$
– W-t-P
Jan 11 at 8:23
1
$begingroup$
There was a stack exchange question similar to this one except that it read $frac{n!-1}{2n+1}$ which can be tackled by modifying wilson's theorem. Note that if 3n+1 is composite there is a non-trivial divisor of 3n+1 at most $(3n+1)^{0.5}$ which is less than $n$ for sufficiently large n but $n! - 1$ is co prime to all numbers less than or equal to $n$
$endgroup$
– acreativename
Jan 11 at 8:28
1
$begingroup$
If $3n+1$ is composite and divides $n!-1$.
$endgroup$
– acreativename
Jan 11 at 8:40
1
$begingroup$
I think the problem with $2n+1$ is much more tractable. Computations show that there are 19 primes $p$ up to $10^6$ with $((p-1)/3)!equiv 1pmod p$, the largest of them being $p=909,151$.
$endgroup$
– W-t-P
Jan 11 at 8:55
1
$begingroup$
@acreativename Are you content with the first few positive integers $n$ satisfying your condition, or do you want to classify those integers ?
$endgroup$
– Peter
Jan 11 at 10:43