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Is it necessary for a number of the form $4k^2+1$ to have at least one prime factor of the form $4n+1$?
3
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Is it necessary for a number of the form $4k^2+1$ to have at least one prime factor of the form $4n+1$?
I was trying to prove that there are infinitely many primes of the form $4n+1$, but to prove it, I need to prove the above statement true.
number-theory prime-numbers quadratic-residues
edited Jan 13 at 16:15
amWhy
1
asked Jan 13 at 15:49
Vibhav AggarwalVibhav Aggarwal
161
$endgroup$
add a comment |
3
$begingroup$
Is it necessary for a number of the form $4k^2+1$ to have at least one prime factor of the form $4n+1$?
I was trying to prove that there are infinitely many primes of the form $4n+1$, but to prove it, I need to prove the above statement true.
number-theory prime-numbers quadratic-residues
edited Jan 13 at 16:15
amWhy
1
asked Jan 13 at 15:49
Vibhav AggarwalVibhav Aggarwal
161
$endgroup$
$begingroup$
Even more : EVERY prime factor of such a number must be of the form $4n+1$
$endgroup$
– Peter
Jan 13 at 15:53
$begingroup$
@Peter I would really appreciate if you provide a proof for that.
$endgroup$
– Vibhav Aggarwal
Jan 13 at 15:55
3
$begingroup$
Considering quadratic residues, the proof is trivial. Every prime factor $p$ must be odd, and since the number is of the form $n^2+1$ , $-1$ is a so-called quadratic residue mod $p$ , whenever $p$ divides $4k^2+1$ , we can conclude immediately that $p$ must be congruent to $1$ modulo $4$.
$endgroup$
– Peter
Jan 13 at 16:05
$begingroup$
Cf. here
$endgroup$
– J. W. Tanner
Jan 13 at 16:11
$begingroup$
Also there are infinitely many primes of the form $4n+1$ trivially by Dirichlet's theorem on arithmetic progressions.
$endgroup$
– Tejas Rao
Jan 16 at 17:12
add a comment |
3
3
3
$begingroup$
Is it necessary for a number of the form $4k^2+1$ to have at least one prime factor of the form $4n+1$?
I was trying to prove that there are infinitely many primes of the form $4n+1$, but to prove it, I need to prove the above statement true.
number-theory prime-numbers quadratic-residues
edited Jan 13 at 16:15
amWhy
1
asked Jan 13 at 15:49
Vibhav AggarwalVibhav Aggarwal
161
$endgroup$
Is it necessary for a number of the form $4k^2+1$ to have at least one prime factor of the form $4n+1$?
I was trying to prove that there are infinitely many primes of the form $4n+1$, but to prove it, I need to prove the above statement true.
number-theory prime-numbers quadratic-residues
number-theory prime-numbers quadratic-residues
edited Jan 13 at 16:15
amWhy
1
asked Jan 13 at 15:49
Vibhav AggarwalVibhav Aggarwal
161
edited Jan 13 at 16:15
amWhy
1
asked Jan 13 at 15:49
Vibhav AggarwalVibhav Aggarwal
161
edited Jan 13 at 16:15
amWhy
1
edited Jan 13 at 16:15
amWhy
1
edited Jan 13 at 16:15
amWhy
1
1
asked Jan 13 at 15:49
Vibhav AggarwalVibhav Aggarwal
161
asked Jan 13 at 15:49
Vibhav AggarwalVibhav Aggarwal
161
asked Jan 13 at 15:49
Vibhav AggarwalVibhav Aggarwal
161
161
$begingroup$
Even more : EVERY prime factor of such a number must be of the form $4n+1$
$endgroup$
– Peter
Jan 13 at 15:53
$begingroup$
@Peter I would really appreciate if you provide a proof for that.
$endgroup$
– Vibhav Aggarwal
Jan 13 at 15:55
3
$begingroup$
Considering quadratic residues, the proof is trivial. Every prime factor $p$ must be odd, and since the number is of the form $n^2+1$ , $-1$ is a so-called quadratic residue mod $p$ , whenever $p$ divides $4k^2+1$ , we can conclude immediately that $p$ must be congruent to $1$ modulo $4$.
$endgroup$
– Peter
Jan 13 at 16:05
$begingroup$
Cf. here
$endgroup$
– J. W. Tanner
Jan 13 at 16:11
$begingroup$
Also there are infinitely many primes of the form $4n+1$ trivially by Dirichlet's theorem on arithmetic progressions.
$endgroup$
– Tejas Rao
Jan 16 at 17:12
add a comment |
$begingroup$
Even more : EVERY prime factor of such a number must be of the form $4n+1$
$endgroup$
– Peter
Jan 13 at 15:53
$begingroup$
@Peter I would really appreciate if you provide a proof for that.
$endgroup$
– Vibhav Aggarwal
Jan 13 at 15:55
3
$begingroup$
Considering quadratic residues, the proof is trivial. Every prime factor $p$ must be odd, and since the number is of the form $n^2+1$ , $-1$ is a so-called quadratic residue mod $p$ , whenever $p$ divides $4k^2+1$ , we can conclude immediately that $p$ must be congruent to $1$ modulo $4$.
$endgroup$
– Peter
Jan 13 at 16:05
$begingroup$
Cf. here
$endgroup$
– J. W. Tanner
Jan 13 at 16:11
$begingroup$
Also there are infinitely many primes of the form $4n+1$ trivially by Dirichlet's theorem on arithmetic progressions.
$endgroup$
– Tejas Rao
Jan 16 at 17:12
$begingroup$
Even more : EVERY prime factor of such a number must be of the form $4n+1$
$endgroup$
– Peter
Jan 13 at 15:53
$begingroup$
Even more : EVERY prime factor of such a number must be of the form $4n+1$
$endgroup$
– Peter
Jan 13 at 15:53
$begingroup$
@Peter I would really appreciate if you provide a proof for that.
$endgroup$
– Vibhav Aggarwal
Jan 13 at 15:55
$begingroup$
@Peter I would really appreciate if you provide a proof for that.
$endgroup$
– Vibhav Aggarwal
Jan 13 at 15:55
3
3
$begingroup$
Considering quadratic residues, the proof is trivial. Every prime factor $p$ must be odd, and since the number is of the form $n^2+1$ , $-1$ is a so-called quadratic residue mod $p$ , whenever $p$ divides $4k^2+1$ , we can conclude immediately that $p$ must be congruent to $1$ modulo $4$.
$endgroup$
– Peter
Jan 13 at 16:05
$begingroup$
Considering quadratic residues, the proof is trivial. Every prime factor $p$ must be odd, and since the number is of the form $n^2+1$ , $-1$ is a so-called quadratic residue mod $p$ , whenever $p$ divides $4k^2+1$ , we can conclude immediately that $p$ must be congruent to $1$ modulo $4$.
$endgroup$
– Peter
Jan 13 at 16:05
$begingroup$
Cf. here
$endgroup$
– J. W. Tanner
Jan 13 at 16:11
$begingroup$
Cf. here
$endgroup$
– J. W. Tanner
Jan 13 at 16:11
$begingroup$
Also there are infinitely many primes of the form $4n+1$ trivially by Dirichlet's theorem on arithmetic progressions.
$endgroup$
– Tejas Rao
Jan 16 at 17:12
$begingroup$
Also there are infinitely many primes of the form $4n+1$ trivially by Dirichlet's theorem on arithmetic progressions.
$endgroup$
– Tejas Rao
Jan 16 at 17:12
add a comment |
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$begingroup$
Even more : EVERY prime factor of such a number must be of the form $4n+1$
$endgroup$
– Peter
Jan 13 at 15:53
$begingroup$
@Peter I would really appreciate if you provide a proof for that.
$endgroup$
– Vibhav Aggarwal
Jan 13 at 15:55
3
$begingroup$
Considering quadratic residues, the proof is trivial. Every prime factor $p$ must be odd, and since the number is of the form $n^2+1$ , $-1$ is a so-called quadratic residue mod $p$ , whenever $p$ divides $4k^2+1$ , we can conclude immediately that $p$ must be congruent to $1$ modulo $4$.
$endgroup$
– Peter
Jan 13 at 16:05
$begingroup$
Cf. here
$endgroup$
– J. W. Tanner
Jan 13 at 16:11
$begingroup$
Also there are infinitely many primes of the form $4n+1$ trivially by Dirichlet's theorem on arithmetic progressions.
$endgroup$
– Tejas Rao
Jan 16 at 17:12