Circle $x^2 + y^2 = n!$ doesn't hit any lattice points for any $n$ except for $0$, $1$, $2$ and $6$ or does...
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I stumbled across the following problem in high school:$$
x^2 + y^2 = n!
$$
I tested it within my laptop capabilities, watched a 3b1b video Pi in prime regularities, where he explains how to find the number of integer solutions based on prime factors. There doesn't seem to be any above $30!$. Maybe I'm wrong and there are infinitely many exceptions like $2$ and $6$, maybe the proof is too difficult for me to grasp or... I hope I'm just too blind to see the obvious.
nt.number-theory analytic-number-theory prime-numbers factorization
edited Mar 30 at 21:13
TheSimpliFire
12310
asked Mar 30 at 12:27
BetydligBetydlig
964
$endgroup$
add a comment |
$begingroup$
I stumbled across the following problem in high school:$$
x^2 + y^2 = n!
$$
I tested it within my laptop capabilities, watched a 3b1b video Pi in prime regularities, where he explains how to find the number of integer solutions based on prime factors. There doesn't seem to be any above $30!$. Maybe I'm wrong and there are infinitely many exceptions like $2$ and $6$, maybe the proof is too difficult for me to grasp or... I hope I'm just too blind to see the obvious.
nt.number-theory analytic-number-theory prime-numbers factorization
edited Mar 30 at 21:13
TheSimpliFire
12310
asked Mar 30 at 12:27
BetydligBetydlig
964
$endgroup$
$begingroup$
Hi and welcome to MO. What is your question?
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– Amir Sagiv
Mar 30 at 12:31
1
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Hi. The question is: are there any integers above 6 for which this equation has integer pairs (x,y) as solutions.
$endgroup$
– Betydlig
Mar 30 at 12:35
21
$begingroup$
At least for sufficiently large $n$, there will be a prime $p equiv 3 bmod 4$ such that $p le n < 2p$. Then $p$ divides $n!$ exactly once, hence $n!$ cannot be a sum of two squares.
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– Michael Stoll
Mar 30 at 12:45
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If you like my answer, please accept it officially (so that it turns green). Thanks in advance!
$endgroup$
– GH from MO
2 days ago
add a comment |
$begingroup$
I stumbled across the following problem in high school:$$
x^2 + y^2 = n!
$$
I tested it within my laptop capabilities, watched a 3b1b video Pi in prime regularities, where he explains how to find the number of integer solutions based on prime factors. There doesn't seem to be any above $30!$. Maybe I'm wrong and there are infinitely many exceptions like $2$ and $6$, maybe the proof is too difficult for me to grasp or... I hope I'm just too blind to see the obvious.
nt.number-theory analytic-number-theory prime-numbers factorization
edited Mar 30 at 21:13
TheSimpliFire
12310
asked Mar 30 at 12:27
BetydligBetydlig
964
$endgroup$
I stumbled across the following problem in high school:$$
x^2 + y^2 = n!
$$
I tested it within my laptop capabilities, watched a 3b1b video Pi in prime regularities, where he explains how to find the number of integer solutions based on prime factors. There doesn't seem to be any above $30!$. Maybe I'm wrong and there are infinitely many exceptions like $2$ and $6$, maybe the proof is too difficult for me to grasp or... I hope I'm just too blind to see the obvious.
nt.number-theory analytic-number-theory prime-numbers factorization
nt.number-theory analytic-number-theory prime-numbers factorization
edited Mar 30 at 21:13
TheSimpliFire
12310
asked Mar 30 at 12:27
BetydligBetydlig
964
edited Mar 30 at 21:13
TheSimpliFire
12310
asked Mar 30 at 12:27
BetydligBetydlig
964
edited Mar 30 at 21:13
TheSimpliFire
12310
edited Mar 30 at 21:13
TheSimpliFire
12310
edited Mar 30 at 21:13
TheSimpliFire
12310
12310
asked Mar 30 at 12:27
BetydligBetydlig
964
asked Mar 30 at 12:27
BetydligBetydlig
964
asked Mar 30 at 12:27
BetydligBetydlig
964
964
$begingroup$
Hi and welcome to MO. What is your question?
$endgroup$
– Amir Sagiv
Mar 30 at 12:31
1
$begingroup$
Hi. The question is: are there any integers above 6 for which this equation has integer pairs (x,y) as solutions.
$endgroup$
– Betydlig
Mar 30 at 12:35
21
$begingroup$
At least for sufficiently large $n$, there will be a prime $p equiv 3 bmod 4$ such that $p le n < 2p$. Then $p$ divides $n!$ exactly once, hence $n!$ cannot be a sum of two squares.
$endgroup$
– Michael Stoll
Mar 30 at 12:45
$begingroup$
If you like my answer, please accept it officially (so that it turns green). Thanks in advance!
$endgroup$
– GH from MO
2 days ago
add a comment |
$begingroup$
Hi and welcome to MO. What is your question?
$endgroup$
– Amir Sagiv
Mar 30 at 12:31
1
$begingroup$
Hi. The question is: are there any integers above 6 for which this equation has integer pairs (x,y) as solutions.
$endgroup$
– Betydlig
Mar 30 at 12:35
21
$begingroup$
At least for sufficiently large $n$, there will be a prime $p equiv 3 bmod 4$ such that $p le n < 2p$. Then $p$ divides $n!$ exactly once, hence $n!$ cannot be a sum of two squares.
$endgroup$
– Michael Stoll
Mar 30 at 12:45
$begingroup$
If you like my answer, please accept it officially (so that it turns green). Thanks in advance!
$endgroup$
– GH from MO
2 days ago
$begingroup$
Hi and welcome to MO. What is your question?
$endgroup$
– Amir Sagiv
Mar 30 at 12:31
$begingroup$
Hi and welcome to MO. What is your question?
$endgroup$
– Amir Sagiv
Mar 30 at 12:31
1
1
$begingroup$
Hi. The question is: are there any integers above 6 for which this equation has integer pairs (x,y) as solutions.
$endgroup$
– Betydlig
Mar 30 at 12:35
$begingroup$
Hi. The question is: are there any integers above 6 for which this equation has integer pairs (x,y) as solutions.
$endgroup$
– Betydlig
Mar 30 at 12:35
21
21
$begingroup$
At least for sufficiently large $n$, there will be a prime $p equiv 3 bmod 4$ such that $p le n < 2p$. Then $p$ divides $n!$ exactly once, hence $n!$ cannot be a sum of two squares.
$endgroup$
– Michael Stoll
Mar 30 at 12:45
$begingroup$
At least for sufficiently large $n$, there will be a prime $p equiv 3 bmod 4$ such that $p le n < 2p$. Then $p$ divides $n!$ exactly once, hence $n!$ cannot be a sum of two squares.
$endgroup$
– Michael Stoll
Mar 30 at 12:45
$begingroup$
If you like my answer, please accept it officially (so that it turns green). Thanks in advance!
$endgroup$
– GH from MO
2 days ago
$begingroup$
If you like my answer, please accept it officially (so that it turns green). Thanks in advance!
$endgroup$
– GH from MO
2 days ago
add a comment |
1 Answer
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For $ngeq 7$, Erdős proved in 1932 that there is a prime $n/2<pleq n$ of the form $p=4k+3$. From this he deduces (in the same paper) that $1!$, $2!$, $6!$ are the only factorials which can be written as a sum of two squares.
answered Mar 30 at 13:42
GH from MOGH from MO
59.3k5148227
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1 Answer
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$begingroup$
For $ngeq 7$, Erdős proved in 1932 that there is a prime $n/2<pleq n$ of the form $p=4k+3$. From this he deduces (in the same paper) that $1!$, $2!$, $6!$ are the only factorials which can be written as a sum of two squares.
answered Mar 30 at 13:42
GH from MOGH from MO
59.3k5148227
$endgroup$
add a comment |
$begingroup$
For $ngeq 7$, Erdős proved in 1932 that there is a prime $n/2<pleq n$ of the form $p=4k+3$. From this he deduces (in the same paper) that $1!$, $2!$, $6!$ are the only factorials which can be written as a sum of two squares.
answered Mar 30 at 13:42
GH from MOGH from MO
59.3k5148227
$endgroup$
add a comment |
$begingroup$
For $ngeq 7$, Erdős proved in 1932 that there is a prime $n/2<pleq n$ of the form $p=4k+3$. From this he deduces (in the same paper) that $1!$, $2!$, $6!$ are the only factorials which can be written as a sum of two squares.
answered Mar 30 at 13:42
GH from MOGH from MO
59.3k5148227
$endgroup$
For $ngeq 7$, Erdős proved in 1932 that there is a prime $n/2<pleq n$ of the form $p=4k+3$. From this he deduces (in the same paper) that $1!$, $2!$, $6!$ are the only factorials which can be written as a sum of two squares.
answered Mar 30 at 13:42
GH from MOGH from MO
59.3k5148227
answered Mar 30 at 13:42
GH from MOGH from MO
59.3k5148227
answered Mar 30 at 13:42
GH from MOGH from MO
59.3k5148227
answered Mar 30 at 13:42
GH from MOGH from MO
59.3k5148227
59.3k5148227
add a comment |
add a comment |
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$begingroup$
Hi and welcome to MO. What is your question?
$endgroup$
– Amir Sagiv
Mar 30 at 12:31
1
$begingroup$
Hi. The question is: are there any integers above 6 for which this equation has integer pairs (x,y) as solutions.
$endgroup$
– Betydlig
Mar 30 at 12:35
21
$begingroup$
At least for sufficiently large $n$, there will be a prime $p equiv 3 bmod 4$ such that $p le n < 2p$. Then $p$ divides $n!$ exactly once, hence $n!$ cannot be a sum of two squares.
$endgroup$
– Michael Stoll
Mar 30 at 12:45
$begingroup$
If you like my answer, please accept it officially (so that it turns green). Thanks in advance!
$endgroup$
– GH from MO
2 days ago