on the quadratic form $5x^2-y^2$
$begingroup$
Consider the following subsets of $mathbb{Z}$:
$A={frac{5p^2-y^2}{4}, | , p, y ,,text{odd positive integers}, p,, text{prime}}$ and
$B={frac{5x^2-y^2}{4}, | , x, y ,,text{odd positive integers}}$.
Are there elements in $B$ which are not in $A$?
number-theory quadratic-forms
asked Jan 20 at 19:27
user84909user84909
1908
$endgroup$
|
show 4 more comments
$begingroup$
Consider the following subsets of $mathbb{Z}$:
$A={frac{5p^2-y^2}{4}, | , p, y ,,text{odd positive integers}, p,, text{prime}}$ and
$B={frac{5x^2-y^2}{4}, | , x, y ,,text{odd positive integers}}$.
Are there elements in $B$ which are not in $A$?
number-theory quadratic-forms
asked Jan 20 at 19:27
user84909user84909
1908
$endgroup$
$begingroup$
there's a problem with $1$ because it is not prime. Hence $1in B$ but I think $1notin A$. Apart from that very good question.
$endgroup$
– Dog_69
Jan 20 at 19:35
$begingroup$
But you haven't prove that $1notin A$, have you? @Dog_69
$endgroup$
– greedoid
Jan 20 at 19:36
$begingroup$
No, I haven't. I'm pretty sure but I don't have any proof.
$endgroup$
– Dog_69
Jan 20 at 19:40
3
$begingroup$
Taking $p=5$ and $y=11$ shows that $1in A$.
$endgroup$
– user84909
Jan 20 at 19:41
1
$begingroup$
What is a minumum number in B that you can't prove it is in A?
$endgroup$
– greedoid
Jan 20 at 19:42
|
show 4 more comments
$begingroup$
Consider the following subsets of $mathbb{Z}$:
$A={frac{5p^2-y^2}{4}, | , p, y ,,text{odd positive integers}, p,, text{prime}}$ and
$B={frac{5x^2-y^2}{4}, | , x, y ,,text{odd positive integers}}$.
Are there elements in $B$ which are not in $A$?
number-theory quadratic-forms
asked Jan 20 at 19:27
user84909user84909
1908
$endgroup$
Consider the following subsets of $mathbb{Z}$:
$A={frac{5p^2-y^2}{4}, | , p, y ,,text{odd positive integers}, p,, text{prime}}$ and
$B={frac{5x^2-y^2}{4}, | , x, y ,,text{odd positive integers}}$.
Are there elements in $B$ which are not in $A$?
number-theory quadratic-forms
number-theory quadratic-forms
asked Jan 20 at 19:27
user84909user84909
1908
asked Jan 20 at 19:27
user84909user84909
1908
asked Jan 20 at 19:27
user84909user84909
1908
asked Jan 20 at 19:27
user84909user84909
1908
asked Jan 20 at 19:27
user84909user84909
1908
1908
$begingroup$
there's a problem with $1$ because it is not prime. Hence $1in B$ but I think $1notin A$. Apart from that very good question.
$endgroup$
– Dog_69
Jan 20 at 19:35
$begingroup$
But you haven't prove that $1notin A$, have you? @Dog_69
$endgroup$
– greedoid
Jan 20 at 19:36
$begingroup$
No, I haven't. I'm pretty sure but I don't have any proof.
$endgroup$
– Dog_69
Jan 20 at 19:40
3
$begingroup$
Taking $p=5$ and $y=11$ shows that $1in A$.
$endgroup$
– user84909
Jan 20 at 19:41
1
$begingroup$
What is a minumum number in B that you can't prove it is in A?
$endgroup$
– greedoid
Jan 20 at 19:42
|
show 4 more comments
$begingroup$
there's a problem with $1$ because it is not prime. Hence $1in B$ but I think $1notin A$. Apart from that very good question.
$endgroup$
– Dog_69
Jan 20 at 19:35
$begingroup$
But you haven't prove that $1notin A$, have you? @Dog_69
$endgroup$
– greedoid
Jan 20 at 19:36
$begingroup$
No, I haven't. I'm pretty sure but I don't have any proof.
$endgroup$
– Dog_69
Jan 20 at 19:40
3
$begingroup$
Taking $p=5$ and $y=11$ shows that $1in A$.
$endgroup$
– user84909
Jan 20 at 19:41
1
$begingroup$
What is a minumum number in B that you can't prove it is in A?
$endgroup$
– greedoid
Jan 20 at 19:42
$begingroup$
there's a problem with $1$ because it is not prime. Hence $1in B$ but I think $1notin A$. Apart from that very good question.
$endgroup$
– Dog_69
Jan 20 at 19:35
$begingroup$
there's a problem with $1$ because it is not prime. Hence $1in B$ but I think $1notin A$. Apart from that very good question.
$endgroup$
– Dog_69
Jan 20 at 19:35
$begingroup$
But you haven't prove that $1notin A$, have you? @Dog_69
$endgroup$
– greedoid
Jan 20 at 19:36
$begingroup$
But you haven't prove that $1notin A$, have you? @Dog_69
$endgroup$
– greedoid
Jan 20 at 19:36
$begingroup$
No, I haven't. I'm pretty sure but I don't have any proof.
$endgroup$
– Dog_69
Jan 20 at 19:40
$begingroup$
No, I haven't. I'm pretty sure but I don't have any proof.
$endgroup$
– Dog_69
Jan 20 at 19:40
3
3
$begingroup$
Taking $p=5$ and $y=11$ shows that $1in A$.
$endgroup$
– user84909
Jan 20 at 19:41
$begingroup$
Taking $p=5$ and $y=11$ shows that $1in A$.
$endgroup$
– user84909
Jan 20 at 19:41
1
1
$begingroup$
What is a minumum number in B that you can't prove it is in A?
$endgroup$
– greedoid
Jan 20 at 19:42
$begingroup$
What is a minumum number in B that you can't prove it is in A?
$endgroup$
– greedoid
Jan 20 at 19:42
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Note that $frac{5cdot9^2-3^2}{4}=99$, and if $frac{5p^2-y^2}{4}=99$ for some prime number $p$, then
$$5p^2-y^2=396equiv0pmod{3},$$
from which it follows that $pequiv yequiv0pmod{3}$. Because $p$ is prime it follows that $p=3$, and hence
$$y^2=5p^2-396=-351,$$
a contradiction. So $99in B$ but $99notin A$.
answered Jan 20 at 21:15
ServaesServaes
25.7k33996
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Note that $frac{5cdot9^2-3^2}{4}=99$, and if $frac{5p^2-y^2}{4}=99$ for some prime number $p$, then
$$5p^2-y^2=396equiv0pmod{3},$$
from which it follows that $pequiv yequiv0pmod{3}$. Because $p$ is prime it follows that $p=3$, and hence
$$y^2=5p^2-396=-351,$$
a contradiction. So $99in B$ but $99notin A$.
answered Jan 20 at 21:15
ServaesServaes
25.7k33996
$endgroup$
add a comment |
$begingroup$
Note that $frac{5cdot9^2-3^2}{4}=99$, and if $frac{5p^2-y^2}{4}=99$ for some prime number $p$, then
$$5p^2-y^2=396equiv0pmod{3},$$
from which it follows that $pequiv yequiv0pmod{3}$. Because $p$ is prime it follows that $p=3$, and hence
$$y^2=5p^2-396=-351,$$
a contradiction. So $99in B$ but $99notin A$.
answered Jan 20 at 21:15
ServaesServaes
25.7k33996
$endgroup$
add a comment |
$begingroup$
Note that $frac{5cdot9^2-3^2}{4}=99$, and if $frac{5p^2-y^2}{4}=99$ for some prime number $p$, then
$$5p^2-y^2=396equiv0pmod{3},$$
from which it follows that $pequiv yequiv0pmod{3}$. Because $p$ is prime it follows that $p=3$, and hence
$$y^2=5p^2-396=-351,$$
a contradiction. So $99in B$ but $99notin A$.
answered Jan 20 at 21:15
ServaesServaes
25.7k33996
$endgroup$
Note that $frac{5cdot9^2-3^2}{4}=99$, and if $frac{5p^2-y^2}{4}=99$ for some prime number $p$, then
$$5p^2-y^2=396equiv0pmod{3},$$
from which it follows that $pequiv yequiv0pmod{3}$. Because $p$ is prime it follows that $p=3$, and hence
$$y^2=5p^2-396=-351,$$
a contradiction. So $99in B$ but $99notin A$.
answered Jan 20 at 21:15
ServaesServaes
25.7k33996
answered Jan 20 at 21:15
ServaesServaes
25.7k33996
answered Jan 20 at 21:15
ServaesServaes
25.7k33996
answered Jan 20 at 21:15
ServaesServaes
25.7k33996
25.7k33996
add a comment |
add a comment |
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$begingroup$
there's a problem with $1$ because it is not prime. Hence $1in B$ but I think $1notin A$. Apart from that very good question.
$endgroup$
– Dog_69
Jan 20 at 19:35
$begingroup$
But you haven't prove that $1notin A$, have you? @Dog_69
$endgroup$
– greedoid
Jan 20 at 19:36
$begingroup$
No, I haven't. I'm pretty sure but I don't have any proof.
$endgroup$
– Dog_69
Jan 20 at 19:40
3
$begingroup$
Taking $p=5$ and $y=11$ shows that $1in A$.
$endgroup$
– user84909
Jan 20 at 19:41
1
$begingroup$
What is a minumum number in B that you can't prove it is in A?
$endgroup$
– greedoid
Jan 20 at 19:42