If $p$ is prime, then $x^2 +5y^2 = p iff pequiv 1,9 $ mod $(20)$.
$begingroup$
Let $pneq 2,5$ be prime. I wish to show that: $x^2 +5y^2 = p Leftrightarrow pequiv 1,9 $ mod $(20)$.
I proved to $Rightarrow$ part, means $x^2 +5y^2=p Rightarrow pequiv 1,9 $ mod $(20)$.
For $Leftarrow$ , $pequiv 1,9(20) Rightarrow pequiv 1(4)$ , $pequiv1 ,4 (5)$ thus $(frac{4}{p})=1,(frac{-1}{p}) =1$ (using legendre symbols) , also $(frac{5}{p})=_{pequiv1(4)}(frac{p}{5})$ and $pequiv1(5)$ so $(frac{5}{p})=1$ , so $(frac{-20}{p})=(frac{5}{p})(frac{4}{p})(frac{-1}{p}) = 1$. So $-20$ is a quadratic residue mod $p$.
Yet I don't succeed to go on from this point (I don't know even if its possible to do so).
number-theory elementary-number-theory algebraic-number-theory quadratic-forms quadratic-residues
asked Dec 31 '18 at 5:55
dandan
503513
$endgroup$
add a comment |
$begingroup$
Let $pneq 2,5$ be prime. I wish to show that: $x^2 +5y^2 = p Leftrightarrow pequiv 1,9 $ mod $(20)$.
I proved to $Rightarrow$ part, means $x^2 +5y^2=p Rightarrow pequiv 1,9 $ mod $(20)$.
For $Leftarrow$ , $pequiv 1,9(20) Rightarrow pequiv 1(4)$ , $pequiv1 ,4 (5)$ thus $(frac{4}{p})=1,(frac{-1}{p}) =1$ (using legendre symbols) , also $(frac{5}{p})=_{pequiv1(4)}(frac{p}{5})$ and $pequiv1(5)$ so $(frac{5}{p})=1$ , so $(frac{-20}{p})=(frac{5}{p})(frac{4}{p})(frac{-1}{p}) = 1$. So $-20$ is a quadratic residue mod $p$.
Yet I don't succeed to go on from this point (I don't know even if its possible to do so).
number-theory elementary-number-theory algebraic-number-theory quadratic-forms quadratic-residues
asked Dec 31 '18 at 5:55
dandan
503513
$endgroup$
$begingroup$
See Cox primes of the form x^2+ny^2
$endgroup$
– Kenny Lau
Jan 2 at 10:09
add a comment |
$begingroup$
Let $pneq 2,5$ be prime. I wish to show that: $x^2 +5y^2 = p Leftrightarrow pequiv 1,9 $ mod $(20)$.
I proved to $Rightarrow$ part, means $x^2 +5y^2=p Rightarrow pequiv 1,9 $ mod $(20)$.
For $Leftarrow$ , $pequiv 1,9(20) Rightarrow pequiv 1(4)$ , $pequiv1 ,4 (5)$ thus $(frac{4}{p})=1,(frac{-1}{p}) =1$ (using legendre symbols) , also $(frac{5}{p})=_{pequiv1(4)}(frac{p}{5})$ and $pequiv1(5)$ so $(frac{5}{p})=1$ , so $(frac{-20}{p})=(frac{5}{p})(frac{4}{p})(frac{-1}{p}) = 1$. So $-20$ is a quadratic residue mod $p$.
Yet I don't succeed to go on from this point (I don't know even if its possible to do so).
number-theory elementary-number-theory algebraic-number-theory quadratic-forms quadratic-residues
asked Dec 31 '18 at 5:55
dandan
503513
$endgroup$
Let $pneq 2,5$ be prime. I wish to show that: $x^2 +5y^2 = p Leftrightarrow pequiv 1,9 $ mod $(20)$.
I proved to $Rightarrow$ part, means $x^2 +5y^2=p Rightarrow pequiv 1,9 $ mod $(20)$.
For $Leftarrow$ , $pequiv 1,9(20) Rightarrow pequiv 1(4)$ , $pequiv1 ,4 (5)$ thus $(frac{4}{p})=1,(frac{-1}{p}) =1$ (using legendre symbols) , also $(frac{5}{p})=_{pequiv1(4)}(frac{p}{5})$ and $pequiv1(5)$ so $(frac{5}{p})=1$ , so $(frac{-20}{p})=(frac{5}{p})(frac{4}{p})(frac{-1}{p}) = 1$. So $-20$ is a quadratic residue mod $p$.
Yet I don't succeed to go on from this point (I don't know even if its possible to do so).
number-theory elementary-number-theory algebraic-number-theory quadratic-forms quadratic-residues
number-theory elementary-number-theory algebraic-number-theory quadratic-forms quadratic-residues
asked Dec 31 '18 at 5:55
dandan
503513
asked Dec 31 '18 at 5:55
dandan
503513
edited Jan 1 at 12:09
dan
asked Dec 31 '18 at 5:55
dandan
503513
asked Dec 31 '18 at 5:55
dandan
503513
asked Dec 31 '18 at 5:55
dandan
503513
503513
$begingroup$
See Cox primes of the form x^2+ny^2
$endgroup$
– Kenny Lau
Jan 2 at 10:09
add a comment |
$begingroup$
See Cox primes of the form x^2+ny^2
$endgroup$
– Kenny Lau
Jan 2 at 10:09
$begingroup$
See Cox primes of the form x^2+ny^2
$endgroup$
– Kenny Lau
Jan 2 at 10:09
$begingroup$
See Cox primes of the form x^2+ny^2
$endgroup$
– Kenny Lau
Jan 2 at 10:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $pequiv1$ or $9pmod{20}$. Then $-5$ is a quadratic residue modulo $p$,
so there are integers $r$ and $s$ with $$r^2+ps=-5.$$
This means that the integer quadratic form
$$f(X,Y)=pX^2+2rXY+sY^2$$
has discriminant $-20$ and is also positive definite. Then $p$
is represented by $f$. Now $f$ is equivalent to a reduced form. There
are two reduced forms of discriminant $-20$: $g_1(X,Y)=X^2+5Y^2$
and $g_2(X,Y)=2X^2+2XY+3Y^2$. But $g_2$ cannot represent any number
congruent to $1$ or $9$ modulo $20$. Therefore $g_1$ represents $p$.
answered Dec 31 '18 at 7:41
Lord Shark the UnknownLord Shark the Unknown
102k1059132
$endgroup$
$begingroup$
Sorry but I am not familiar with the terminology of "reduced form" and $f$ which represents prime. May you please suggest a good source for me to read about it?
$endgroup$
– dan
Dec 31 '18 at 8:36
2
$begingroup$
Try Davenport's The Higher Arithmetic.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 8:40
$begingroup$
For questions of this type,you'll find all you need in D. Cox's marvelous book, "Primes of the form $x^2+ny^2"$
$endgroup$
– nguyen quang do
Jan 2 at 17:33
add a comment |
$begingroup$
An alternative solution: the class number of $K=mathbb{Q}(sqrt{-5})$ is $2$, its Hilbert class field is $L=mathbb{Q}(sqrt{-5},sqrt{-1})$. A prime $pneq 2,5$ can be written as $p=x^2+5y^2$, iff $p$ splits into two principal prime ideals in $K$, iff $p$ splits completely in $L$, iff $p$ splits in both $mathbb{Q}(sqrt{5})$ and $mathbb{Q}(sqrt{-1})$, iff
$$left(frac{-1}{p}right) = 1 qquad left(frac{5}{p}right)= left(frac{p}{5}right) = 1$$
which happens iff $pequiv 1,9 pmod{20}$.
The solution using reduced binary quadratic form works here, because when $D=-20$ there are only one form $x^2+5y^2$ in the principal genus. This also happens for some other small $|D|$.
answered Dec 31 '18 at 12:54
piscopisco
11.6k21742
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $pequiv1$ or $9pmod{20}$. Then $-5$ is a quadratic residue modulo $p$,
so there are integers $r$ and $s$ with $$r^2+ps=-5.$$
This means that the integer quadratic form
$$f(X,Y)=pX^2+2rXY+sY^2$$
has discriminant $-20$ and is also positive definite. Then $p$
is represented by $f$. Now $f$ is equivalent to a reduced form. There
are two reduced forms of discriminant $-20$: $g_1(X,Y)=X^2+5Y^2$
and $g_2(X,Y)=2X^2+2XY+3Y^2$. But $g_2$ cannot represent any number
congruent to $1$ or $9$ modulo $20$. Therefore $g_1$ represents $p$.
answered Dec 31 '18 at 7:41
Lord Shark the UnknownLord Shark the Unknown
102k1059132
$endgroup$
$begingroup$
Sorry but I am not familiar with the terminology of "reduced form" and $f$ which represents prime. May you please suggest a good source for me to read about it?
$endgroup$
– dan
Dec 31 '18 at 8:36
2
$begingroup$
Try Davenport's The Higher Arithmetic.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 8:40
$begingroup$
For questions of this type,you'll find all you need in D. Cox's marvelous book, "Primes of the form $x^2+ny^2"$
$endgroup$
– nguyen quang do
Jan 2 at 17:33
add a comment |
$begingroup$
Let $pequiv1$ or $9pmod{20}$. Then $-5$ is a quadratic residue modulo $p$,
so there are integers $r$ and $s$ with $$r^2+ps=-5.$$
This means that the integer quadratic form
$$f(X,Y)=pX^2+2rXY+sY^2$$
has discriminant $-20$ and is also positive definite. Then $p$
is represented by $f$. Now $f$ is equivalent to a reduced form. There
are two reduced forms of discriminant $-20$: $g_1(X,Y)=X^2+5Y^2$
and $g_2(X,Y)=2X^2+2XY+3Y^2$. But $g_2$ cannot represent any number
congruent to $1$ or $9$ modulo $20$. Therefore $g_1$ represents $p$.
answered Dec 31 '18 at 7:41
Lord Shark the UnknownLord Shark the Unknown
102k1059132
$endgroup$
$begingroup$
Sorry but I am not familiar with the terminology of "reduced form" and $f$ which represents prime. May you please suggest a good source for me to read about it?
$endgroup$
– dan
Dec 31 '18 at 8:36
2
$begingroup$
Try Davenport's The Higher Arithmetic.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 8:40
$begingroup$
For questions of this type,you'll find all you need in D. Cox's marvelous book, "Primes of the form $x^2+ny^2"$
$endgroup$
– nguyen quang do
Jan 2 at 17:33
add a comment |
$begingroup$
Let $pequiv1$ or $9pmod{20}$. Then $-5$ is a quadratic residue modulo $p$,
so there are integers $r$ and $s$ with $$r^2+ps=-5.$$
This means that the integer quadratic form
$$f(X,Y)=pX^2+2rXY+sY^2$$
has discriminant $-20$ and is also positive definite. Then $p$
is represented by $f$. Now $f$ is equivalent to a reduced form. There
are two reduced forms of discriminant $-20$: $g_1(X,Y)=X^2+5Y^2$
and $g_2(X,Y)=2X^2+2XY+3Y^2$. But $g_2$ cannot represent any number
congruent to $1$ or $9$ modulo $20$. Therefore $g_1$ represents $p$.
answered Dec 31 '18 at 7:41
Lord Shark the UnknownLord Shark the Unknown
102k1059132
$endgroup$
Let $pequiv1$ or $9pmod{20}$. Then $-5$ is a quadratic residue modulo $p$,
so there are integers $r$ and $s$ with $$r^2+ps=-5.$$
This means that the integer quadratic form
$$f(X,Y)=pX^2+2rXY+sY^2$$
has discriminant $-20$ and is also positive definite. Then $p$
is represented by $f$. Now $f$ is equivalent to a reduced form. There
are two reduced forms of discriminant $-20$: $g_1(X,Y)=X^2+5Y^2$
and $g_2(X,Y)=2X^2+2XY+3Y^2$. But $g_2$ cannot represent any number
congruent to $1$ or $9$ modulo $20$. Therefore $g_1$ represents $p$.
answered Dec 31 '18 at 7:41
Lord Shark the UnknownLord Shark the Unknown
102k1059132
answered Dec 31 '18 at 7:41
Lord Shark the UnknownLord Shark the Unknown
102k1059132
answered Dec 31 '18 at 7:41
Lord Shark the UnknownLord Shark the Unknown
102k1059132
answered Dec 31 '18 at 7:41
Lord Shark the UnknownLord Shark the Unknown
102k1059132
102k1059132
$begingroup$
Sorry but I am not familiar with the terminology of "reduced form" and $f$ which represents prime. May you please suggest a good source for me to read about it?
$endgroup$
– dan
Dec 31 '18 at 8:36
2
$begingroup$
Try Davenport's The Higher Arithmetic.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 8:40
$begingroup$
For questions of this type,you'll find all you need in D. Cox's marvelous book, "Primes of the form $x^2+ny^2"$
$endgroup$
– nguyen quang do
Jan 2 at 17:33
add a comment |
$begingroup$
Sorry but I am not familiar with the terminology of "reduced form" and $f$ which represents prime. May you please suggest a good source for me to read about it?
$endgroup$
– dan
Dec 31 '18 at 8:36
2
$begingroup$
Try Davenport's The Higher Arithmetic.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 8:40
$begingroup$
For questions of this type,you'll find all you need in D. Cox's marvelous book, "Primes of the form $x^2+ny^2"$
$endgroup$
– nguyen quang do
Jan 2 at 17:33
$begingroup$
Sorry but I am not familiar with the terminology of "reduced form" and $f$ which represents prime. May you please suggest a good source for me to read about it?
$endgroup$
– dan
Dec 31 '18 at 8:36
$begingroup$
Sorry but I am not familiar with the terminology of "reduced form" and $f$ which represents prime. May you please suggest a good source for me to read about it?
$endgroup$
– dan
Dec 31 '18 at 8:36
2
2
$begingroup$
Try Davenport's The Higher Arithmetic.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 8:40
$begingroup$
Try Davenport's The Higher Arithmetic.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 8:40
$begingroup$
For questions of this type,you'll find all you need in D. Cox's marvelous book, "Primes of the form $x^2+ny^2"$
$endgroup$
– nguyen quang do
Jan 2 at 17:33
$begingroup$
For questions of this type,you'll find all you need in D. Cox's marvelous book, "Primes of the form $x^2+ny^2"$
$endgroup$
– nguyen quang do
Jan 2 at 17:33
add a comment |
$begingroup$
An alternative solution: the class number of $K=mathbb{Q}(sqrt{-5})$ is $2$, its Hilbert class field is $L=mathbb{Q}(sqrt{-5},sqrt{-1})$. A prime $pneq 2,5$ can be written as $p=x^2+5y^2$, iff $p$ splits into two principal prime ideals in $K$, iff $p$ splits completely in $L$, iff $p$ splits in both $mathbb{Q}(sqrt{5})$ and $mathbb{Q}(sqrt{-1})$, iff
$$left(frac{-1}{p}right) = 1 qquad left(frac{5}{p}right)= left(frac{p}{5}right) = 1$$
which happens iff $pequiv 1,9 pmod{20}$.
The solution using reduced binary quadratic form works here, because when $D=-20$ there are only one form $x^2+5y^2$ in the principal genus. This also happens for some other small $|D|$.
answered Dec 31 '18 at 12:54
piscopisco
11.6k21742
$endgroup$
add a comment |
$begingroup$
An alternative solution: the class number of $K=mathbb{Q}(sqrt{-5})$ is $2$, its Hilbert class field is $L=mathbb{Q}(sqrt{-5},sqrt{-1})$. A prime $pneq 2,5$ can be written as $p=x^2+5y^2$, iff $p$ splits into two principal prime ideals in $K$, iff $p$ splits completely in $L$, iff $p$ splits in both $mathbb{Q}(sqrt{5})$ and $mathbb{Q}(sqrt{-1})$, iff
$$left(frac{-1}{p}right) = 1 qquad left(frac{5}{p}right)= left(frac{p}{5}right) = 1$$
which happens iff $pequiv 1,9 pmod{20}$.
The solution using reduced binary quadratic form works here, because when $D=-20$ there are only one form $x^2+5y^2$ in the principal genus. This also happens for some other small $|D|$.
answered Dec 31 '18 at 12:54
piscopisco
11.6k21742
$endgroup$
add a comment |
$begingroup$
An alternative solution: the class number of $K=mathbb{Q}(sqrt{-5})$ is $2$, its Hilbert class field is $L=mathbb{Q}(sqrt{-5},sqrt{-1})$. A prime $pneq 2,5$ can be written as $p=x^2+5y^2$, iff $p$ splits into two principal prime ideals in $K$, iff $p$ splits completely in $L$, iff $p$ splits in both $mathbb{Q}(sqrt{5})$ and $mathbb{Q}(sqrt{-1})$, iff
$$left(frac{-1}{p}right) = 1 qquad left(frac{5}{p}right)= left(frac{p}{5}right) = 1$$
which happens iff $pequiv 1,9 pmod{20}$.
The solution using reduced binary quadratic form works here, because when $D=-20$ there are only one form $x^2+5y^2$ in the principal genus. This also happens for some other small $|D|$.
answered Dec 31 '18 at 12:54
piscopisco
11.6k21742
$endgroup$
An alternative solution: the class number of $K=mathbb{Q}(sqrt{-5})$ is $2$, its Hilbert class field is $L=mathbb{Q}(sqrt{-5},sqrt{-1})$. A prime $pneq 2,5$ can be written as $p=x^2+5y^2$, iff $p$ splits into two principal prime ideals in $K$, iff $p$ splits completely in $L$, iff $p$ splits in both $mathbb{Q}(sqrt{5})$ and $mathbb{Q}(sqrt{-1})$, iff
$$left(frac{-1}{p}right) = 1 qquad left(frac{5}{p}right)= left(frac{p}{5}right) = 1$$
which happens iff $pequiv 1,9 pmod{20}$.
The solution using reduced binary quadratic form works here, because when $D=-20$ there are only one form $x^2+5y^2$ in the principal genus. This also happens for some other small $|D|$.
answered Dec 31 '18 at 12:54
piscopisco
11.6k21742
edited Dec 31 '18 at 17:30
answered Dec 31 '18 at 12:54
piscopisco
11.6k21742
answered Dec 31 '18 at 12:54
piscopisco
11.6k21742
answered Dec 31 '18 at 12:54
piscopisco
11.6k21742
11.6k21742
add a comment |
add a comment |
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$begingroup$
See Cox primes of the form x^2+ny^2
$endgroup$
– Kenny Lau
Jan 2 at 10:09