If $p$ is prime, then $x^2 +5y^2 = p iff pequiv 1,9 $ mod $(20)$.












11












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    See Cox primes of the form x^2+ny^2
    $endgroup$
    – Kenny Lau
    Jan 2 at 10:09
















11












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    See Cox primes of the form x^2+ny^2
    $endgroup$
    – Kenny Lau
    Jan 2 at 10:09














11












11








11


4



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 12:09







dan

















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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















5












$begingroup$

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 $p$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















5












$begingroup$

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 $|D|$.






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    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 $p$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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
















    5












    $begingroup$

    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 $p$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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














    5












    5








    5





    $begingroup$

    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 $p$.






    share|cite|improve this answer









    $endgroup$



    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 $p$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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


















    • $begingroup$
      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











    5












    $begingroup$

    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 $|D|$.






    share|cite|improve this answer











    $endgroup$


















      5












      $begingroup$

      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 $|D|$.






      share|cite|improve this answer











      $endgroup$
















        5












        5








        5





        $begingroup$

        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 $|D|$.






        share|cite|improve this answer











        $endgroup$



        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 $|D|$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 31 '18 at 17:30

























        answered Dec 31 '18 at 12:54









        piscopisco

        11.6k21742




        11.6k21742






























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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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