What is the positive integral solution of $x^2+y^2= qz^2$?
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Is there a standard method to find positive integral solutions of the equation $x^2+y^2= qz^2$, where $q$ is a positive integral constant? If yes, then would someone please show it to me? Actually I had come across a problem in which we had to find integral solutions of $a^2+b^2=2c^2$ which was done by changing it to $(a-2/2)^2+(b-2/2)^2= c^2$ and using the parametric form of Pythogorean triplets but what if we had other numbers such as 4 or 5 in place of q?
number-theory diophantine-equations
asked Dec 19 '18 at 7:10
Shashwat1337Shashwat1337
959
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show 1 more comment
1
$begingroup$
Is there a standard method to find positive integral solutions of the equation $x^2+y^2= qz^2$, where $q$ is a positive integral constant? If yes, then would someone please show it to me? Actually I had come across a problem in which we had to find integral solutions of $a^2+b^2=2c^2$ which was done by changing it to $(a-2/2)^2+(b-2/2)^2= c^2$ and using the parametric form of Pythogorean triplets but what if we had other numbers such as 4 or 5 in place of q?
number-theory diophantine-equations
asked Dec 19 '18 at 7:10
Shashwat1337Shashwat1337
959
$endgroup$
$begingroup$
The left hand side looks like an ellipse, by swapping x and y you would get another ellipse, so I think you get a family of solutions.
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– Emil
Dec 19 '18 at 7:26
1
$begingroup$
en.wikipedia.org/wiki/Hasse_principle
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– metamorphy
Dec 19 '18 at 8:04
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If you can find one solution then the method of parametrizing Pythagorean triples (draw lines with rational slope thru $(1,0)$ and find the other point of intersection with the unit circle) generalizes to this case as well. You will use the ellipse $x^2+py^2=q$ instead.
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– Jyrki Lahtonen
Dec 19 '18 at 13:37
1
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@Jyrki the rational slope method does give all rational solutions; multiplying out it gives a Pythagorean triple type of formula. If you then say to divide out by the gcd of x,y,z you do get all primitive triples. More work: for such a parametrization, with coprime $u,v$ the set of gcd's of $x,y,z$ is finite. For each, a new integer parametrization may be produced.. compare answer and my comment at math.stackexchange.com/questions/3044517/…
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– Will Jagy
Dec 19 '18 at 17:54
1
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artofproblemsolving.com/community/…
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– individ
Dec 21 '18 at 5:50
|
show 1 more comment
1
1
1
1
$begingroup$
Is there a standard method to find positive integral solutions of the equation $x^2+y^2= qz^2$, where $q$ is a positive integral constant? If yes, then would someone please show it to me? Actually I had come across a problem in which we had to find integral solutions of $a^2+b^2=2c^2$ which was done by changing it to $(a-2/2)^2+(b-2/2)^2= c^2$ and using the parametric form of Pythogorean triplets but what if we had other numbers such as 4 or 5 in place of q?
number-theory diophantine-equations
asked Dec 19 '18 at 7:10
Shashwat1337Shashwat1337
959
$endgroup$
Is there a standard method to find positive integral solutions of the equation $x^2+y^2= qz^2$, where $q$ is a positive integral constant? If yes, then would someone please show it to me? Actually I had come across a problem in which we had to find integral solutions of $a^2+b^2=2c^2$ which was done by changing it to $(a-2/2)^2+(b-2/2)^2= c^2$ and using the parametric form of Pythogorean triplets but what if we had other numbers such as 4 or 5 in place of q?
number-theory diophantine-equations
number-theory diophantine-equations
asked Dec 19 '18 at 7:10
Shashwat1337Shashwat1337
959
asked Dec 19 '18 at 7:10
Shashwat1337Shashwat1337
959
edited Dec 20 '18 at 18:53
Shashwat1337
asked Dec 19 '18 at 7:10
Shashwat1337Shashwat1337
959
asked Dec 19 '18 at 7:10
Shashwat1337Shashwat1337
959
asked Dec 19 '18 at 7:10
Shashwat1337Shashwat1337
959
959
$begingroup$
The left hand side looks like an ellipse, by swapping x and y you would get another ellipse, so I think you get a family of solutions.
$endgroup$
– Emil
Dec 19 '18 at 7:26
1
$begingroup$
en.wikipedia.org/wiki/Hasse_principle
$endgroup$
– metamorphy
Dec 19 '18 at 8:04
$begingroup$
If you can find one solution then the method of parametrizing Pythagorean triples (draw lines with rational slope thru $(1,0)$ and find the other point of intersection with the unit circle) generalizes to this case as well. You will use the ellipse $x^2+py^2=q$ instead.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 13:37
1
$begingroup$
@Jyrki the rational slope method does give all rational solutions; multiplying out it gives a Pythagorean triple type of formula. If you then say to divide out by the gcd of x,y,z you do get all primitive triples. More work: for such a parametrization, with coprime $u,v$ the set of gcd's of $x,y,z$ is finite. For each, a new integer parametrization may be produced.. compare answer and my comment at math.stackexchange.com/questions/3044517/…
$endgroup$
– Will Jagy
Dec 19 '18 at 17:54
1
$begingroup$
artofproblemsolving.com/community/…
$endgroup$
– individ
Dec 21 '18 at 5:50
|
show 1 more comment
$begingroup$
The left hand side looks like an ellipse, by swapping x and y you would get another ellipse, so I think you get a family of solutions.
$endgroup$
– Emil
Dec 19 '18 at 7:26
1
$begingroup$
en.wikipedia.org/wiki/Hasse_principle
$endgroup$
– metamorphy
Dec 19 '18 at 8:04
$begingroup$
If you can find one solution then the method of parametrizing Pythagorean triples (draw lines with rational slope thru $(1,0)$ and find the other point of intersection with the unit circle) generalizes to this case as well. You will use the ellipse $x^2+py^2=q$ instead.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 13:37
1
$begingroup$
@Jyrki the rational slope method does give all rational solutions; multiplying out it gives a Pythagorean triple type of formula. If you then say to divide out by the gcd of x,y,z you do get all primitive triples. More work: for such a parametrization, with coprime $u,v$ the set of gcd's of $x,y,z$ is finite. For each, a new integer parametrization may be produced.. compare answer and my comment at math.stackexchange.com/questions/3044517/…
$endgroup$
– Will Jagy
Dec 19 '18 at 17:54
1
$begingroup$
artofproblemsolving.com/community/…
$endgroup$
– individ
Dec 21 '18 at 5:50
$begingroup$
The left hand side looks like an ellipse, by swapping x and y you would get another ellipse, so I think you get a family of solutions.
$endgroup$
– Emil
Dec 19 '18 at 7:26
$begingroup$
The left hand side looks like an ellipse, by swapping x and y you would get another ellipse, so I think you get a family of solutions.
$endgroup$
– Emil
Dec 19 '18 at 7:26
1
1
$begingroup$
en.wikipedia.org/wiki/Hasse_principle
$endgroup$
– metamorphy
Dec 19 '18 at 8:04
$begingroup$
en.wikipedia.org/wiki/Hasse_principle
$endgroup$
– metamorphy
Dec 19 '18 at 8:04
$begingroup$
If you can find one solution then the method of parametrizing Pythagorean triples (draw lines with rational slope thru $(1,0)$ and find the other point of intersection with the unit circle) generalizes to this case as well. You will use the ellipse $x^2+py^2=q$ instead.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 13:37
$begingroup$
If you can find one solution then the method of parametrizing Pythagorean triples (draw lines with rational slope thru $(1,0)$ and find the other point of intersection with the unit circle) generalizes to this case as well. You will use the ellipse $x^2+py^2=q$ instead.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 13:37
1
1
$begingroup$
@Jyrki the rational slope method does give all rational solutions; multiplying out it gives a Pythagorean triple type of formula. If you then say to divide out by the gcd of x,y,z you do get all primitive triples. More work: for such a parametrization, with coprime $u,v$ the set of gcd's of $x,y,z$ is finite. For each, a new integer parametrization may be produced.. compare answer and my comment at math.stackexchange.com/questions/3044517/…
$endgroup$
– Will Jagy
Dec 19 '18 at 17:54
$begingroup$
@Jyrki the rational slope method does give all rational solutions; multiplying out it gives a Pythagorean triple type of formula. If you then say to divide out by the gcd of x,y,z you do get all primitive triples. More work: for such a parametrization, with coprime $u,v$ the set of gcd's of $x,y,z$ is finite. For each, a new integer parametrization may be produced.. compare answer and my comment at math.stackexchange.com/questions/3044517/…
$endgroup$
– Will Jagy
Dec 19 '18 at 17:54
1
1
$begingroup$
artofproblemsolving.com/community/…
$endgroup$
– individ
Dec 21 '18 at 5:50
$begingroup$
artofproblemsolving.com/community/…
$endgroup$
– individ
Dec 21 '18 at 5:50
|
show 1 more comment
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$begingroup$
The left hand side looks like an ellipse, by swapping x and y you would get another ellipse, so I think you get a family of solutions.
$endgroup$
– Emil
Dec 19 '18 at 7:26
1
$begingroup$
en.wikipedia.org/wiki/Hasse_principle
$endgroup$
– metamorphy
Dec 19 '18 at 8:04
$begingroup$
If you can find one solution then the method of parametrizing Pythagorean triples (draw lines with rational slope thru $(1,0)$ and find the other point of intersection with the unit circle) generalizes to this case as well. You will use the ellipse $x^2+py^2=q$ instead.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 13:37
1
$begingroup$
@Jyrki the rational slope method does give all rational solutions; multiplying out it gives a Pythagorean triple type of formula. If you then say to divide out by the gcd of x,y,z you do get all primitive triples. More work: for such a parametrization, with coprime $u,v$ the set of gcd's of $x,y,z$ is finite. For each, a new integer parametrization may be produced.. compare answer and my comment at math.stackexchange.com/questions/3044517/…
$endgroup$
– Will Jagy
Dec 19 '18 at 17:54
1
$begingroup$
artofproblemsolving.com/community/…
$endgroup$
– individ
Dec 21 '18 at 5:50