Parametrization of pythagorean-like equation [duplicate]Diophantine equation $a^2+b^2=c^2+d^2$From the result...
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Parametrization of pythagorean-like equation [duplicate]
Diophantine equation $a^2+b^2=c^2+d^2$From the result of Hilbert's tenth problem, does it follow that there exist infinitely numbers of equation…Help solving this Diophantine equationFind largest $k$ such that the diophantine equation $ax+by=k$ does not have nonnegative solution.General quadratic diophantine equation.What are some applications of parametrization of curves and surfaces?Parametrization of solutions of diophantine equationThe diophantine equation $z^2=a^2+bx^2+cy^2$Parametrization of $x^2+ay^2=z^k$, where $gcd(x,y,z)=1$In the Diophantine equation, 5x + 17y = cPythagorean like Diophantine Equation
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Diophantine equation $a^2+b^2=c^2+d^2$
7 answers
Is there any known complete parametrization of the Diophantine equation
$$
A^{2} + B^{2} = C^{2} + D^{2}
$$
where $A, B, C, D$ are (positive) rational numbers, or equivalently, integers?
number-theory diophantine-equations
asked 2 days ago
Seewoo LeeSeewoo Lee
7,057927
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marked as duplicate by Dietrich Burde number-theory
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This question already has an answer here:
Diophantine equation $a^2+b^2=c^2+d^2$
7 answers
Is there any known complete parametrization of the Diophantine equation
$$
A^{2} + B^{2} = C^{2} + D^{2}
$$
where $A, B, C, D$ are (positive) rational numbers, or equivalently, integers?
number-theory diophantine-equations
asked 2 days ago
Seewoo LeeSeewoo Lee
7,057927
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marked as duplicate by Dietrich Burde number-theory
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– individ
2 days ago
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$begingroup$
This question already has an answer here:
Diophantine equation $a^2+b^2=c^2+d^2$
7 answers
Is there any known complete parametrization of the Diophantine equation
$$
A^{2} + B^{2} = C^{2} + D^{2}
$$
where $A, B, C, D$ are (positive) rational numbers, or equivalently, integers?
number-theory diophantine-equations
asked 2 days ago
Seewoo LeeSeewoo Lee
7,057927
$endgroup$
This question already has an answer here:
Diophantine equation $a^2+b^2=c^2+d^2$
7 answers
Is there any known complete parametrization of the Diophantine equation
$$
A^{2} + B^{2} = C^{2} + D^{2}
$$
where $A, B, C, D$ are (positive) rational numbers, or equivalently, integers?
This question already has an answer here:
Diophantine equation $a^2+b^2=c^2+d^2$
7 answers
number-theory diophantine-equations
number-theory diophantine-equations
asked 2 days ago
Seewoo LeeSeewoo Lee
7,057927
asked 2 days ago
Seewoo LeeSeewoo Lee
7,057927
asked 2 days ago
Seewoo LeeSeewoo Lee
7,057927
asked 2 days ago
Seewoo LeeSeewoo Lee
7,057927
asked 2 days ago
Seewoo LeeSeewoo Lee
7,057927
7,057927
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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1 Answer
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Yes, there are. You can put
$$A=ms+nt\B=nt-ms\C=ms-nt\D=mt+ns$$ where $m,n,s,t$ are arbitrary.
answered 2 days ago
PiquitoPiquito
18k31539
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$begingroup$
In your "soloution" it seems impossible for both B and C to be positive.
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– Peter Green
2 days ago
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It should be $(A,B,C,D) = (p r + q s , q r - p s , p r - q s , p s + q r),$
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– Dietrich Burde
2 days ago
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@Peter Green: It is not "my solution" but a quite known parameterization in which I have had an obvious lapse.
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– Piquito
yesterday
$begingroup$
@Dietrich Burde.- Look please at my comment answering to Peter Green.
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– Piquito
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, there are. You can put
$$A=ms+nt\B=nt-ms\C=ms-nt\D=mt+ns$$ where $m,n,s,t$ are arbitrary.
answered 2 days ago
PiquitoPiquito
18k31539
$endgroup$
$begingroup$
In your "soloution" it seems impossible for both B and C to be positive.
$endgroup$
– Peter Green
2 days ago
$begingroup$
It should be $(A,B,C,D) = (p r + q s , q r - p s , p r - q s , p s + q r),$
$endgroup$
– Dietrich Burde
2 days ago
$begingroup$
@Peter Green: It is not "my solution" but a quite known parameterization in which I have had an obvious lapse.
$endgroup$
– Piquito
yesterday
$begingroup$
@Dietrich Burde.- Look please at my comment answering to Peter Green.
$endgroup$
– Piquito
yesterday
add a comment |
$begingroup$
Yes, there are. You can put
$$A=ms+nt\B=nt-ms\C=ms-nt\D=mt+ns$$ where $m,n,s,t$ are arbitrary.
answered 2 days ago
PiquitoPiquito
18k31539
$endgroup$
$begingroup$
In your "soloution" it seems impossible for both B and C to be positive.
$endgroup$
– Peter Green
2 days ago
$begingroup$
It should be $(A,B,C,D) = (p r + q s , q r - p s , p r - q s , p s + q r),$
$endgroup$
– Dietrich Burde
2 days ago
$begingroup$
@Peter Green: It is not "my solution" but a quite known parameterization in which I have had an obvious lapse.
$endgroup$
– Piquito
yesterday
$begingroup$
@Dietrich Burde.- Look please at my comment answering to Peter Green.
$endgroup$
– Piquito
yesterday
add a comment |
$begingroup$
Yes, there are. You can put
$$A=ms+nt\B=nt-ms\C=ms-nt\D=mt+ns$$ where $m,n,s,t$ are arbitrary.
answered 2 days ago
PiquitoPiquito
18k31539
$endgroup$
Yes, there are. You can put
$$A=ms+nt\B=nt-ms\C=ms-nt\D=mt+ns$$ where $m,n,s,t$ are arbitrary.
answered 2 days ago
PiquitoPiquito
18k31539
edited 2 days ago
answered 2 days ago
PiquitoPiquito
18k31539
answered 2 days ago
PiquitoPiquito
18k31539
answered 2 days ago
PiquitoPiquito
18k31539
18k31539
$begingroup$
In your "soloution" it seems impossible for both B and C to be positive.
$endgroup$
– Peter Green
2 days ago
$begingroup$
It should be $(A,B,C,D) = (p r + q s , q r - p s , p r - q s , p s + q r),$
$endgroup$
– Dietrich Burde
2 days ago
$begingroup$
@Peter Green: It is not "my solution" but a quite known parameterization in which I have had an obvious lapse.
$endgroup$
– Piquito
yesterday
$begingroup$
@Dietrich Burde.- Look please at my comment answering to Peter Green.
$endgroup$
– Piquito
yesterday
add a comment |
$begingroup$
In your "soloution" it seems impossible for both B and C to be positive.
$endgroup$
– Peter Green
2 days ago
$begingroup$
It should be $(A,B,C,D) = (p r + q s , q r - p s , p r - q s , p s + q r),$
$endgroup$
– Dietrich Burde
2 days ago
$begingroup$
@Peter Green: It is not "my solution" but a quite known parameterization in which I have had an obvious lapse.
$endgroup$
– Piquito
yesterday
$begingroup$
@Dietrich Burde.- Look please at my comment answering to Peter Green.
$endgroup$
– Piquito
yesterday
$begingroup$
In your "soloution" it seems impossible for both B and C to be positive.
$endgroup$
– Peter Green
2 days ago
$begingroup$
In your "soloution" it seems impossible for both B and C to be positive.
$endgroup$
– Peter Green
2 days ago
$begingroup$
It should be $(A,B,C,D) = (p r + q s , q r - p s , p r - q s , p s + q r),$
$endgroup$
– Dietrich Burde
2 days ago
$begingroup$
It should be $(A,B,C,D) = (p r + q s , q r - p s , p r - q s , p s + q r),$
$endgroup$
– Dietrich Burde
2 days ago
$begingroup$
@Peter Green: It is not "my solution" but a quite known parameterization in which I have had an obvious lapse.
$endgroup$
– Piquito
yesterday
$begingroup$
@Peter Green: It is not "my solution" but a quite known parameterization in which I have had an obvious lapse.
$endgroup$
– Piquito
yesterday
$begingroup$
@Dietrich Burde.- Look please at my comment answering to Peter Green.
$endgroup$
– Piquito
yesterday
$begingroup$
@Dietrich Burde.- Look please at my comment answering to Peter Green.
$endgroup$
– Piquito
yesterday
add a comment |
1
$begingroup$
math.stackexchange.com/questions/153603/…
$endgroup$
– individ
2 days ago