How to solve this Diophantine equation?
Can anyone say how one can find solutions to the Diophantine equation $$x^3+y^4=z^2$$ in General? Only a few triples of numbers have been found, and most likely this equation has infinitely many solutions.
Examples of triples: $(6,5,29),(2,1,3),(9,6,45)$...
calculus diophantine-equations natural-numbers
edited 2 days ago
Martín Vacas Vignolo
3,710623
asked 2 days ago
Yan Dashkow
241
New contributor
Yan Dashkow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
Can anyone say how one can find solutions to the Diophantine equation $$x^3+y^4=z^2$$ in General? Only a few triples of numbers have been found, and most likely this equation has infinitely many solutions.
Examples of triples: $(6,5,29),(2,1,3),(9,6,45)$...
calculus diophantine-equations natural-numbers
edited 2 days ago
Martín Vacas Vignolo
3,710623
asked 2 days ago
Yan Dashkow
241
New contributor
Yan Dashkow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Can anyone say how one can find solutions to the Diophantine equation $$x^3+y^4=z^2$$ in General? Only a few triples of numbers have been found, and most likely this equation has infinitely many solutions.
Examples of triples: $(6,5,29),(2,1,3),(9,6,45)$...
calculus diophantine-equations natural-numbers
edited 2 days ago
Martín Vacas Vignolo
3,710623
asked 2 days ago
Yan Dashkow
241
New contributor
Yan Dashkow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Can anyone say how one can find solutions to the Diophantine equation $$x^3+y^4=z^2$$ in General? Only a few triples of numbers have been found, and most likely this equation has infinitely many solutions.
Examples of triples: $(6,5,29),(2,1,3),(9,6,45)$...
calculus diophantine-equations natural-numbers
calculus diophantine-equations natural-numbers
edited 2 days ago
Martín Vacas Vignolo
3,710623
asked 2 days ago
Yan Dashkow
241
New contributor
Yan Dashkow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Martín Vacas Vignolo
3,710623
asked 2 days ago
Yan Dashkow
241
New contributor
Yan Dashkow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Martín Vacas Vignolo
3,710623
edited 2 days ago
Martín Vacas Vignolo
3,710623
edited 2 days ago
Martín Vacas Vignolo
3,710623
3,710623
asked 2 days ago
Yan Dashkow
241
New contributor
Yan Dashkow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Yan Dashkow
241
asked 2 days ago
Yan Dashkow
241
241
New contributor
Yan Dashkow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Yan Dashkow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
5 Answers
5
active
oldest
votes
This is a case of the generalized Fermat equation
$$
x^p+y^q=z^r.
$$
For $(p,q,r)=(3,4,2)$ we have $frac{1}{p}+frac{1}{q}+frac{1}{r}>1$, which is the spherical case. Here we have infinitely many integer solutions for this triple. The solutions are given by a finite set of polynomial parametrisations of the equation, see the following paper:
F. Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.
Further Reference: The generalized Fermat equation.
answered 2 days ago
Dietrich Burde
77.6k64386
add a comment |
Here is one simple parameterization. We have,
$$x^4 +(y^2-1)^3 = (y^3+3y)^2$$
given the Pell equation $x^2-3y^2 =1$.
answered 2 days ago
Tito Piezas III
26.8k365169
@AlexD $r=2$ in this question.
– Dietrich Burde
2 days ago
Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$?
– Alex D
yesterday
@AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$.
– Dietrich Burde
yesterday
Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)?
– Yan Dashkow
yesterday
@YanDashkow We find them using one of the parametrizations, see Beukers article.
– Dietrich Burde
yesterday
|
show 1 more comment
Above equation shown below has parameterization:
$x^3+y^4=z^2$
The below parameterization has no restriction such as the
Pell equation condition demonstrated by Tito Piezas.
$x=(p)^2(-q)^3$
$y=(p)(q)^2(k-1)$
$z=(p)^2(q)^4(2k-3)$
where, $p=(k-2)$ and $q=(k^2-2)$
For $k=3$ we get :
$(-343)^3+(98)^4=(7203)^2$
answered 2 days ago
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
– Yan Dashkow
yesterday
add a comment |
"OP" asked for parametric solution for $(x,y,z)=(6,5,29)$ in $x^3+y^4=z^2$
Solution is:
$x=3p^3(8k^2-40k+50)$
$y=p^2(20k^2-104k+135)$
$z=p^4(2k-5)^2(116k^2-540k+621)$
Where, $p=(4k^2-27)$
For $k=(13/5)$, we get after removing common factors:
$6^3+5^4=29^2$
answered yesterday
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Wait, why can't k be whole?
– Yan Dashkow
yesterday
add a comment |
"OP" enquired about integer coefficent's for the parametric
solution for the equation $(x^2+y^4=z^2)$. "OP" just needs
to substitute $k=(m/n)$ in the parametrization & the resulting
solution after removing common factors is given below.
$x=6(u^3)(v^2)$
$y=(u^2)(v)(10m-27n)$
$z=(u^4)(v^2)(116m^2-540mn+621n^2)$
And $u=(4m^2-27n^2)$ & $v=(2m-5n)$
For $(m,n)=(13,5)$ we get:
$6^3+5^4=29^2$
answered 21 hours ago
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is a case of the generalized Fermat equation
$$
x^p+y^q=z^r.
$$
For $(p,q,r)=(3,4,2)$ we have $frac{1}{p}+frac{1}{q}+frac{1}{r}>1$, which is the spherical case. Here we have infinitely many integer solutions for this triple. The solutions are given by a finite set of polynomial parametrisations of the equation, see the following paper:
F. Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.
Further Reference: The generalized Fermat equation.
answered 2 days ago
Dietrich Burde
77.6k64386
add a comment |
This is a case of the generalized Fermat equation
$$
x^p+y^q=z^r.
$$
For $(p,q,r)=(3,4,2)$ we have $frac{1}{p}+frac{1}{q}+frac{1}{r}>1$, which is the spherical case. Here we have infinitely many integer solutions for this triple. The solutions are given by a finite set of polynomial parametrisations of the equation, see the following paper:
F. Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.
Further Reference: The generalized Fermat equation.
answered 2 days ago
Dietrich Burde
77.6k64386
add a comment |
This is a case of the generalized Fermat equation
$$
x^p+y^q=z^r.
$$
For $(p,q,r)=(3,4,2)$ we have $frac{1}{p}+frac{1}{q}+frac{1}{r}>1$, which is the spherical case. Here we have infinitely many integer solutions for this triple. The solutions are given by a finite set of polynomial parametrisations of the equation, see the following paper:
F. Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.
Further Reference: The generalized Fermat equation.
answered 2 days ago
Dietrich Burde
77.6k64386
This is a case of the generalized Fermat equation
$$
x^p+y^q=z^r.
$$
For $(p,q,r)=(3,4,2)$ we have $frac{1}{p}+frac{1}{q}+frac{1}{r}>1$, which is the spherical case. Here we have infinitely many integer solutions for this triple. The solutions are given by a finite set of polynomial parametrisations of the equation, see the following paper:
F. Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.
Further Reference: The generalized Fermat equation.
answered 2 days ago
Dietrich Burde
77.6k64386
edited 2 days ago
answered 2 days ago
Dietrich Burde
77.6k64386
answered 2 days ago
Dietrich Burde
77.6k64386
answered 2 days ago
Dietrich Burde
77.6k64386
77.6k64386
add a comment |
add a comment |
Here is one simple parameterization. We have,
$$x^4 +(y^2-1)^3 = (y^3+3y)^2$$
given the Pell equation $x^2-3y^2 =1$.
answered 2 days ago
Tito Piezas III
26.8k365169
@AlexD $r=2$ in this question.
– Dietrich Burde
2 days ago
Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$?
– Alex D
yesterday
@AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$.
– Dietrich Burde
yesterday
Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)?
– Yan Dashkow
yesterday
@YanDashkow We find them using one of the parametrizations, see Beukers article.
– Dietrich Burde
yesterday
|
show 1 more comment
Here is one simple parameterization. We have,
$$x^4 +(y^2-1)^3 = (y^3+3y)^2$$
given the Pell equation $x^2-3y^2 =1$.
answered 2 days ago
Tito Piezas III
26.8k365169
@AlexD $r=2$ in this question.
– Dietrich Burde
2 days ago
Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$?
– Alex D
yesterday
@AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$.
– Dietrich Burde
yesterday
Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)?
– Yan Dashkow
yesterday
@YanDashkow We find them using one of the parametrizations, see Beukers article.
– Dietrich Burde
yesterday
|
show 1 more comment
Here is one simple parameterization. We have,
$$x^4 +(y^2-1)^3 = (y^3+3y)^2$$
given the Pell equation $x^2-3y^2 =1$.
answered 2 days ago
Tito Piezas III
26.8k365169
Here is one simple parameterization. We have,
$$x^4 +(y^2-1)^3 = (y^3+3y)^2$$
given the Pell equation $x^2-3y^2 =1$.
answered 2 days ago
Tito Piezas III
26.8k365169
edited 2 days ago
answered 2 days ago
Tito Piezas III
26.8k365169
answered 2 days ago
Tito Piezas III
26.8k365169
answered 2 days ago
Tito Piezas III
26.8k365169
26.8k365169
@AlexD $r=2$ in this question.
– Dietrich Burde
2 days ago
Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$?
– Alex D
yesterday
@AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$.
– Dietrich Burde
yesterday
Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)?
– Yan Dashkow
yesterday
@YanDashkow We find them using one of the parametrizations, see Beukers article.
– Dietrich Burde
yesterday
|
show 1 more comment
@AlexD $r=2$ in this question.
– Dietrich Burde
2 days ago
Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$?
– Alex D
yesterday
@AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$.
– Dietrich Burde
yesterday
Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)?
– Yan Dashkow
yesterday
@YanDashkow We find them using one of the parametrizations, see Beukers article.
– Dietrich Burde
yesterday
@AlexD $r=2$ in this question.
– Dietrich Burde
2 days ago
@AlexD $r=2$ in this question.
– Dietrich Burde
2 days ago
Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$?
– Alex D
yesterday
Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$?
– Alex D
yesterday
@AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$.
– Dietrich Burde
yesterday
@AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$.
– Dietrich Burde
yesterday
Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)?
– Yan Dashkow
yesterday
Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)?
– Yan Dashkow
yesterday
@YanDashkow We find them using one of the parametrizations, see Beukers article.
– Dietrich Burde
yesterday
@YanDashkow We find them using one of the parametrizations, see Beukers article.
– Dietrich Burde
yesterday
|
show 1 more comment
Above equation shown below has parameterization:
$x^3+y^4=z^2$
The below parameterization has no restriction such as the
Pell equation condition demonstrated by Tito Piezas.
$x=(p)^2(-q)^3$
$y=(p)(q)^2(k-1)$
$z=(p)^2(q)^4(2k-3)$
where, $p=(k-2)$ and $q=(k^2-2)$
For $k=3$ we get :
$(-343)^3+(98)^4=(7203)^2$
answered 2 days ago
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
– Yan Dashkow
yesterday
add a comment |
Above equation shown below has parameterization:
$x^3+y^4=z^2$
The below parameterization has no restriction such as the
Pell equation condition demonstrated by Tito Piezas.
$x=(p)^2(-q)^3$
$y=(p)(q)^2(k-1)$
$z=(p)^2(q)^4(2k-3)$
where, $p=(k-2)$ and $q=(k^2-2)$
For $k=3$ we get :
$(-343)^3+(98)^4=(7203)^2$
answered 2 days ago
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
– Yan Dashkow
yesterday
add a comment |
Above equation shown below has parameterization:
$x^3+y^4=z^2$
The below parameterization has no restriction such as the
Pell equation condition demonstrated by Tito Piezas.
$x=(p)^2(-q)^3$
$y=(p)(q)^2(k-1)$
$z=(p)^2(q)^4(2k-3)$
where, $p=(k-2)$ and $q=(k^2-2)$
For $k=3$ we get :
$(-343)^3+(98)^4=(7203)^2$
answered 2 days ago
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Above equation shown below has parameterization:
$x^3+y^4=z^2$
The below parameterization has no restriction such as the
Pell equation condition demonstrated by Tito Piezas.
$x=(p)^2(-q)^3$
$y=(p)(q)^2(k-1)$
$z=(p)^2(q)^4(2k-3)$
where, $p=(k-2)$ and $q=(k^2-2)$
For $k=3$ we get :
$(-343)^3+(98)^4=(7203)^2$
answered 2 days ago
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Sam
1
answered 2 days ago
Sam
1
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
– Yan Dashkow
yesterday
add a comment |
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
– Yan Dashkow
yesterday
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
– Yan Dashkow
yesterday
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
– Yan Dashkow
yesterday
add a comment |
"OP" asked for parametric solution for $(x,y,z)=(6,5,29)$ in $x^3+y^4=z^2$
Solution is:
$x=3p^3(8k^2-40k+50)$
$y=p^2(20k^2-104k+135)$
$z=p^4(2k-5)^2(116k^2-540k+621)$
Where, $p=(4k^2-27)$
For $k=(13/5)$, we get after removing common factors:
$6^3+5^4=29^2$
answered yesterday
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Wait, why can't k be whole?
– Yan Dashkow
yesterday
add a comment |
"OP" asked for parametric solution for $(x,y,z)=(6,5,29)$ in $x^3+y^4=z^2$
Solution is:
$x=3p^3(8k^2-40k+50)$
$y=p^2(20k^2-104k+135)$
$z=p^4(2k-5)^2(116k^2-540k+621)$
Where, $p=(4k^2-27)$
For $k=(13/5)$, we get after removing common factors:
$6^3+5^4=29^2$
answered yesterday
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Wait, why can't k be whole?
– Yan Dashkow
yesterday
add a comment |
"OP" asked for parametric solution for $(x,y,z)=(6,5,29)$ in $x^3+y^4=z^2$
Solution is:
$x=3p^3(8k^2-40k+50)$
$y=p^2(20k^2-104k+135)$
$z=p^4(2k-5)^2(116k^2-540k+621)$
Where, $p=(4k^2-27)$
For $k=(13/5)$, we get after removing common factors:
$6^3+5^4=29^2$
answered yesterday
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
"OP" asked for parametric solution for $(x,y,z)=(6,5,29)$ in $x^3+y^4=z^2$
Solution is:
$x=3p^3(8k^2-40k+50)$
$y=p^2(20k^2-104k+135)$
$z=p^4(2k-5)^2(116k^2-540k+621)$
Where, $p=(4k^2-27)$
For $k=(13/5)$, we get after removing common factors:
$6^3+5^4=29^2$
answered yesterday
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Sam
1
New contributor
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answered yesterday
Sam
1
answered yesterday
Sam
1
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
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Wait, why can't k be whole?
– Yan Dashkow
yesterday
add a comment |
Wait, why can't k be whole?
– Yan Dashkow
yesterday
Wait, why can't k be whole?
– Yan Dashkow
yesterday
Wait, why can't k be whole?
– Yan Dashkow
yesterday
add a comment |
"OP" enquired about integer coefficent's for the parametric
solution for the equation $(x^2+y^4=z^2)$. "OP" just needs
to substitute $k=(m/n)$ in the parametrization & the resulting
solution after removing common factors is given below.
$x=6(u^3)(v^2)$
$y=(u^2)(v)(10m-27n)$
$z=(u^4)(v^2)(116m^2-540mn+621n^2)$
And $u=(4m^2-27n^2)$ & $v=(2m-5n)$
For $(m,n)=(13,5)$ we get:
$6^3+5^4=29^2$
answered 21 hours ago
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
"OP" enquired about integer coefficent's for the parametric
solution for the equation $(x^2+y^4=z^2)$. "OP" just needs
to substitute $k=(m/n)$ in the parametrization & the resulting
solution after removing common factors is given below.
$x=6(u^3)(v^2)$
$y=(u^2)(v)(10m-27n)$
$z=(u^4)(v^2)(116m^2-540mn+621n^2)$
And $u=(4m^2-27n^2)$ & $v=(2m-5n)$
For $(m,n)=(13,5)$ we get:
$6^3+5^4=29^2$
answered 21 hours ago
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
"OP" enquired about integer coefficent's for the parametric
solution for the equation $(x^2+y^4=z^2)$. "OP" just needs
to substitute $k=(m/n)$ in the parametrization & the resulting
solution after removing common factors is given below.
$x=6(u^3)(v^2)$
$y=(u^2)(v)(10m-27n)$
$z=(u^4)(v^2)(116m^2-540mn+621n^2)$
And $u=(4m^2-27n^2)$ & $v=(2m-5n)$
For $(m,n)=(13,5)$ we get:
$6^3+5^4=29^2$
answered 21 hours ago
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
"OP" enquired about integer coefficent's for the parametric
solution for the equation $(x^2+y^4=z^2)$. "OP" just needs
to substitute $k=(m/n)$ in the parametrization & the resulting
solution after removing common factors is given below.
$x=6(u^3)(v^2)$
$y=(u^2)(v)(10m-27n)$
$z=(u^4)(v^2)(116m^2-540mn+621n^2)$
And $u=(4m^2-27n^2)$ & $v=(2m-5n)$
For $(m,n)=(13,5)$ we get:
$6^3+5^4=29^2$
answered 21 hours ago
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 21 hours ago
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 21 hours ago
Sam
1
answered 21 hours ago
Sam
1
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
Yan Dashkow is a new contributor. Be nice, and check out our Code of Conduct.
Yan Dashkow is a new contributor. Be nice, and check out our Code of Conduct.
Yan Dashkow is a new contributor. Be nice, and check out our Code of Conduct.
Yan Dashkow is a new contributor. Be nice, and check out our Code of Conduct.
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