How to solve this Diophantine equation?
$begingroup$
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 Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,816623
asked Dec 29 '18 at 12:43
Yan DashkowYan Dashkow
241
$endgroup$
add a comment |
$begingroup$
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 Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,816623
asked Dec 29 '18 at 12:43
Yan DashkowYan Dashkow
241
$endgroup$
add a comment |
$begingroup$
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 Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,816623
asked Dec 29 '18 at 12:43
Yan DashkowYan Dashkow
241
$endgroup$
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 Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,816623
asked Dec 29 '18 at 12:43
Yan DashkowYan Dashkow
241
edited Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,816623
asked Dec 29 '18 at 12:43
Yan DashkowYan Dashkow
241
edited Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,816623
edited Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,816623
edited Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,816623
3,816623
asked Dec 29 '18 at 12:43
Yan DashkowYan Dashkow
241
asked Dec 29 '18 at 12:43
Yan DashkowYan Dashkow
241
asked Dec 29 '18 at 12:43
Yan DashkowYan Dashkow
241
241
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
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 Dec 29 '18 at 13:14
Dietrich BurdeDietrich Burde
81.1k648106
$endgroup$
add a comment |
$begingroup$
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 Dec 29 '18 at 14:03
Tito Piezas IIITito Piezas III
27.8k368178
$endgroup$
$begingroup$
@AlexD $r=2$ in this question.
$endgroup$
– Dietrich Burde
Dec 29 '18 at 23:34
$begingroup$
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 $z$?
$endgroup$
– Alex D
Dec 30 '18 at 11:21
$begingroup$
@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 $X,Y,Z$.
$endgroup$
– Dietrich Burde
Dec 30 '18 at 11:47
$begingroup$
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)?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 11:55
$begingroup$
@YanDashkow We find them using one of the parametrizations, see Beukers article.
$endgroup$
– Dietrich Burde
Dec 30 '18 at 13:31
|
show 1 more comment
$begingroup$
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 Dec 30 '18 at 8:19
SamSam
1
$endgroup$
$begingroup$
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 10:55
add a comment |
$begingroup$
"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 Dec 30 '18 at 18:09
SamSam
1
$endgroup$
$begingroup$
Wait, why can't k be whole?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 19:30
add a comment |
$begingroup$
"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 Dec 31 '18 at 11:49
SamSam
1
$endgroup$
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
$begingroup$
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 Dec 29 '18 at 13:14
Dietrich BurdeDietrich Burde
81.1k648106
$endgroup$
add a comment |
$begingroup$
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 Dec 29 '18 at 13:14
Dietrich BurdeDietrich Burde
81.1k648106
$endgroup$
add a comment |
$begingroup$
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 Dec 29 '18 at 13:14
Dietrich BurdeDietrich Burde
81.1k648106
$endgroup$
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 Dec 29 '18 at 13:14
Dietrich BurdeDietrich Burde
81.1k648106
edited Dec 29 '18 at 13:46
answered Dec 29 '18 at 13:14
Dietrich BurdeDietrich Burde
81.1k648106
answered Dec 29 '18 at 13:14
Dietrich BurdeDietrich Burde
81.1k648106
answered Dec 29 '18 at 13:14
Dietrich BurdeDietrich Burde
81.1k648106
81.1k648106
add a comment |
add a comment |
$begingroup$
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 Dec 29 '18 at 14:03
Tito Piezas IIITito Piezas III
27.8k368178
$endgroup$
$begingroup$
@AlexD $r=2$ in this question.
$endgroup$
– Dietrich Burde
Dec 29 '18 at 23:34
$begingroup$
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 $z$?
$endgroup$
– Alex D
Dec 30 '18 at 11:21
$begingroup$
@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 $X,Y,Z$.
$endgroup$
– Dietrich Burde
Dec 30 '18 at 11:47
$begingroup$
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)?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 11:55
$begingroup$
@YanDashkow We find them using one of the parametrizations, see Beukers article.
$endgroup$
– Dietrich Burde
Dec 30 '18 at 13:31
|
show 1 more comment
$begingroup$
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 Dec 29 '18 at 14:03
Tito Piezas IIITito Piezas III
27.8k368178
$endgroup$
$begingroup$
@AlexD $r=2$ in this question.
$endgroup$
– Dietrich Burde
Dec 29 '18 at 23:34
$begingroup$
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 $z$?
$endgroup$
– Alex D
Dec 30 '18 at 11:21
$begingroup$
@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 $X,Y,Z$.
$endgroup$
– Dietrich Burde
Dec 30 '18 at 11:47
$begingroup$
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)?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 11:55
$begingroup$
@YanDashkow We find them using one of the parametrizations, see Beukers article.
$endgroup$
– Dietrich Burde
Dec 30 '18 at 13:31
|
show 1 more comment
$begingroup$
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 Dec 29 '18 at 14:03
Tito Piezas IIITito Piezas III
27.8k368178
$endgroup$
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 Dec 29 '18 at 14:03
Tito Piezas IIITito Piezas III
27.8k368178
edited Dec 30 '18 at 2:21
answered Dec 29 '18 at 14:03
Tito Piezas IIITito Piezas III
27.8k368178
answered Dec 29 '18 at 14:03
Tito Piezas IIITito Piezas III
27.8k368178
answered Dec 29 '18 at 14:03
Tito Piezas IIITito Piezas III
27.8k368178
27.8k368178
$begingroup$
@AlexD $r=2$ in this question.
$endgroup$
– Dietrich Burde
Dec 29 '18 at 23:34
$begingroup$
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 $z$?
$endgroup$
– Alex D
Dec 30 '18 at 11:21
$begingroup$
@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 $X,Y,Z$.
$endgroup$
– Dietrich Burde
Dec 30 '18 at 11:47
$begingroup$
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)?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 11:55
$begingroup$
@YanDashkow We find them using one of the parametrizations, see Beukers article.
$endgroup$
– Dietrich Burde
Dec 30 '18 at 13:31
|
show 1 more comment
$begingroup$
@AlexD $r=2$ in this question.
$endgroup$
– Dietrich Burde
Dec 29 '18 at 23:34
$begingroup$
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 $z$?
$endgroup$
– Alex D
Dec 30 '18 at 11:21
$begingroup$
@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 $X,Y,Z$.
$endgroup$
– Dietrich Burde
Dec 30 '18 at 11:47
$begingroup$
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)?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 11:55
$begingroup$
@YanDashkow We find them using one of the parametrizations, see Beukers article.
$endgroup$
– Dietrich Burde
Dec 30 '18 at 13:31
$begingroup$
@AlexD $r=2$ in this question.
$endgroup$
– Dietrich Burde
Dec 29 '18 at 23:34
$begingroup$
@AlexD $r=2$ in this question.
$endgroup$
– Dietrich Burde
Dec 29 '18 at 23:34
$begingroup$
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 $z$?
$endgroup$
– Alex D
Dec 30 '18 at 11:21
$begingroup$
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 $z$?
$endgroup$
– Alex D
Dec 30 '18 at 11:21
$begingroup$
@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 $X,Y,Z$.
$endgroup$
– Dietrich Burde
Dec 30 '18 at 11:47
$begingroup$
@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 $X,Y,Z$.
$endgroup$
– Dietrich Burde
Dec 30 '18 at 11:47
$begingroup$
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)?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 11:55
$begingroup$
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)?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 11:55
$begingroup$
@YanDashkow We find them using one of the parametrizations, see Beukers article.
$endgroup$
– Dietrich Burde
Dec 30 '18 at 13:31
$begingroup$
@YanDashkow We find them using one of the parametrizations, see Beukers article.
$endgroup$
– Dietrich Burde
Dec 30 '18 at 13:31
|
show 1 more comment
$begingroup$
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 Dec 30 '18 at 8:19
SamSam
1
$endgroup$
$begingroup$
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 10:55
add a comment |
$begingroup$
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 Dec 30 '18 at 8:19
SamSam
1
$endgroup$
$begingroup$
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 10:55
add a comment |
$begingroup$
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 Dec 30 '18 at 8:19
SamSam
1
$endgroup$
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 Dec 30 '18 at 8:19
SamSam
1
answered Dec 30 '18 at 8:19
SamSam
1
answered Dec 30 '18 at 8:19
SamSam
1
answered Dec 30 '18 at 8:19
SamSam
1
1
$begingroup$
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 10:55
add a comment |
$begingroup$
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 10:55
$begingroup$
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 10:55
$begingroup$
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 10:55
add a comment |
$begingroup$
"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 Dec 30 '18 at 18:09
SamSam
1
$endgroup$
$begingroup$
Wait, why can't k be whole?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 19:30
add a comment |
$begingroup$
"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 Dec 30 '18 at 18:09
SamSam
1
$endgroup$
$begingroup$
Wait, why can't k be whole?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 19:30
add a comment |
$begingroup$
"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 Dec 30 '18 at 18:09
SamSam
1
$endgroup$
"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 Dec 30 '18 at 18:09
SamSam
1
answered Dec 30 '18 at 18:09
SamSam
1
answered Dec 30 '18 at 18:09
SamSam
1
answered Dec 30 '18 at 18:09
SamSam
1
1
$begingroup$
Wait, why can't k be whole?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 19:30
add a comment |
$begingroup$
Wait, why can't k be whole?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 19:30
$begingroup$
Wait, why can't k be whole?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 19:30
$begingroup$
Wait, why can't k be whole?
$endgroup$
– Yan Dashkow
Dec 30 '18 at 19:30
add a comment |
$begingroup$
"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 Dec 31 '18 at 11:49
SamSam
1
$endgroup$
add a comment |
$begingroup$
"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 Dec 31 '18 at 11:49
SamSam
1
$endgroup$
add a comment |
$begingroup$
"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 Dec 31 '18 at 11:49
SamSam
1
$endgroup$
"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 Dec 31 '18 at 11:49
SamSam
1
answered Dec 31 '18 at 11:49
SamSam
1
answered Dec 31 '18 at 11:49
SamSam
1
answered Dec 31 '18 at 11:49
SamSam
1
1
add a comment |
add a comment |
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