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 1 hour ago
Martín Vacas Vignolo
3,530522
asked 1 hour ago
Yan Dashkow
191
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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 1 hour ago
Martín Vacas Vignolo
3,530522
asked 1 hour ago
Yan Dashkow
191
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 1 hour ago
Martín Vacas Vignolo
3,530522
asked 1 hour ago
Yan Dashkow
191
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 1 hour ago
Martín Vacas Vignolo
3,530522
asked 1 hour ago
Yan Dashkow
191
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 1 hour ago
Martín Vacas Vignolo
3,530522
asked 1 hour ago
Yan Dashkow
191
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 1 hour ago
Martín Vacas Vignolo
3,530522
edited 1 hour ago
Martín Vacas Vignolo
3,530522
edited 1 hour ago
Martín Vacas Vignolo
3,530522
3,530522
asked 1 hour ago
Yan Dashkow
191
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 1 hour ago
Yan Dashkow
191
asked 1 hour ago
Yan Dashkow
191
191
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.
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.
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 |
add a comment |
2 Answers
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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 51 mins ago
Dietrich Burde
77.5k64386
add a comment |
Here is one simple parameterization. We have,
$$p^4 +(q^2-1)^3 = (q^3+3q)^2$$
given the Pell equation $p^2-3q^2 =1$.
answered 2 mins ago
Tito Piezas III
26.8k364169
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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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 51 mins ago
Dietrich Burde
77.5k64386
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 51 mins ago
Dietrich Burde
77.5k64386
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 51 mins ago
Dietrich Burde
77.5k64386
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 51 mins ago
Dietrich Burde
77.5k64386
edited 19 mins ago
answered 51 mins ago
Dietrich Burde
77.5k64386
answered 51 mins ago
Dietrich Burde
77.5k64386
answered 51 mins ago
Dietrich Burde
77.5k64386
77.5k64386
add a comment |
add a comment |
Here is one simple parameterization. We have,
$$p^4 +(q^2-1)^3 = (q^3+3q)^2$$
given the Pell equation $p^2-3q^2 =1$.
answered 2 mins ago
Tito Piezas III
26.8k364169
add a comment |
Here is one simple parameterization. We have,
$$p^4 +(q^2-1)^3 = (q^3+3q)^2$$
given the Pell equation $p^2-3q^2 =1$.
answered 2 mins ago
Tito Piezas III
26.8k364169
add a comment |
Here is one simple parameterization. We have,
$$p^4 +(q^2-1)^3 = (q^3+3q)^2$$
given the Pell equation $p^2-3q^2 =1$.
answered 2 mins ago
Tito Piezas III
26.8k364169
Here is one simple parameterization. We have,
$$p^4 +(q^2-1)^3 = (q^3+3q)^2$$
given the Pell equation $p^2-3q^2 =1$.
answered 2 mins ago
Tito Piezas III
26.8k364169
answered 2 mins ago
Tito Piezas III
26.8k364169
answered 2 mins ago
Tito Piezas III
26.8k364169
answered 2 mins ago
Tito Piezas III
26.8k364169
26.8k364169
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.
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